AP Chemistry flashcards
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Unit 1: Atomic Structure and Properties
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Topic 1.1 · Work it outHow many water molecules are in 9.01 g of water? The molar mass of H₂O is 18.02 g/mol.
3.01 × 10²³ molecules. n = m/M = (9.01 g)/(18.02 g/mol) = 0.500 mol, and (0.500 mol)(6.022 × 10²³ mol⁻¹) = 3.01 × 10²³ molecules. Grams cancel in the first step and moles cancel in the second.
Review: 1.1 Moles and Molar Mass
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Topic 1.1 · Spot the mistake"One mole of O₂ and one mole of H₂ have the same mass, since a mole is a mole." Is that right?
No. They hold the same number of molecules, 6.022 × 10²³, but an O₂ molecule is much heavier than an H₂ molecule. A mole of O₂ has a mass of 32.00 g and a mole of H₂ only 2.016 g, because the molar mass in grams matches the mass of one molecule in amu.
Review: 1.1 Moles and Molar Mass
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Topic 1.1 · Work it outWhat is the mass of 2.50 × 10²² atoms of aluminum? The molar mass of Al is 26.98 g/mol.
1.12 g. n = (2.50 × 10²² atoms)/(6.022 × 10²³ mol⁻¹) = 0.0415 mol, and m = nM = (0.0415 mol)(26.98 g/mol) = 1.12 g. Divide by Avogadro's number to get moles, then multiply by the molar mass to get grams.
Review: 1.1 Moles and Molar Mass
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Topic 1.2 · Work it outThe mass spectrum of boron shows two peaks: 10.01 amu with 19.9% relative abundance and 11.01 amu with 80.1%. Estimate the average atomic mass of boron.
10.81 amu. (10.01 amu)(0.199) + (11.01 amu)(0.801) = 1.99 amu + 8.82 amu = 10.81 amu. The weighted average sits close to 11 because the heavier isotope is far more abundant.
Review: 1.2 Mass Spectra of Elements
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Topic 1.2 · Say it in contextThe mass spectrum of a sample of chlorine atoms, detected as singly charged ions, has peaks at 35 amu and 37 amu. The 35 amu peak is three times as tall. What does this tell you?
Chlorine has two isotopes, chlorine-35 and chlorine-37, and chlorine-35 is three times as abundant. The average atomic mass is about (3 × 35 + 1 × 37)/4 = 35.5 amu, closer to 35 because most of the atoms are the lighter isotope.
Review: 1.2 Mass Spectra of Elements
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Topic 1.3 · Work it outA compound is 40.0% carbon, 6.71% hydrogen and 53.3% oxygen by mass. Find its empirical formula. Use C 12.01, H 1.008 and O 16.00 g/mol.
CH₂O. In 100 g there are 40.0/12.01 = 3.33 mol C, 6.71/1.008 = 6.66 mol H and 53.3/16.00 = 3.33 mol O. Dividing each by the smallest, 3.33, gives 1 : 2.00 : 1, the lowest whole-number ratio of atoms.
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Topic 1.3 · Spot the mistake"Glucose, C₆H₁₂O₆, and formaldehyde, CH₂O, have the same empirical formula, so they are the same compound." Is that right?
No. An empirical formula gives only the lowest whole-number ratio of atoms, and both formulas reduce to CH₂O. A glucose molecule holds 6 C, 12 H and 6 O atoms, so the two compounds have different molecules, molar masses and properties.
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Topic 1.4 · Work it outA 2.50 g sample of table salt mixed with sand contains 1.20 g of chlorine, and the sand contains none. What percent of the sample is NaCl? Use Na 22.99 and Cl 35.45 g/mol. These numbers are a teaching example.
79.1% NaCl. All the chlorine comes from NaCl, whose molar mass is 58.44 g/mol, so the mass of NaCl is (1.20 g)(58.44/35.45) = 1.978 g. Then (1.978 g)/(2.50 g) × 100 = 79.1%.
Review: 1.4 Composition of Mixtures
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Topic 1.5 · Work it outTwo charged particles attract each other with force F. If both charges are doubled and the distance between them is also doubled, what is the new force?
F, unchanged. By Coulomb's law, F ∝ q₁q₂/r². Doubling both charges multiplies the force by 2 × 2 = 4, and doubling the distance divides it by 2² = 4, so the two effects cancel.
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Topic 1.5 · Spot the mistake"A 1s electron and a 3s electron in a sodium atom are held equally tightly, since both are attracted by the same 11 protons." Is that right?
No. The 1s electron is much closer to the nucleus and feels almost the full nuclear charge, while the 3s electron is farther away and shielded by 10 inner electrons. By Coulomb's law the 1s electron is held far more strongly, so removing it takes much more energy.
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Topic 1.6 · Say it in contextA photoelectron spectrum has three peaks with relative heights 2 : 2 : 3, listed from highest to lowest binding energy. Which element is it?
Nitrogen, 1s² 2s² 2p³. Each peak is one subshell, in order from the most tightly held, 1s, to the least, 2p, and each height is proportional to the number of electrons in that subshell. The heights add to 7 electrons.
Review: 1.6 Photoelectron Spectroscopy
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Topic 1.6 · Spot the mistake"In a photoelectron spectrum, the tallest peak belongs to the electrons that are hardest to remove." Is that right?
No. A peak's height shows how many electrons are in that subshell. How hard they are to remove shows in its position: the higher the binding energy, the more energy it takes. In argon, the 1s peak has the highest binding energy, but the 2p and 3p peaks, with 6 electrons each, are taller.
Review: 1.6 Photoelectron Spectroscopy
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Topic 1.7 · Say it in contextFirst ionization energies are 520 kJ/mol for Li, 496 kJ/mol for Na and 419 kJ/mol for K. Predict whether the value for rubidium is above or below 419 kJ/mol, and explain the trend.
Below 419 kJ/mol; the measured value is 403 kJ/mol. Going down the group, the valence s electron is in a shell farther from the nucleus and is shielded by more inner electrons. By Coulomb's law it is held less tightly, so it is easier to remove.
Review: 1.7 Periodic Trends
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Topic 1.7 · Spot the mistake"A chloride ion, Cl⁻, is smaller than a chlorine atom, because the nucleus pulls the extra electron in." Is that right?
No. Cl⁻ is larger than a Cl atom. It has the same 17 protons but one more electron, so repulsion among the electrons increases and each electron feels less net pull from the nucleus. The electron cloud spreads out.
Review: 1.7 Periodic Trends
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Topic 1.8 · Which one fitsSodium and oxygen form Na₂O. Predict the formula of the compound formed by potassium and sulfur, and say why.
K₂S. Potassium is in the same group as sodium, and sulfur is in the same group as oxygen, so K has as many valence electrons as Na and S as many as O. K forms K⁺ and S forms S²⁻, so two K⁺ ions balance one S²⁻ ion.
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Topic 1.8 · Spot the mistake"Magnesium and nitrogen form MgN, one atom of each." Is that right?
No. Magnesium has 2 valence electrons and forms Mg²⁺, while nitrogen has 5 and gains 3 to form N³⁻. Three Mg²⁺ ions (a total of 6+) balance two N³⁻ ions (6−), so the formula is Mg₃N₂.
Unit 2: Compound Structure and Properties
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Topic 2.1 · Which one fitsA white solid melts at 801 °C, does not conduct electricity as a solid, and conducts when melted. What type of bonding does it most likely have?
Ionic bonding. The high melting point points to strong attractions, and conducting only when molten means the solid holds charged particles that cannot move until it melts. Properties like these are the best evidence for the type of bonding in a compound.
Review: 2.1 Types of Chemical Bonds
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Topic 2.1 · Spot the mistake"C-H bonds are polar covalent, because carbon is more electronegative than hydrogen." Is that right?
No. Carbon is only slightly more electronegative than hydrogen, so the electrons are shared almost equally and C-H bonds are treated as effectively nonpolar. A bond such as O-H, between atoms with a much larger difference in electronegativity, is polar.
Review: 2.1 Types of Chemical Bonds
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Topic 2.2 · Say it in contextA graph of potential energy against the distance between two hydrogen atoms has its lowest point at 74 pm and −436 kJ/mol. What do these two numbers mean?
74 pm is the bond length of H₂, the separation at which the potential energy is lowest. 436 kJ/mol is the bond energy, the energy needed to separate the atoms in a mole of H₂ molecules completely, back up to zero potential energy.
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Topic 2.2 · Which one fitsWhich has the stronger attractions between its ions, NaF or MgO?
MgO. Coulombic attraction grows with the charge on each ion, and Mg²⁺ and O²⁻ carry twice the charges of Na⁺ and F⁻. The ions in MgO are also slightly closer together, so both effects point the same way, and MgO has a much higher melting point.
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Topic 2.3 · Spot the mistake"In solid sodium chloride, the Na⁺ ions cluster together in one region and the Cl⁻ ions in another." Is that right?
No. The ions alternate in a regular three-dimensional array, so each Na⁺ is surrounded by Cl⁻ ions and each Cl⁻ by Na⁺ ions. That arrangement maximizes the attractions between opposite charges while keeping like charges apart.
Review: 2.3 Structure of Ionic Solids
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Topic 2.4 · Which one fitsSteel forms when small carbon atoms are added to iron. Which kind of alloy is this, and how should a particle model show it?
An interstitial alloy. Carbon atoms are much smaller than iron atoms, so the model shows them in the spaces between the iron atoms of the lattice, not in place of iron atoms.
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Topic 2.5 · Work it outHow many valence electrons go into the Lewis diagram of the carbonate ion, CO₃²⁻?
