AP Chemistry

Lesson 1.1 · Unit 1 · CED topic 1.1

Moles and molar mass

Atoms are far too small to count, but a balance can weigh them by the trillions. The mole is the bridge: it turns a mass you can measure into a number of particles you can reason about, and back again. Nearly every calculation in this course passes through it.

Set every problem up as one chain with the units written on every factor. If the units don't cancel to what the question asks for, the setup is wrong no matter how reasonable the number looks. Graders award the setup point even when a later arithmetic slip costs the answer point, so always show the chain.

Lesson 1.2 · Unit 1 · CED topic 1.2

Mass spectrometry of elements

Most elements are a blend of isotopes: atoms with the same number of protons but different numbers of neutrons. A mass spectrometer sorts a sample's atoms by mass and reports how many of each it found. The atomic masses printed on the periodic table are weighted averages of exactly these spectra.

Exam tip: when asked to estimate the average from a spectrum without exact numbers, state that the average must lie between the lightest and heaviest peak and be pulled toward the tallest one. That sentence earns the reasoning point.

Lesson 1.3 · Unit 1 · CED topics 1.3–1.4

Elemental composition of pure substances and mixtures

A pure compound has a fixed recipe: every molecule of glucose is C₆H₁₂O₆, so every sample of glucose is 40.00% carbon by mass. Mixtures don't work that way: the ratio of components can be anything. That difference is what lets a percent composition identify an unknown compound.

A ratio like 2.50 or 1.33 is not rounding error: multiply through by 2 or 3. Only ratios within about 0.05 of a whole number should be rounded.

Lesson 1.4 · Unit 1 · CED topic 1.5

Atomic structure and electron configuration

Everything an atom does chemically comes from where its electrons sit and how tightly the nucleus holds them. Configuration is the bookkeeping; Coulomb's law is the physics.

Lesson 1.5 · Unit 1 · CED topic 1.6

Photoelectron spectroscopy

Electron configurations can be measured. Photoelectron spectroscopy (PES) fires high-energy photons at a sample, knocks electrons out, and records how much energy each one needed: a direct picture of the shells and subshells.

Exam tip: a PES explanation names the subshell, its distance from the nucleus, and the nuclear charge or shielding it feels. "It's in a higher energy level" alone doesn't score.

Lesson 1.6 · Unit 1 · CED topics 1.7–1.8

Periodic trends and valence electrons

Every periodic trend is Coulomb's law: how much charge the valence electrons feel, and how far away they are. Argue from those two quantities and you can explain any trend: including the exceptions.

Unit 1 practice · 10 problems

Unit 1 practice: Atomic Structure and Properties

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer: the reveal shows the key steps, not just the number. Calculator allowed; use \(N_A = 6.022 \times 10^{23}\) and periodic-table masses.

1. How many moles of calcium nitrate, Ca(NO₃)₂, are in a 45.0 g sample? How many nitrate ions does it contain?

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   Molar mass: \(40.08 + 2(14.01) + 6(16.00) = 164.10\) g/mol. \(n = 45.0 / 164.10 = 0.274\) mol. Each formula unit has two NO₃⁻: \(0.274 \times 2 \times 6.022 \times 10^{23} = 3.30 \times 10^{23}\) nitrate ions.

2. What is the mass, in grams, of \(4.25 \times 10^{23}\) molecules of water?

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   \(\dfrac{4.25 \times 10^{23}}{6.022 \times 10^{23}} = 0.706\) mol; \(0.706\ \mathrm{mol} \times 18.02\ \mathrm{g/mol} = 12.7\) g. Chain: molecules → mol → g.

3. A mass spectrum of an element shows peaks at 27.977 amu (92.23%), 28.976 amu (4.67%), and 29.974 amu (3.10%). Compute the average atomic mass and identify the element.

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   \(0.9223(27.977) + 0.0467(28.976) + 0.0310(29.974) = 25.80 + 1.353 + 0.929 = 28.09\) amu. The periodic table gives 28.09 for silicon. Check: the average sits just above the dominant 28 peak, pulled slightly up by the two heavier isotopes.

4. A mass spectrum of an element shows two peaks: one at 63 amu that is three times as tall as the other at 65 amu. The average atomic mass is closest to

   63.0 63.5 64.0 64.5

   63.0 would be the average only if every atom had mass 63, but one atom in four is the 65 isotope, so the weighted average must sit above 63. The heavier peak is small, but it is not zero.

   Heights 3 : 1 mean fractions 0.75 and 0.25: \(0.75(63) + 0.25(65) = 63.5\) amu: copper. The average always lies between the peaks and is pulled toward the taller one.

   64.0 is the simple midpoint of 63 and 65, which treats the two isotopes as equally abundant. The 3 : 1 peak-height ratio means abundances of 75% and 25%, so the average is pulled toward 63.

   64.5 would require the 65 peak to be the taller one (75% of the atoms at 65 amu). This choice reverses which isotope is more abundant.

5. Iron(III) oxide, Fe₂O₃, is the main ore of iron. What is the mass percent of iron, and what mass of iron can be obtained from 25.0 g of pure Fe₂O₃?