- Carbon brings 4 and each of the three oxygen atoms brings 6, which gives 22, and the 2− charge adds 2 more electrons. Every bond and lone pair in the finished diagram must account for exactly these 24 electrons.
Review: 2.5 Lewis Diagrams
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Topic 2.5 · Check the conditionsA draft Lewis diagram of CO₂ has a single bond from C to each O and three lone pairs on each O, using all 16 valence electrons. Does it satisfy the octet rule? If not, how do you fix it?
No. Each O has an octet, but C has only 4 electrons around it. Move one lone pair from each O into a bond to make O=C=O, with two double bonds. Every atom then has an octet, and all 16 electrons are still used.
Review: 2.5 Lewis Diagrams
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Topic 2.6 · Work it outA Lewis diagram of CO₂ has a triple bond from C to one O, which keeps one lone pair, and a single bond to the other O, which keeps three. Find each formal charge. Is this better than O=C=O?
Triple-bonded O: 6 − 2 − 3 = +1. C: 4 − 0 − 4 = 0. Single-bonded O: 6 − 6 − 1 = −1. No: in O=C=O every formal charge is 0, so it is the better model. Formal charge is valence electrons minus lone-pair electrons minus bonds.
Review: 2.6 Resonance and Formal Charge
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Topic 2.6 · Say it in contextOzone, O₃, can be drawn with its double bond on either side of the central O. Experiments show both O-O bonds have the same length, between that of a single and a double bond. What does this mean?
Neither drawing is right on its own. Ozone is described by resonance between the two equivalent structures, so each O-O bond has a bond order of 1.5, and both bonds have the same length.
Review: 2.6 Resonance and Formal Charge
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Topic 2.6 · Spot the mistake"With enough tries, any molecule can be given a Lewis diagram with an octet on every atom." Is that right?
No. A molecule with an odd number of valence electrons cannot give every atom an octet. NO₂ has 5 + 2(6) = 17 valence electrons, so one atom is left with an unpaired electron. This is one of the limits of the Lewis model.
Review: 2.6 Resonance and Formal Charge
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Topic 2.7 · Which one fitsIn sulfur tetrafluoride, SF₄, the sulfur atom has four bonding pairs and one lone pair. What is the molecular geometry?
Seesaw. Five electron domains repel each other into a trigonal bipyramidal arrangement. One of the five is a lone pair, so the four F atoms alone form a seesaw shape.
Review: 2.7 VSEPR and Hybridization
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Topic 2.7 · Spot the mistake"CO₂ has polar C=O bonds, so CO₂ is a polar molecule." Is that right?
No. CO₂ is linear, so its two equal bond dipoles point in opposite directions and cancel, and the molecule has no net dipole moment. Polar bonds make a polar molecule only when the geometry does not cancel them, as in bent H₂O.
Review: 2.7 VSEPR and Hybridization
Unit 3: Properties of Substances and Mixtures
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Topic 3.1 · Which one fitsWhat is the strongest intermolecular force between molecules of methanol, CH₃OH?
Hydrogen bonding. Methanol has an H atom bonded directly to O, a highly electronegative atom, and that H is attracted to the O of a neighboring molecule. Dispersion and dipole-dipole forces are also present, but hydrogen bonding is the strongest of them here.
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Topic 3.1 · Spot the mistake"Fluoromethane, CH₃F, forms hydrogen bonds between its molecules because it contains both H and F." Is that right?
No. Hydrogen bonding needs an H atom bonded directly to N, O or F. In CH₃F every H is bonded to carbon, so no H carries a large partial positive charge. CH₃F molecules attract each other through dipole-dipole and dispersion forces.
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Topic 3.2 · Which one fitsA solid made of silicon and oxygen is very hard, melts above 1600 °C, and does not conduct electricity. What type of solid is it?
A covalent network solid, silicon dioxide. Its atoms are joined by covalent bonds in a three-dimensional network with fixed bond angles, which makes it rigid and hard and gives it a high melting point. It has no mobile ions or electrons, so it does not conduct.
Review: 3.2 Properties of Solids
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Topic 3.2 · Say it in contextSolid KCl does not conduct electricity, but KCl dissolved in water does: KCl(s) → K⁺(aq) + Cl⁻(aq). Why the difference?
In the solid, the ions are locked in place in the lattice and cannot move. Dissolving frees them, and moving ions carry the current. An ionic solid conducts only when its ions are mobile, as when it is melted or dissolved.
Review: 3.2 Properties of Solids
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Topic 3.2 · Spot the mistake"Graphite is soft, so the bonds between its carbon atoms must be weak." Is that right?
No. Within each layer, carbon atoms are joined by strong covalent bonds, which is why graphite has a high melting point. Graphite is soft because adjacent layers are held together much more weakly and can slide past each other.
Review: 3.2 Properties of Solids
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Topic 3.3 · Say it in contextAt 100 °C and 1 atm, one mole of liquid water takes up about 19 mL, but one mole of steam takes up about 30.6 L. The molar volumes of ice and liquid water differ by less than 10%. Why such a gap for the gas?
In solids and liquids the particles are in close contact at all times, so their molar volumes are similar. In a gas the particles are far apart and in constant motion, with little effect from attractions, so a mole spreads over a volume more than a thousand times larger.
Review: 3.3 Solids, Liquids, and Gases
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Topic 3.4 · Work it outWhat pressure does 0.250 mol of an ideal gas exert in a 5.00 L container at 298 K? Use R = 0.08206 L·atm/(mol·K).
1.22 atm. P = nRT/V = (0.250 mol)(0.08206 L·atm/(mol·K))(298 K)/(5.00 L) = 1.22 atm. Moles, liters and kelvins cancel, leaving atmospheres.
Review: 3.4 Ideal Gas Law
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Topic 3.4 · Work it outA flask holds 0.300 mol of N₂ and 0.100 mol of O₂ at a total pressure of 2.00 atm. What is the partial pressure of O₂?
0.500 atm. The mole fraction of O₂ is 0.100/(0.300 + 0.100) = 0.250, so P(O₂) = X(O₂) × P(total) = (0.250)(2.00 atm) = 0.500 atm. Each gas contributes pressure in proportion to its share of the moles.
Review: 3.4 Ideal Gas Law
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Topic 3.4 · Say it in contextFor a gas sample at constant pressure, a graph of volume against temperature in kelvins is a straight line through the origin. What does that tell you?
Volume is directly proportional to Kelvin temperature when n and P are fixed. PV = nRT rearranges to V = (nR/P)T, a line through the origin with slope nR/P, so doubling the Kelvin temperature doubles the volume.
Review: 3.4 Ideal Gas Law
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Topic 3.5 · Work it outHelium (4.003 g/mol) and argon (39.95 g/mol) are at the same temperature. Compare their average kinetic energies, and find how many times faster a typical helium atom moves.
Their average kinetic energies are equal, because average kinetic energy depends only on the Kelvin temperature. From KE = ½mv², speed is proportional to 1/√m, so helium moves √(39.95/4.003) = 3.16 times as fast.
Review: 3.5 Kinetic Molecular Theory
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Topic 3.5 · Spot the mistake"Heating a gas from 20 °C to 40 °C doubles the average kinetic energy of its particles." Is that right?
No. Average kinetic energy is proportional to Kelvin temperature, not Celsius. The temperature goes from 293 K to 313 K, so the average kinetic energy rises by a factor of 313/293 = 1.07, about 7%.
Review: 3.5 Kinetic Molecular Theory
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Topic 3.6 · Check the conditionsWould the ideal gas law describe CO₂ better at 1 atm and 500 K or at 40 atm and 300 K? Explain.
At 1 atm and 500 K. There the molecules are far apart and fast, so attractions and the molecules' own volume barely matter. At 40 atm and 300 K the gas is much nearer to condensing, so attractions between molecules pull the real behavior away from ideal.
Review: 3.6 Deviation from Ideal Gas Law
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Topic 3.7 · Work it out5.85 g of NaCl (58.44 g/mol) is dissolved in enough water to make 250.0 mL of solution. What is the molarity?
0.400 M. n = (5.85 g)/(58.44 g/mol) = 0.100 mol, and M = n/V = (0.100 mol)/(0.2500 L) = 0.400 M. Convert the volume to liters of solution before dividing.
Review: 3.7 Solutions and Mixtures
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Topic 3.7 · Spot the mistake"To make 1.00 L of 0.100 M sugar solution, dissolve 0.100 mol of sugar in 1.00 L of water." Is that right?
No. Molarity is moles of solute per liter of solution, not per liter of water. Adding the sugar to 1.00 L of water gives more than 1.00 L of solution. Dissolve the sugar in less water, then add water until the total volume is 1.00 L.
Review: 3.7 Solutions and Mixtures
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Topic 3.8 · Say it in contextA particle drawing of aqueous CaCl₂ shows 3 Ca²⁺ ions. How many Cl⁻ ions should it show, and which end of each nearby water molecule should face each kind of ion?
6 Cl⁻ ions, so the charges balance. The oxygen end of water, with its partial negative charge, faces each Ca²⁺, and the hydrogen ends, with their partial positive charges, face each Cl⁻. The drawing shows both the ion ratio and the ion-dipole attractions.
Review: 3.8 Representations of Solutions
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Topic 3.9 · Work it outA chromatogram is run on polar paper with a solvent much less polar than the paper. The solvent front moved 8.0 cm and dye A moved 3.2 cm. Dye B has an Rf of 0.75. Find dye A's Rf and say which dye is more polar.
Rf = 3.2 cm ÷ 8.0 cm = 0.40. Dye A is more polar. The polar paper holds a more polar dye back more strongly, so it travels a shorter distance with the less polar solvent and has the smaller Rf.
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Topic 3.10 · Which one fitsWill iodine, I₂, dissolve better in water or in hexane, C₆H₁₄? Explain.
In hexane. I₂ and hexane are both nonpolar and attract each other through dispersion forces, so they mix readily. Water molecules are held to each other by hydrogen bonds, and I₂ cannot attract water strongly enough to break in among them.
Review: 3.10 Solubility
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Topic 3.11 · Which one fitsA molecule absorbs microwave radiation. What kind of transition is that? What kind does infrared cause, and what kind does ultraviolet or visible light cause?
Microwaves cause transitions between rotational levels. Infrared causes transitions between vibrational levels, and ultraviolet or visible light causes transitions between electronic energy levels. Each region matches a different kind of molecular motion or electronic change.
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Topic 3.12 · Work it outWhat is the energy of one photon of violet light with a wavelength of 434 nm? Use h = 6.626 × 10⁻³⁴ J·s and c = 2.998 × 10⁸ m/s.
4.58 × 10⁻¹⁹ J. Convert 434 nm to 4.34 × 10⁻⁷ m, then E = hc/λ = (6.626 × 10⁻³⁴ J·s)(2.998 × 10⁸ m/s) ÷ (4.34 × 10⁻⁷ m) = 4.58 × 10⁻¹⁹ J. This combines c = λν with E = hν.
Review: 3.12 Properties of Photons
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Topic 3.12 · Work it outAn atom emits a photon, and the atom's energy drops by 3.03 × 10⁻¹⁹ J. What is the wavelength of the photon? Use h = 6.626 × 10⁻³⁴ J·s and c = 2.998 × 10⁸ m/s.
656 nm. The photon carries exactly the energy the atom lost, 3.03 × 10⁻¹⁹ J. λ = hc/E = (6.626 × 10⁻³⁴ J·s)(2.998 × 10⁸ m/s) ÷ (3.03 × 10⁻¹⁹ J) = 6.56 × 10⁻⁷ m, which is 656 nm.
Review: 3.12 Properties of Photons
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Topic 3.13 · Work it outA solution has an absorbance of 0.450 in a 1.00 cm cuvette. At this wavelength the absorbing species has a molar absorptivity of 1.50 × 10⁴ M⁻¹ cm⁻¹. What is its concentration?
3.00 × 10⁻⁵ M. From A = εbc, c = A/(εb) = 0.450 ÷ ((1.50 × 10⁴ M⁻¹ cm⁻¹)(1.00 cm)) = 3.00 × 10⁻⁵ M. Absorbance has no units, so the result comes out in moles per liter.
Review: 3.13 Beer-Lambert Law
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Topic 3.13 · Check the conditionsA student makes a calibration curve with standards in 1.00 cm cuvettes, then measures an unknown in a 2.00 cm cuvette and reads its concentration off the curve. Is the result valid?
No. The curve holds only for the path length used to make it. Doubling b doubles the absorbance at the same concentration, so the reading will be twice the true concentration. Path length and wavelength must stay fixed for absorbance to track concentration alone.
Review: 3.13 Beer-Lambert Law
Unit 4: Chemical Reactions
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Topic 4.1 · Which one fitsTwo clear, colorless solutions are mixed, and a cloudy yellow solid appears at once. Is this evidence of a physical change or a chemical change?
A chemical change. Formation of a precipitate is evidence that a new substance has formed: a solid that neither starting solution contained, with a different composition from what was there before.
Review: 4.1 Introduction for Reactions
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Topic 4.1 · Spot the mistake"Boiling water gives off bubbles of gas, and forming a gas is a sign of a chemical change, so boiling is a chemical change." Is that right?
No. The bubbles are water vapor, still H₂O, so the composition has not changed. A change of phase is a physical change. Gas formation points to a chemical change only when the gas is a new substance, such as CO₂ from vinegar and baking soda.
Review: 4.1 Introduction for Reactions
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Topic 4.2 · Spot the mistakeA student mixes Pb(NO₃)₂(aq) with KI(aq), solid PbI₂ forms, and the student writes the net ionic equation Pb²⁺(aq) + I⁻(aq) → PbI₂(s). Is that right?
No. It does not balance: the right side has 2 I atoms and the left has 1, and the charge is 1+ on the left but 0 on the right. The correct equation is Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s). K⁺ and NO₃⁻ are spectator ions and are left out.
Review: 4.2 Net Ionic Equations
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Topic 4.2 · Which one fitsAqueous Na₂CO₃ and CaCl₂ are mixed and solid CaCO₃ forms. Which form of equation shows only the species that actually react, and what is it?
The net ionic equation: Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s). Na⁺ and Cl⁻ are spectator ions, unchanged on both sides of the complete ionic equation, so they are left out. Atoms balance, and the total charge is 0 on each side.
Review: 4.2 Net Ionic Equations
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Topic 4.3 · Work it outA particle diagram shows 4 H₂ molecules and 3 O₂ molecules before the reaction 2 H₂ + O₂ → 2 H₂O runs to completion. What should the particle diagram after the reaction show?
4 H₂O molecules and 1 leftover O₂ molecule. Four H₂ molecules need only two O₂ molecules, so one O₂ is unused. Both diagrams then show 8 H atoms and 6 O atoms, because atoms are conserved.
Review: 4.3 Representations of Reactions
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Topic 4.4 · Which one fitsMelting ice and decomposing water into H₂ and O₂ both take in energy. Which one is a chemical process, and what is the difference at the level of bonds?
Decomposing water is chemical: it breaks O-H covalent bonds and forms H-H and O=O bonds, making new substances. Melting only changes the intermolecular interactions between H₂O molecules, which stay intact, so it is a physical process.
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Topic 4.4 · Spot the mistake"Dissolving NaCl in water is clearly a physical change, because no bonds are broken." Is that right?
No. Dissolving breaks the ionic bonds between Na⁺ and Cl⁻ and forms ion-dipole interactions with water. Because bonds are broken, a plausible argument can be made for calling it either a physical or a chemical process. The claim that no bonds break is wrong.
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Topic 4.5 · Work it outCH₄ + 2 O₂ → CO₂ + 2 H₂O. What mass of CO₂ forms when 4.00 g of CH₄ burns completely in excess oxygen? Molar masses: CH₄ 16.04 g/mol, CO₂ 44.01 g/mol.
(4.00 g)/(16.04 g/mol) = 0.2494 mol CH₄. The coefficients give 1 mol CO₂ per mole of CH₄, so 0.2494 mol CO₂ forms, and (0.2494 mol)(44.01 g/mol) = 11.0 g CO₂, to three significant figures.
Review: 4.5 Stoichiometry
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Topic 4.5 · Work it out3.00 mol of H₂ and 2.00 mol of O₂ react by 2 H₂ + O₂ → 2 H₂O until one of them runs out. How many moles of H₂O form, and what is left over?
H₂ runs out first, because 3.00 mol H₂ needs only 1.50 mol O₂. So 3.00 mol H₂O forms (2 mol H₂O per 2 mol H₂), and 2.00 − 1.50 = 0.50 mol O₂ is left over.
Review: 4.5 Stoichiometry
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Topic 4.5 · Work it outZn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g). Excess zinc reacts with 50.0 mL of 2.00 M HCl. What volume of H₂ forms at 298 K and 1.00 atm? Use R = 0.08206 L·atm/(mol·K).
(0.0500 L)(2.00 mol/L) = 0.100 mol HCl, which gives 0.0500 mol H₂ (2 mol HCl per mole of H₂). V = nRT/P = (0.0500 mol)(0.08206 L·atm/(mol·K))(298 K)/(1.00 atm) = 1.22 L.
Review: 4.5 Stoichiometry
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Topic 4.6 · Work it out25.00 mL of HCl solution is titrated with 0.1000 M NaOH, and the equivalence point is reached after 18.40 mL of NaOH. HCl + NaOH → NaCl + H₂O. What is the concentration of the HCl?
(0.01840 L)(0.1000 mol/L) = 1.840 × 10⁻³ mol NaOH. The 1:1 ratio means 1.840 × 10⁻³ mol HCl was present at the equivalence point, so [HCl] = (1.840 × 10⁻³ mol)/(0.02500 L) = 0.07360 M.
Review: 4.6 Introduction to Titration
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Topic 4.6 · Work it out20.0 mL of H₂SO₄ solution needs 30.0 mL of 0.100 M NaOH to reach the equivalence point. H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O. What is the concentration of the H₂SO₄?
(0.0300 L)(0.100 mol/L) = 3.00 × 10⁻³ mol NaOH. Two NaOH react with each H₂SO₄, so the moles are not equal at equivalence: there was 1.50 × 10⁻³ mol H₂SO₄, and (1.50 × 10⁻³ mol)/(0.0200 L) = 0.0750 M.
Review: 4.6 Introduction to Titration
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Topic 4.6 · Say it in contextIn a titration of an acid with NaOH, a student keeps adding NaOH well past the point where the indicator first changes color, then records that volume. How does this affect the calculated amount of acid?
It makes it too high. The endpoint, the color change, signals the equivalence point, where the acid has just been used up. NaOH added after that reacted with nothing, but the calculation counts it as if it had reacted with acid.
Review: 4.6 Introduction to Titration
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Topic 4.7 · Which one fitsClassify Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) as acid-base, redox or precipitation, and give the evidence.
Redox. Oxidation numbers change: zinc goes from 0 to +2, so it is oxidized and loses two electrons, and copper goes from +2 to 0, so it is reduced. No proton is transferred and no insoluble ionic compound forms, so it is neither of the other two types.
Review: 4.7 Types of Chemical Reactions
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Topic 4.7 · Check the conditionsSolutions of Na₂SO₄ and KNO₃ are mixed. Do the conditions for a precipitation reaction hold?
No. The only new combinations of ions are K₂SO₄ and NaNO₃, and all sodium, potassium and nitrate salts are soluble in water. No insoluble compound can form, so the ions stay dissolved and no precipitate appears.
Review: 4.7 Types of Chemical Reactions
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Topic 4.7 · Spot the mistake"Burning methane, CH₄ + 2 O₂ → CO₂ + 2 H₂O, is not a redox reaction, because no ions are involved." Is that right?
No. Redox is identified by changes in oxidation numbers, not by the presence of ions. Carbon goes from −4 in CH₄ to +4 in CO₂, so it is oxidized, and oxygen goes from 0 in O₂ to −2, so it is reduced. Combustion is a subclass of redox.
Review: 4.7 Types of Chemical Reactions
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Topic 4.8 · Which one fitsIn NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), which species is the Brønsted-Lowry acid, which is the base, and what are the two conjugate acid-base pairs?
H₂O is the acid, since it donates a proton, and NH₃ is the base, since it accepts one. The conjugate pairs are NH₄⁺/NH₃ and H₂O/OH⁻. The two members of each pair differ by one H⁺.
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Topic 4.8 · Spot the mistake"In HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq), water acts as a base, so water is a base." Is that right?
No. Water's structure lets it accept a proton, as it does from HCl, and also donate one, as it does to NH₃. Whether water acts as an acid or a base depends on the species it reacts with in the solution.
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Topic 4.9 · Work it outCombine the half-reactions Al(s) → Al³⁺(aq) + 3 e⁻ and Cu²⁺(aq) + 2 e⁻ → Cu(s) into one balanced redox equation.
Make the electrons lost equal the electrons gained: multiply the aluminum half-reaction by 2 and the copper one by 3, so 6 electrons move. Adding them and canceling the electrons gives 2 Al(s) + 3 Cu²⁺(aq) → 2 Al³⁺(aq) + 3 Cu(s), with a charge of 6+ on each side.
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Topic 4.9 · Spot the mistake"Ag⁺(aq) + Cu(s) → Ag(s) + Cu²⁺(aq) is balanced, because each side has one Ag and one Cu." Is that right?
No. The atoms balance but the charge does not: 1+ on the left, 2+ on the right. Copper loses 2 electrons while each Ag⁺ gains only 1, so two Ag⁺ are needed: 2 Ag⁺(aq) + Cu(s) → 2 Ag(s) + Cu²⁺(aq).
Unit 5: Kinetics
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Topic 5.1 · Work it outIn 2 N₂O₅(g) → 4 NO₂(g) + O₂(g), N₂O₅ is being used up at 0.0120 M/s at one moment. How fast are NO₂ and O₂ forming at that moment?
The rates follow the coefficients: every 2 mol of N₂O₅ that react make 4 mol NO₂ and 1 mol O₂. NO₂ forms at (0.0120 M/s)(4/2) = 0.0240 M/s, and O₂ forms at (0.0120 M/s)(1/2) = 0.00600 M/s.
Review: 5.1 Reaction Rates
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Topic 5.1 · Which one fitsPowdered zinc reacts with hydrochloric acid much faster than a single zinc pellet of the same mass in the same acid at the same temperature. Which factor explains the difference?
Surface area. The powder exposes far more zinc atoms to the acid at once, so more collisions between the acid and zinc happen each second and the rate is higher. Concentration, temperature and the amount of zinc are the same in both.
Review: 5.1 Reaction Rates
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Topic 5.2 · Work it outFor A + B → products, initial rates: [A] 0.100 M, [B] 0.100 M, rate 2.00 × 10⁻³ M/s; [A] 0.200 M, [B] 0.100 M, rate 8.00 × 10⁻³ M/s; [A] 0.100 M, [B] 0.200 M, rate 2.00 × 10⁻³ M/s. Find the rate law and k.
Doubling [A] multiplies the rate by 4 = 2², so the reaction is second order in A. Doubling [B] leaves the rate unchanged, so it is zeroth order in B. Rate = k[A]², and k = (2.00 × 10⁻³ M/s)/(0.100 M)² = 0.200 M⁻¹·s⁻¹.
Review: 5.2 Introduction to Rate Law
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Topic 5.2 · Spot the mistake"For 2 NO(g) + 2 H₂(g) → N₂(g) + 2 H₂O(g), the rate law must be rate = k[NO]²[H₂]², from the coefficients." Is that right?
No. The orders in a rate law come from experiment, such as comparing initial rates, not from the coefficients of an overall equation. Experiments give rate = k[NO]²[H₂] for this reaction, first order in H₂, not second.
Review: 5.2 Introduction to Rate Law
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Topic 5.3 · Say it in contextFor the decomposition of A, plots of [A], ln[A] and 1/[A] against time are made. Only the ln[A] plot is a straight line, with a slope of −0.0450 s⁻¹. What is the order in A, and what is k?
First order, because a straight line for ln[A] against time means first order in A. The first-order equation ln[A]t − ln[A]0 = −kt makes the slope equal to −k, so k = 0.0450 s⁻¹.
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Topic 5.3 · Work it outA first-order reaction has a rate constant k = 0.0350 min⁻¹. What is its half-life?
t½ = 0.693/k = 0.693/(0.0350 min⁻¹) = 19.8 min. A first-order half-life is constant, so the reactant concentration halves every 19.8 min, however much is left.
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Topic 5.3 · Work it outA radioactive isotope decays by first-order kinetics with a half-life of 10.0 min. A sample starts with 80.0 mg of the isotope. How much is left after 30.0 min?
30.0 min is three half-lives, and each one halves the amount: 80.0 mg, then 40.0 mg, then 20.0 mg, then 10.0 mg. Counting half-lives works because a first-order half-life stays the same as the amount falls.
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Topic 5.4 · Which one fitsNO(g) + O₃(g) → NO₂(g) + O₂(g) is an elementary step. What is its rate law, and why can you write it straight from the equation?
Rate = k[NO][O₃]. An elementary step shows exactly which particles collide, here one NO and one O₃, so each is first order and the step is second order overall. That shortcut does not work for an overall equation, whose orders must be found by experiment.
Review: 5.4 Elementary Reactions
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Topic 5.4 · Check the conditionsSomeone proposes that 2 H₂(g) + 2 NO(g) → N₂(g) + 2 H₂O(g) happens in a single elementary step. Is that a reasonable assumption?
No. As one step it would need four molecules to collide at the same moment. Elementary reactions in which even three particles collide at once are rare, so this reaction almost certainly goes through a mechanism of several steps.
Review: 5.4 Elementary Reactions
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Topic 5.5 · Say it in contextTwo molecules collide with more than enough energy to get over the activation energy, yet no products form. What explains this?
Their orientation was wrong. A successful collision needs both enough energy to overcome the activation energy and an orientation that lets the bonds rearrange in the required way. If the reacting atoms do not meet, the molecules bounce apart unchanged.
Review: 5.5 Collision Model
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Topic 5.5 · Spot the mistake"Raising the temperature speeds up a reaction mainly because the particles collide more often." Is that right?
No. Collisions do become somewhat more frequent, but the main effect shows on the Maxwell-Boltzmann distribution: at a higher temperature it spreads toward higher energies, so a larger fraction of collisions have at least the activation energy and more of them succeed.
Review: 5.5 Collision Model
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Topic 5.6 · Work it outA reaction energy profile (invented values) shows the reactants at 40 kJ/mol, the transition state at 125 kJ/mol and the products at 15 kJ/mol. Find the activation energy of the forward reaction and the overall energy change.
Activation energy = transition state − reactants = 125 − 40 = 85 kJ/mol. Overall change = products − reactants = 15 − 40 = −25 kJ/mol, so the reaction is exothermic. The reverse reaction would need 125 − 15 = 110 kJ/mol.
Review: 5.6 Reaction Energy Profile
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Topic 5.7 · Which one fitsA proposed mechanism: step 1, NO₂ + NO₂ → NO₃ + NO; step 2, NO₃ + CO → NO₂ + CO₂. What is the overall reaction, and which species is an intermediate?
Adding the steps and canceling what appears on both sides gives NO₂ + CO → NO + CO₂. NO₃ is the intermediate: step 1 makes it and step 2 uses it up, so it is present only while the reaction is occurring.
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Topic 5.7 · Check the conditionsIn a proposed mechanism, species X is used up in step 1 and made again in step 2, and X does not appear in the overall equation. Is X an intermediate or a catalyst?
A catalyst. A catalyst is consumed in an earlier step and regenerated in a later one, so its net amount stays the same. An intermediate runs the other way: one step produces it and a later step consumes it.
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Topic 5.8 · Which one fitsFor NO₂ + CO → NO + CO₂, step 1, NO₂ + NO₂ → NO₃ + NO, is slow and step 2, NO₃ + CO → NO₂ + CO₂, is fast. What rate law does this mechanism predict?
Rate = k[NO₂]². The slow first step limits the rate, and its rate law follows from its molecularity: two NO₂ molecules collide. CO takes part only in the fast second step, so it does not appear in the rate law even though it is a reactant.
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Topic 5.8 · Spot the mistakeMechanism for 2 H₂O₂ → 2 H₂O + O₂: slow step 1, H₂O₂ + I⁻ → H₂O + IO⁻; fast step 2, H₂O₂ + IO⁻ → H₂O + O₂ + I⁻. A student says, "The rate law is rate = k[H₂O₂]²." Is that right?
No. The slow first step sets the rate law: one H₂O₂ and one I⁻ collide, so rate = k[H₂O₂][I⁻]. I⁻ is in the rate law but not in the overall equation, because it is a catalyst used up in the slow step and regenerated in the fast one.
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Topic 5.9 · Work it outMechanism for 2 NO + Br₂ → 2 NOBr: step 1, fast and reversible, NO + Br₂ ⇌ NOBr₂; step 2, slow, NOBr₂ + NO → 2 NOBr. Use the pre-equilibrium approximation to find the rate law.
Step 2 sets the rate: rate = k₂[NOBr₂][NO]. NOBr₂ is an intermediate, so replace it. In the fast equilibrium the forward and reverse rates are equal: k₁[NO][Br₂] = k₋₁[NOBr₂], so [NOBr₂] = (k₁/k₋₁)[NO][Br₂]. Then rate = k[NO]²[Br₂], with k = k₂k₁/k₋₁.
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Topic 5.9 · Check the conditionsMechanism for 2 NO(g) + O₂(g) → 2 NO₂(g): step 1, fast and reversible, 2 NO ⇌ N₂O₂; step 2, slow, N₂O₂ + O₂ → 2 NO₂. A student writes rate = k₂[N₂O₂][O₂] from the slow step and stops there. Is that a finished rate law?
No. N₂O₂ is an intermediate, present only while the reaction runs, so the rate law must not be left in terms of it. Because the first step is not rate limiting, use the pre-equilibrium approximation: k₁[NO]² = k₋₁[N₂O₂], so [N₂O₂] = (k₁/k₋₁)[NO]². Then rate = k[NO]²[O₂], with k = k₂k₁/k₋₁.
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Topic 5.10 · Work it outA two-step energy profile (invented values): reactants 0 kJ/mol, first transition state 50 kJ/mol, intermediate 20 kJ/mol, second transition state 95 kJ/mol, products −30 kJ/mol. Find each step's activation energy, the overall energy change and the slower step.
Step 1: 50 − 0 = 50 kJ/mol. Step 2: 95 − 20 = 75 kJ/mol. Overall: −30 − 0 = −30 kJ/mol, so the reaction is exothermic. Step 2 has the larger activation energy, so it is the slower, rate-limiting step. The valley at 20 kJ/mol is the intermediate.
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Topic 5.11 · Spot the mistake"Adding a catalyst makes a reaction release more energy, which is why it goes faster." Is that right?
No. A catalyst provides a reaction path with a lower activation energy, so more collisions are effective and the rate rises. The reactants and products are the same, so the overall energy change is the same with or without the catalyst.
Review: 5.11 Catalysis
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Topic 5.11 · Which one fitsChlorine atoms speed up the loss of ozone: step 1, Cl + O₃ → ClO + O₂; step 2, ClO + O → Cl + O₂. Which species is the catalyst, and which is the intermediate?
Cl is the catalyst: step 1 consumes it and step 2 regenerates it, so its net amount is unchanged. ClO is the intermediate, made in step 1 and used up in step 2. Adding the steps gives the overall reaction O₃ + O → 2 O₂.
Review: 5.11 Catalysis
Unit 6: Thermochemistry
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Topic 6.1 · Say it in contextA solid dissolves in water in a coffee-cup calorimeter, and the temperature of the mixture falls from 22.0 °C to 16.5 °C. Is the dissolution exothermic or endothermic?
Endothermic. The temperature fell, so the dissolving process absorbed thermal energy from the water around it. The energy the process gained is the energy the water lost.
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Topic 6.1 · Spot the mistake"An ice cube melting in my hand feels cold, so melting gives off cold and is exothermic." Is that right?
No. Melting is endothermic. The ice absorbs energy from your hand by heat transfer, so your hand loses energy and its temperature drops, which you feel as cold. Energy flows out of your hand; nothing called cold flows in.
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Topic 6.2 · Say it in contextAn energy diagram for a reaction shows the products drawn higher than the reactants. What does that say about the reaction and its surroundings?
The reaction is endothermic: the system ends with more energy than it started with, so ΔH is positive. That energy comes from the surroundings, which lose energy and cool unless energy is supplied from outside.
Review: 6.2 Energy Diagrams
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Topic 6.3 · Say it in contextA block of copper at 100 °C is dropped into water at 20 °C in an insulated cup. Describe, in terms of particles, what happens until nothing more changes.
The copper atoms have the greater average kinetic energy, so collisions where the two touch transfer energy from the copper to the water. The copper cools and the water warms until both have the same average kinetic energy, and so the same temperature. That is thermal equilibrium.
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Topic 6.4 · Work it outHow much heat does 50.0 g of water absorb when it warms from 22.0 °C to 34.5 °C? Use c = 4.18 J/(g·°C).
ΔT = 34.5 °C − 22.0 °C = 12.5 °C. q = mcΔT = (50.0 g)(4.18 J/(g·°C))(12.5 °C) = 2.61 × 10³ J, or 2.61 kJ. The value is positive because the water gained energy.
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Topic 6.4 · Work it outA 60.0 g piece of an unknown metal at 100.0 °C is placed in 50.0 g of water at 20.0 °C, and both end at 30.0 °C. Find the metal's specific heat. Water: c = 4.18 J/(g·°C). Assume no energy is lost to the surroundings.
The water gains q = (50.0 g)(4.18 J/(g·°C))(10.0 °C) = 2.09 × 10³ J, so the metal loses 2.09 × 10³ J. c = q/(mΔT) = (2.09 × 10³ J)/((60.0 g)(70.0 °C)) = 0.498 J/(g·°C).
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Topic 6.4 · Spot the mistake"If equal masses of water and copper absorb the same amount of energy, their temperatures rise by the same number of degrees." Is that right?
No. ΔT = q/(mc), so the substance with the lower specific heat capacity warms more. Copper's is 0.385 J/(g·°C) and water's is 4.18 J/(g·°C), so the copper's temperature rises about 11 times as much as the water's.
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Topic 6.5 · Work it outHow much energy is needed to melt 45.0 g of ice at 0 °C? The molar enthalpy of fusion of water is 6.01 kJ/mol, and its molar mass is 18.02 g/mol.
(45.0 g)/(18.02 g/mol) = 2.497 mol. q = nΔH = (2.497 mol)(6.01 kJ/mol) = 15.0 kJ. The temperature stays at 0 °C while the ice melts, even though the system is absorbing energy.
Review: 6.5 Energy of Phase Changes
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Topic 6.5 · Work it outThe molar enthalpy of vaporization of water is +40.7 kJ/mol. What is q when 2.00 mol of steam condenses to liquid water at 100 °C?
Condensing is the reverse of vaporizing, so its molar enthalpy is −40.7 kJ/mol. q = (2.00 mol)(−40.7 kJ/mol) = −81.4 kJ. The system releases 81.4 kJ, the same amount it would absorb to vaporize 2.00 mol.
Review: 6.5 Energy of Phase Changes
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Topic 6.6 · Work it outCH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l), with a molar enthalpy of reaction of −890. kJ per mole of CH₄. How much heat is released when 12.0 g of CH₄ burns? Molar mass of CH₄: 16.04 g/mol.
(12.0 g)/(16.04 g/mol) = 0.7481 mol CH₄. q = nΔH = (0.7481 mol)(−890. kJ/mol) = −666 kJ, so 666 kJ is released. The negative sign means thermal energy leaves the reacting system for the surroundings.
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Topic 6.6 · Say it in contextA reaction in a coffee-cup calorimeter makes the solution warm up. Where did that thermal energy come from, at the particle level?
From chemical potential energy. Breaking and forming bonds leaves the products with lower potential energy than the reactants, and the difference becomes kinetic energy of the particles, which shows up as a rise in temperature. The reaction is exothermic, and the warm solution then transfers energy to its surroundings.
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Topic 6.7 · Work it outEstimate ΔH for H₂(g) + Br₂(g) → 2 HBr(g) from average bond energies: H-H 436 kJ/mol, Br-Br 193 kJ/mol, H-Br 366 kJ/mol.
Broken: 436 + 193 = 629 kJ. Formed: 2 × 366 = 732 kJ. ΔH = broken − formed = 629 − 732 = −103 kJ/mol. More energy is released in forming the bonds than is required to break them, so the reaction is exothermic.
Review: 6.7 Bond Enthalpies
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Topic 6.7 · Spot the mistake"Breaking bonds releases energy, and that is why burning a fuel gives off heat." Is that right?
No. Breaking a bond always requires energy. Burning a fuel releases energy because forming the bonds in the products, CO₂ and H₂O, releases more energy than breaking the bonds in the fuel and oxygen requires.
Review: 6.7 Bond Enthalpies
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Topic 6.8 · Work it outFind ΔH° for C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l). Standard enthalpies of formation: C₃H₈(g) −103.8 kJ/mol, CO₂(g) −393.5 kJ/mol, H₂O(l) −285.8 kJ/mol.
ΔH° = ΣΔH°f(products) − ΣΔH°f(reactants) = [3(−393.5) + 4(−285.8)] − [−103.8 + 5(0)] = −2323.7 − (−103.8) = −2219.9 kJ/mol. O₂(g) is an element in its standard state, so its ΔH°f is zero.
Review: 6.8 Enthalpy of Formation
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Topic 6.8 · Spot the mistake"For 2 H₂(g) + O₂(g) → 2 H₂O(l), ΔH° is just the ΔH°f of H₂O(l), −285.8 kJ/mol." Is that right?
No. Each ΔH°f is multiplied by its coefficient in the equation. Two moles of H₂O form, so ΔH° = 2(−285.8) − 0 = −571.6 kJ/mol. The reactants are elements in their standard states, whose ΔH°f values are zero.
Review: 6.8 Enthalpy of Formation
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Topic 6.9 · Work it outGiven C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ, and 2 CO(g) + O₂(g) → 2 CO₂(g), ΔH = −566.0 kJ, find ΔH for 2 C(s) + O₂(g) → 2 CO(g).
Double the first: 2 C + 2 O₂ → 2 CO₂, ΔH = −787.0 kJ. Reverse the second: 2 CO₂ → 2 CO + O₂, ΔH = +566.0 kJ. Adding them cancels 2 CO₂ and one O₂, giving 2 C + O₂ → 2 CO, with ΔH = −787.0 + 566.0 = −221.0 kJ.
Review: 6.9 Hess’s Law
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Topic 6.9 · Spot the mistake"A → B has ΔH = +40 kJ, so 2 B → 2 A has ΔH = +80 kJ." Is that right?
No. Multiplying a reaction by 2 doubles ΔH, and reversing it changes the sign. Both happen here, so ΔH for 2 B → 2 A is −80 kJ.
Review: 6.9 Hess’s Law
Unit 7: Equilibrium
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Topic 7.1 · Say it in contextA flask of N₂O₄ is sealed. A graph shows [N₂O₄] falling and [NO₂] rising, then both lines go flat while neither is zero. What do the flat lines tell you?
The system has reached equilibrium. Both species are present and their concentrations stay constant because the forward and reverse reactions now run at equal rates, not because the reactions stopped.
Review: 7.1 Introduction to Equilibrium
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Topic 7.1 · Spot the mistake"Once a reaction reaches equilibrium, the molecules stop reacting." Is that right?
No. Equilibrium is dynamic. The forward and reverse reactions keep going at equal rates, so products form exactly as fast as they are used up, and no net change is observable.
Review: 7.1 Introduction to Equilibrium
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Topic 7.2 · Which one fitsA sealed mixture is not yet at equilibrium, and right now the forward reaction is faster than the reverse reaction. Which way is the composition changing, and when does that change stop?
Reactants are being converted to products, because more product forms each second than is used up. The net change stops when the two rates become equal, which is the equilibrium state.
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Topic 7.3 · Work it outFor N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), a mixture has [N₂] = 0.500 M, [H₂] = 0.200 M and [NH₃] = 0.100 M, and Kc = 0.060. These numbers are a teaching example. Find Qc and the direction of net reaction.
Qc = [NH₃]² / ([N₂][H₂]³) = (0.100)² / ((0.500)(0.200)³) = 2.50. Qc is greater than Kc, so the reaction runs in reverse, using up NH₃, until Qc falls to Kc.
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Topic 7.3 · Spot the mistake"For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = P(CaO) × P(CO₂) / P(CaCO₃)." Is that right?
No. Solids are left out of the expression, because their concentrations do not depend on how much solid is present. The expression is Kp = P(CO₂), so only the carbon dioxide pressure matters.
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Topic 7.4 · Work it outH₂(g) + I₂(g) ⇌ 2 HI(g) reaches equilibrium with [H₂] = 0.0200 M, [I₂] = 0.0300 M and [HI] = 0.180 M. These numbers are a teaching example. What is Kc?
Kc = [HI]² / ([H₂][I₂]) = (0.180)² / ((0.0200)(0.0300)) = 0.0324 / 0.000600 = 54.0. Each equilibrium concentration is raised to the power of its coefficient, and K is written as a plain number without units.
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Topic 7.5 · Say it in contextReaction X has K = 3 × 10⁻¹⁸ and reaction Y has K = 5 × 10¹⁵. These numbers are a teaching example. What does each value say about its equilibrium mixture?
X barely proceeds, so its equilibrium mixture is almost all reactants. Y proceeds essentially to completion, so its mixture is almost all products. A K far below 1 favors reactants, and a K far above 1 favors products.
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Topic 7.6 · Work it outA reaction has K = 5.0 × 10⁻³. What is K for the reverse reaction with every coefficient doubled?
Reversing inverts K: 1 / (5.0 × 10⁻³) = 2.0 × 10². Doubling the coefficients squares it: (2.0 × 10²)² = 4.0 × 10⁴. The order of the two steps does not matter.
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Topic 7.6 · Work it outN₂(g) + O₂(g) ⇌ 2 NO(g) has K₁ = 2.0 × 10⁻⁵, and 2 NO(g) + O₂(g) ⇌ 2 NO₂(g) has K₂ = 3.0 × 10². These numbers are a teaching example. What is K for the sum of the two reactions?
Adding the reactions cancels 2 NO and gives N₂(g) + 2 O₂(g) ⇌ 2 NO₂(g). When reactions are added, their K values multiply: K = (2.0 × 10⁻⁵)(3.0 × 10²) = 6.0 × 10⁻³.
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Topic 7.6 · Spot the mistake"If I double every coefficient in a balanced equation, K doubles too." Is that right?
No. Each coefficient is an exponent in the expression, so doubling them squares K. In general, multiplying the coefficients by a factor c raises K to the power c, so K = 10 becomes 100, not 20.
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Topic 7.7 · Work it outPCl₅(g) ⇌ PCl₃(g) + Cl₂(g) has Kc = 1.0 × 10⁻⁵. A flask starts with [PCl₅] = 0.400 M and no products. These numbers are a teaching example. Find [Cl₂] at equilibrium.
Let x = [Cl₂] = [PCl₃], so Kc = x² / (0.400 − x). K is small, so take 0.400 − x ≈ 0.400: x = √(1.0 × 10⁻⁵ × 0.400) = 2.0 × 10⁻³ M. That x is 0.50% of 0.400 M, so the approximation holds.
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Topic 7.7 · Check the conditionsA student solves Kc = x² / (0.0100 − x) with Kc = 4.0 × 10⁻⁴ by dropping x from the denominator, and gets x = 2.0 × 10⁻³ M. Is dropping x justified here?
No. That x is 20% of 0.0100 M, far above the common 5% rule of thumb, so 0.0100 − x is not close to 0.0100. Solving the full quadratic gives x = 1.8 × 10⁻³ M.
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Topic 7.8 · Say it in contextFor A(g) ⇌ B(g), a particle diagram at equilibrium shows 3 A particles and 9 B particles in one box. What is K, and what would a box with 6 A and 6 B mean?
K = [B] / [A] = 9 / 3 = 3, since both share one volume and counts are proportional to concentrations. A box with 6 A and 6 B has Q = 1, below K, so it is not at equilibrium and more B will form.
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Topic 7.9 · Say it in context2 NO₂(g) ⇌ N₂O₄(g) is exothermic as written. NO₂ is brown and N₂O₄ is colorless. A sealed tube of the equilibrium mixture is put in hot water. What happens to the color, and why?
It turns darker brown. Raising the temperature favors the endothermic direction, which here is the reverse reaction, so some N₂O₄ breaks apart into more NO₂.
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Topic 7.9 · Which one fitsAn equilibrium mixture of NO₂ and N₂O₄ in a syringe is squeezed to half its volume at constant temperature. Which way does the equilibrium shift?
Toward N₂O₄, the side with fewer moles of gas. Halving the volume raises the pressure, and turning 2 mol of gas into 1 mol lowers it, which partly relieves the stress.
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Topic 7.10 · Spot the mistake"Adding H₂ to an equilibrium mixture of N₂, H₂ and NH₃ increases K, and that is why more NH₃ forms." Is that right?
No. Adding H₂ changes Q, not K. Q drops below K, so the forward reaction makes NH₃ until Q again equals the same K. At a fixed temperature, K does not change.
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Topic 7.10 · Which one fitsTwo stresses on an equilibrium: removing some of a product, or raising the temperature. Which one changes the value of K?
Raising the temperature. Removing a product changes only Q, and the system shifts until Q returns to the unchanged K. A temperature change moves K itself, and the concentrations redistribute until Q matches the new K.
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Topic 7.11 · Work it outA saturated solution of the salt MX₂ has a molar solubility of 1.50 × 10⁻³ M. These numbers are a teaching example. What is Ksp?
MX₂(s) ⇌ M²⁺(aq) + 2 X⁻(aq), so [M²⁺] = s and [X⁻] = 2s. Ksp = [M²⁺][X⁻]² = s(2s)² = 4s³ = 4(1.50 × 10⁻³)³ = 1.35 × 10⁻⁸.
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Topic 7.11 · Spot the mistake"Ag₂CrO₄ (Ksp = 1.1 × 10⁻¹²) is less soluble than AgCl (Ksp = 1.8 × 10⁻¹⁰), because its Ksp is smaller." Is that right?
No. The salts release different numbers of ions, so compare molar solubilities. For AgCl, s = √(1.8 × 10⁻¹⁰) = 1.3 × 10⁻⁵ M. For Ag₂CrO₄, 4s³ = 1.1 × 10⁻¹², so s = 6.5 × 10⁻⁵ M. Silver chromate is the more soluble salt.
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Topic 7.12 · Work it outAgCl has Ksp = 1.8 × 10⁻¹⁰. What is its molar solubility in 0.10 M NaCl?
[Cl⁻] starts at 0.10 M, so Ksp = s(0.10 + s) ≈ 0.10s, which gives s = 1.8 × 10⁻⁹ M. That s is tiny next to 0.10 M, so dropping it was fair, and it is far below the 1.3 × 10⁻⁵ M AgCl dissolves to in pure water.
Review: 7.12 Common-Ion Effect
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Topic 7.12 · Say it in contextSolid PbI₂ sits in its saturated solution: PbI₂(s) ⇌ Pb²⁺(aq) + 2 I⁻(aq). Some KI is dissolved in the solution. What do you observe, and why?
More yellow PbI₂ precipitates. The added iodide makes Q greater than Ksp, so the equilibrium shifts toward the solid until Q equals Ksp again. A salt is less soluble in a solution that already contains one of its ions.
Review: 7.12 Common-Ion Effect
Unit 8: Acids and Bases
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Topic 8.1 · Work it outA solution at 25 °C has [H₃O⁺] = 2.5 × 10⁻⁴ M. What are its pH and pOH?
pH = −log(2.5 × 10⁻⁴) = 3.60. At 25 °C, pH + pOH = 14.00, so pOH = 10.40. Two significant figures in the concentration give two decimal places in the pH.
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Topic 8.1 · Spot the mistake"Pure water at 50 °C has a pH of about 6.6, so it is slightly acidic." Is that right? At 50 °C, Kw = 5.5 × 10⁻¹⁴.
No. Pure water is neutral at any temperature because [H₃O⁺] = [OH⁻], so pH = pOH. At 50 °C, [H₃O⁺] = √(5.5 × 10⁻¹⁴) = 2.3 × 10⁻⁷ M, so the neutral pH is 6.63. Kw is larger than at 25 °C, so both ions are more concentrated, and the neutral pH is 7.0 only at 25 °C.
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Topic 8.2 · Work it outWhat is the pH of 0.0050 M Ba(OH)₂ at 25 °C?
Ba(OH)₂ is a group II hydroxide, so it dissociates completely and gives two OH⁻ per formula unit: [OH⁻] = 2 × 0.0050 M = 0.010 M. Then pOH = 2.00 and pH = 14.00 − 2.00 = 12.00.
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Topic 8.3 · Work it outWhat is the pH of 0.100 M acetic acid at 25 °C? Ka = 1.8 × 10⁻⁵.
Ka = x² / (0.100 − x) ≈ x² / 0.100, so x = [H₃O⁺] = √(1.8 × 10⁻⁵ × 0.100) = 1.3 × 10⁻³ M and pH = 2.87. The approximation holds, since x is 1.3% of 0.100 M.
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Topic 8.3 · Say it in contextA 0.0500 M solution of a weak acid HA has pH 3.00 at 25 °C. What is its percent ionization, and what does that say about the solution?
[H₃O⁺] = 1.0 × 10⁻³ M, and it equals [A⁻]. Percent ionization = (1.0 × 10⁻³ / 0.0500) × 100 = 2.0%. Only 2 of every 100 molecules ionize, so the vast majority of the acid stays as un-ionized HA.
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Topic 8.3 · Work it outAcetic acid has Ka = 1.8 × 10⁻⁵ at 25 °C. What is Kb for its conjugate base, the acetate ion CH₃COO⁻?
For a conjugate acid-base pair, Ka × Kb = Kw. So Kb = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. The weaker an acid is, the larger the Kb of its conjugate base.
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Topic 8.4 · Work it out25.0 mL of 0.100 M HCl is mixed with 20.0 mL of 0.100 M NaOH at 25 °C. What is the pH of the mixture?
HCl gives 2.50 × 10⁻³ mol H₃O⁺ and NaOH gives 2.00 × 10⁻³ mol OH⁻. They react completely, leaving 5.0 × 10⁻⁴ mol H₃O⁺ in 45.0 mL. [H₃O⁺] = 5.0 × 10⁻⁴ mol / 0.0450 L = 0.011 M, so pH = 1.95.
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Topic 8.4 · Say it in contextEqual moles of acetic acid and NaOH are mixed. Why is the resulting solution slightly basic rather than neutral?
They react completely to make acetate, the conjugate base of a weak acid, and acetate takes protons from water: CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq). That small amount of OH⁻ puts the pH above 7 at 25 °C.
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Topic 8.5 · Work it outA 25.00 mL sample of a monoprotic acid needs 18.40 mL of 0.1000 M NaOH to reach the equivalence point. What is the acid's concentration?
At equivalence, moles of OH⁻ added equal moles of acid present: 0.01840 L × 0.1000 M = 1.840 × 10⁻³ mol. The concentration is 1.840 × 10⁻³ mol / 0.02500 L = 0.07360 M, whether the acid is strong or weak.
Review: 8.5 Acid-Base Titrations
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Topic 8.5 · Say it in contextA weak acid is titrated with NaOH. The pH is 4.74 when half the volume needed to reach equivalence has been added. What does 4.74 tell you?
It is the acid's pKa. At the half-equivalence point half of the HA has become A⁻, so [HA] = [A⁻], and pH = pKa when the pair is at equal concentrations. The Ka is about 1.8 × 10⁻⁵.
Review: 8.5 Acid-Base Titrations
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Topic 8.5 · Which one fitsA solution of a single pure acid is titrated with NaOH. The titration curve shows two distinct equivalence points, the second at twice the volume of NaOH needed for the first. What does that tell you about the acid?
It has at least two acidic protons, removed one at a time, so the expected answer is diprotic (H₃PO₄ can look the same, since its third proton is too weak to show a break). The pH halfway to the first equivalence point gives the first pKa, and the pH halfway between the two gives the second.
Review: 8.5 Acid-Base Titrations
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Topic 8.6 · Which one fitsWhich is the stronger acid, acetic acid (CH₃COOH) or chloroacetic acid (ClCH₂COOH)? Explain from their structures.
Chloroacetic acid. The electronegative chlorine pulls electron density toward itself through the bonds, which stabilizes the negative charge on the conjugate base. A more stable conjugate base means a stronger acid.
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Topic 8.7 · Which one fitsA weak acid HA has pKa = 4.74. In a solution held at pH 3.00, which form is present at the higher concentration, HA or A⁻?
HA. When the solution pH is below the acid's pKa, the protonated form has the higher concentration. Here the ratio [A⁻] / [HA] is about 0.018, so nearly all of it is HA.
Review: 8.7 pH and pKa
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Topic 8.7 · Check the conditionsA weak acid is titrated with NaOH, and the equivalence point is at pH 8.8. The indicators on hand have pKa values of 4.0, 7.0 and 9.0. These numbers are a teaching example. Which indicator should you use?
The one with pKa 9.0. An indicator changes color over a range around its pKa, so its pKa should be close to the pH at the equivalence point. The other two would change color well before equivalence is reached.
Review: 8.7 pH and pKa
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Topic 8.8 · Which one fitsWhich mixture is a buffer: 0.10 M HF with 0.10 M NaF, or 0.10 M HCl with 0.10 M NaCl?
HF with NaF. A buffer needs a large concentration of both members of a conjugate pair: HF reacts with added base and F⁻ reacts with added acid. Cl⁻ is such a weak base that it does not react with added acid, so the HCl mixture cannot buffer.
Review: 8.8 Properties of Buffers
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Topic 8.9 · Work it outA buffer contains 0.300 M CH₃COOH and 0.150 M CH₃COO⁻. The pKa of acetic acid is 4.74. What is the pH?
pH = pKa + log([A⁻] / [HA]) = 4.74 + log(0.150 / 0.300) = 4.74 − 0.30 = 4.44. The pH is below the pKa because the buffer holds more acid than base.
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Topic 8.9 · Spot the mistake"Adding water to double a buffer's volume halves both concentrations, so its pH must change a lot." Is that right?
No. Dilution lowers [HA] and [A⁻] by the same factor, so the ratio [A⁻] / [HA] in the Henderson-Hasselbalch equation stays the same, and so does the pH. What dilution lowers is how much acid or base the buffer can absorb.
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Topic 8.10 · Which one fitsBuffer X is 0.50 M HF and 0.50 M NaF. Buffer Y is 0.050 M HF and 0.050 M NaF. Compare their pH values and their capacities.
They have the same pH, because the ratio [F⁻] / [HF] is 1 in both. Buffer X holds ten times as much of each component, so it can neutralize far more added acid or base before its pH changes sharply.
Review: 8.10 Buffer Capacity
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Topic 8.10 · Say it in contextA buffer contains 0.40 M NH₄⁺ and 0.10 M NH₃. Is it better at absorbing added acid or added base?
Added base. NH₄⁺, the conjugate acid, is the component that reacts with added OH⁻, and there is four times as much of it as of NH₃, which reacts with added acid. A buffer with more conjugate acid than base has greater capacity for added base.
Review: 8.10 Buffer Capacity
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Topic 8.11 · Which one fitsWhich salt becomes more soluble when a strong acid is added, AgCl or Mg(OH)₂?
Mg(OH)₂. Added H₃O⁺ removes OH⁻, a product of Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2 OH⁻(aq), so more solid dissolves. Cl⁻ is the conjugate base of a strong acid and does not react with H₃O⁺, so the solubility of AgCl does not depend on pH.
Review: 8.11 pH and Solubility
Unit 9: Thermodynamics and Electrochemistry
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Topic 9.1 · Which one fitsPredict the sign of ΔS° for 2 H₂(g) + O₂(g) → 2 H₂O(l).
Negative. Three moles of gas become liquid water, so matter becomes much less dispersed: the particles lose freedom of motion and occupy far less volume.
Review: 9.1 Introduction to Entropy
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Topic 9.1 · Spot the mistake"Heating a gas in a rigid container does not change its entropy, because the gas cannot spread into more space." Is that right?
No. Entropy also increases when energy becomes more dispersed. At a higher temperature the distribution of kinetic energies among the gas particles broadens, so the entropy rises even though the volume is fixed.
Review: 9.1 Introduction to Entropy
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Topic 9.2 · Work it outFor N₂(g) + 3 H₂(g) → 2 NH₃(g), use S° values of 191.6 for N₂(g), 130.7 for H₂(g) and 192.5 for NH₃(g), all in J/(mol·K). What is ΔS°?
ΔS° = ΣS°(products) − ΣS°(reactants) = 2(192.5) − [191.6 + 3(130.7)] = 385.0 − 583.7 = −198.7 J/(mol·K). The negative sign fits the reaction, since 4 mol of gas become 2 mol.
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Topic 9.2 · Spot the mistake"Like its ΔHf°, the S° of an element in its standard state, such as O₂(g), is zero, so I can leave it out of ΔS°." Is that right?
No. S° is an absolute entropy, not a change measured from the elements, so every substance has a positive S° at 298 K. O₂(g) has S° of about 205 J/(mol·K), and it must be included when you calculate ΔS°.
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Topic 9.3 · Work it outFor N₂(g) + 3 H₂(g) → 2 NH₃(g), ΔH° = −92.2 kJ/mol and ΔS° = −198.7 J/(mol·K). Find ΔG° at 298 K, and say whether the reaction is thermodynamically favored.
Convert ΔS° to −0.1987 kJ/(mol·K). ΔG° = ΔH° − TΔS° = −92.2 − (298)(−0.1987) = −92.2 + 59.2 = −33.0 kJ/mol. ΔG° is negative, so the reaction is thermodynamically favored at 298 K.
Review: 9.3 Gibbs Free Energy and Thermodynamic Favorability
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Topic 9.3 · Check the conditionsA process has ΔH° > 0 and ΔS° > 0. Under what temperature conditions is it thermodynamically favored?
At high temperature. In ΔG° = ΔH° − TΔS°, the positive ΔH° works against favorability while the −TΔS° term works for it, and that term grows with T. Above some temperature it outweighs ΔH° and ΔG° becomes negative.
Review: 9.3 Gibbs Free Energy and Thermodynamic Favorability
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Topic 9.3 · Work it outUsing ΔGf° values of 86.6 kJ/mol for NO(g) and 51.3 kJ/mol for NO₂(g), find ΔG° for 2 NO(g) + O₂(g) → 2 NO₂(g).
ΔG° = ΣΔGf°(products) − ΣΔGf°(reactants) = 2(51.3) − [2(86.6) + 0] = −70.6 kJ/mol. O₂(g) is an element in its standard state, so its ΔGf° is 0.
Review: 9.3 Gibbs Free Energy and Thermodynamic Favorability
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Topic 9.4 · Say it in contextDiamond turning into graphite has ΔG° < 0 at room temperature and 1 atm, yet a diamond does not change over a lifetime. What explains this?
The conversion is under kinetic control. It is thermodynamically favored, but its activation energy is so high that it does not proceed at a measurable rate. The unchanging diamond is not at equilibrium; it is reacting too slowly to notice.
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Topic 9.5 · Work it outA reaction has ΔG° = −5.70 kJ/mol at 298 K. These numbers are a teaching example. Using R = 8.314 J/(mol·K), what is K?
K = e^(−ΔG°/RT) = e^(5700 / (8.314 × 298)) = e^(2.301) = 9.98. ΔG° is negative, so K is greater than 1 and products are favored at equilibrium.
Review: 9.5 Free Energy and Equilibrium
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Topic 9.5 · Say it in contextA reaction has ΔG° = +50 kJ/mol at 298 K, where RT is about 2.5 kJ/mol. Without a calculator, what can you say about K?
K is far less than 1, so reactants are strongly favored at equilibrium. ΔG° is positive and about 20 times RT, and K depends exponentially on −ΔG°/RT, so K is about e⁻²⁰, on the order of 10⁻⁹.
Review: 9.5 Free Energy and Equilibrium
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Topic 9.6 · Say it in contextWhen a salt dissolves in water, ΔG° reflects three processes at the particle level. Name the three processes, and say why the overall sign of ΔG° is hard to predict.
Breaking the interactions that hold the solid together, reorganizing the solvent (water molecules move apart and rearrange around the dissolved particles), and the new interactions between those particles and the solvent. Each has its own enthalpy and entropy parts, and these partly cancel, so the total is hard to predict from estimates.
Review: 9.6 Free Energy of Dissolution
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Topic 9.7 · Work it outCu₂S(s) → 2 Cu(s) + S(s) has ΔG° = +86.2 kJ/mol, and S(s) + O₂(g) → SO₂(g) has ΔG° = −300.1 kJ/mol. What is ΔG° for the coupled reaction?
Adding them cancels S(s), the shared intermediate: Cu₂S(s) + O₂(g) → 2 Cu(s) + SO₂(g). ΔG° = 86.2 + (−300.1) = −213.9 kJ/mol. The sum is negative, so the favorable step drives the unfavorable one.
Review: 9.7 Coupled Reactions
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Topic 9.7 · Which one fitsPhotosynthesis converts CO₂ and H₂O into glucose, an overall process with ΔG° > 0. What makes it happen?
An external source of energy: absorbed light. Outside energy can make a thermodynamically unfavorable process occur, the same way electrical energy drives an electrolytic cell or recharges a battery.
Review: 9.7 Coupled Reactions
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Topic 9.8 · Which one fitsA cell's reaction has ΔG° > 0, and a power supply forces current through it. Is it galvanic or electrolytic, and at which electrode does oxidation occur?
Electrolytic, because its reaction is thermodynamically unfavored and needs an external source to run. Oxidation still occurs at the anode and reduction at the cathode, as in every electrochemical cell.
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Topic 9.8 · Say it in contextA galvanic cell runs Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). After an hour, which electrode has lost mass, which has gained mass, and which way do electrons move in the wire?
The zinc anode loses mass as Zn atoms are oxidized to Zn²⁺ ions. The copper cathode gains mass as Cu²⁺ ions are reduced and deposit as Cu. Electrons move through the wire from the zinc electrode to the copper electrode.
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Topic 9.8 · Spot the mistake"In a galvanic cell, electrons travel through the salt bridge from one half-cell to the other." Is that right?
No. Electrons travel only through the external wire. The salt bridge carries ions: anions move toward the anode compartment and cations toward the cathode compartment, which keeps both solutions electrically neutral as the cell runs.
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Topic 9.9 · Work it outAg⁺(aq) + e⁻ → Ag(s) has E° = +0.80 V, and Ni²⁺(aq) + 2e⁻ → Ni(s) has E° = −0.25 V. What is E°cell for the galvanic cell made from these half-cells?
Ag⁺ has the higher reduction potential, so it is reduced at the cathode and Ni is oxidized at the anode. E°cell = E°cathode − E°anode = 0.80 − (−0.25) = +1.05 V. Doubling the silver half-reaction to balance electrons does not change its E°.
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Topic 9.9 · Work it outThe cell reaction 2 Ag⁺(aq) + Ni(s) → 2 Ag(s) + Ni²⁺(aq) has E°cell = +1.05 V. Using F = 96,485 C/mol, what is ΔG°?
Two electrons transfer, so n = 2. ΔG° = −nFE° = −(2)(96,485 C/mol)(1.05 V) = −2.03 × 10⁵ J/mol = −203 kJ/mol. The negative ΔG° matches the positive E°: the reaction is thermodynamically favored.
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Topic 9.10 · Say it in contextA cell runs Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with E° = +1.10 V. It is built with 2.0 M Cu²⁺ and 0.010 M Zn²⁺. Is its cell potential larger or smaller than 1.10 V?
Larger. Q = [Zn²⁺] / [Cu²⁺] = 0.010 / 2.0 = 0.0050, which is less than 1, so the cell starts farther from equilibrium than under standard conditions. In E = E° − (RT/nF) ln Q, ln Q is negative, so E is above E°.
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Topic 9.10 · Which one fitsTwo copper electrodes sit in 0.010 M and 1.0 M Cu²⁺ solutions joined by a salt bridge. Which way do electrons flow in the wire?
From the electrode in 0.010 M Cu²⁺ to the electrode in 1.0 M Cu²⁺. The cell runs toward equal concentrations: Cu is oxidized in the dilute half-cell, adding Cu²⁺, and Cu²⁺ is reduced in the concentrated half-cell, removing it.
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Topic 9.10 · Spot the mistake"A battery that has run down to a cell potential of zero has reached standard conditions, Q = 1." Is that right?
No. Standard conditions, Q = 1, give E°, not zero. The magnitude of the cell potential shrinks as the reaction approaches equilibrium and reaches zero when Q = K, so a dead battery is at equilibrium.
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Topic 9.11 · Work it outA current of 1.50 A runs through a Cu²⁺ solution for 30.0 min. What mass of copper deposits? Use F = 96,485 C/mol and 63.55 g/mol for Cu.
q = It = 1.50 A × 1800 s = 2.70 × 10³ C, which is 2.70 × 10³ / 96,485 = 0.02798 mol e⁻. Cu²⁺(aq) + 2e⁻ → Cu(s), so 0.01399 mol Cu deposits: 0.01399 mol × 63.55 g/mol = 0.889 g.
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Topic 9.11 · Work it outHow long must a 0.500 A current run to plate 1.00 g of silver from a solution of Ag⁺? Use F = 96,485 C/mol and 107.87 g/mol for Ag.
1.00 g / 107.87 g/mol = 9.27 × 10⁻³ mol Ag, and Ag⁺(aq) + e⁻ → Ag(s) takes 1 mol e⁻ per mol Ag. q = 9.27 × 10⁻³ mol × 96,485 C/mol = 894 C. t = q / I = 894 C / 0.500 A = 1.79 × 10³ s, about 29.8 min.