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   Molar mass \(= 2(55.85) + 3(16.00) = 159.70\) g/mol. \(\%\mathrm{Fe} = \dfrac{111.70}{159.70} \times 100\% = 69.94\%\). Mass of Fe: \(25.0\ \mathrm{g} \times 0.6994 = 17.5\) g.

6. A hydrocarbon is 92.26% carbon and 7.74% hydrogen by mass, with a molar mass of 78.11 g/mol. Find its empirical and molecular formulas.

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   Per 100 g: C \(92.26/12.01 = 7.682\) mol; H \(7.74/1.008 = 7.679\) mol. Ratio 1 : 1 → empirical formula CH, mass 13.02 g/mol. \(78.11 / 13.02 = 6.00\), so the molecular formula is C₆H₆ (benzene).

7. Write the full and noble-gas electron configurations of selenium (Z = 34). How many unpaired electrons does a Se atom have? Write the configuration of Se²⁻.

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   1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁴ = [Ar]4s²3d¹⁰4p⁴. By Hund's rule the four 4p electrons occupy three orbitals as one pair plus two singles: 2 unpaired. Se²⁻ gains two electrons to fill 4p: [Ar]4s²3d¹⁰4p⁶, the same as krypton.

8. A photoelectron spectrum shows peaks at about 39.6, 2.45, and 1.40 MJ/mol with relative heights 2 : 2 : 3. Identify the element, and explain why the 39.6 MJ/mol peak is so far from the other two.

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   Highest energy first: 1s², 2s², 2p³: seven electrons, so nitrogen. The 1s electrons are in the \(n = 1\) shell, much closer to the nucleus and with no inner electrons to shield them, so they feel nearly the full +7 charge; by Coulomb's law the attraction (\(\propto q_1 q_2 / r^2\)) is far stronger and far more energy is needed to remove them. The 2s and 2p electrons are farther out and shielded by the 1s pair.

9. Ca²⁺, K⁺, and Cl⁻ each have 18 electrons. Rank them from smallest to largest radius and justify with Coulomb's law.

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   Ca²⁺ < K⁺ < Cl⁻. Same number of electrons in the same shells, so shielding is identical; the ion with more protons (Ca 20, K 19, Cl 17) exerts a stronger pull on each electron, drawing the cloud in. Isoelectronic ions shrink as nuclear charge grows.

10. An element has successive ionization energies (kJ/mol) of 578, 1817, 2745, and 11,577. How many valence electrons does it have, which group is it in, and why is the fourth value so much larger than the third?

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    The huge jump comes between the third and fourth electrons, so three valence electrons: group 13 (this is aluminum). The first three electrons come from the \(n = 3\) shell; the fourth must be pulled from the \(n = 2\) core, which is much closer to the nucleus and shielded only by the two 1s electrons, and it is being removed from an ion that already carries a +3 charge.

Lesson 2.1 · Unit 2 · CED topic 2.1

Types of chemical bonds

Atoms bond because the arrangement lowers their potential energy: the nuclei end up attracting more electron density than they repel each other. How the electrons are shared (transferred, split, or pooled) decides the bond type, and the bond type decides most of a substance's bulk properties.

Lesson 2.2 · Unit 2 · CED topic 2.2

Intramolecular force and potential energy

Picture two atoms approaching from far apart. Their potential energy traces a curve with one valley, and that valley is the bond: its position is the bond length, its depth the bond energy.

Exam tip: to justify "shorter is stronger," name both factors: more shared electron density and a shorter distance over which the nuclei attract it.

Lesson 2.3 · Unit 2 · CED topics 2.3–2.4

Structure of ionic solids and metals

Salt shatters when you hit it; copper bends. Both are held together by electrostatic attraction, so the difference is in how the charges are arranged.

Lesson 2.4 · Unit 2 · CED topic 2.5

Lewis diagrams

A Lewis diagram is an electron inventory: every valence electron is a bond line or a lone-pair dot. Count first, draw second, count again.

Lesson 2.5 · Unit 2 · CED topic 2.6

Resonance and formal charge

Sometimes the Lewis procedure gives you a choice of where to put a double bond, and experiment says all the choices are equally right, or equally wrong. Resonance handles the first case; formal charge sorts out the second.

Lesson 2.6 · Unit 2 · CED topic 2.7

VSEPR, hybridization, and molecular polarity

Electron pairs repel, so the pairs around a central atom spread out as far as they can. That idea, VSEPR, predicts the shape of almost any small molecule, and shape decides polarity.

Unit 2 practice · 10 problems

Unit 2 practice: Compound Structure and Properties

Ten problems covering the whole unit. Work each one on paper before revealing the answer. No calculator needed. Electronegativities where needed: K 0.82, Br 2.96, N 3.04, F 3.98.

1. Classify the bonding in KBr, N₂, NF₃, and Ag as ionic, nonpolar covalent, polar covalent, or metallic. Give one piece of evidence for each.

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   KBr: metal + nonmetal, ΔEN = 2.14 → ionic (high-melting crystalline solid, conducts when molten). N₂: identical atoms, ΔEN = 0 → nonpolar covalent. NF₃: two nonmetals, ΔEN = 0.94 → polar covalent, F end δ−. Ag: a metal alone → metallic (conducts as a solid, malleable).

2. Three potential-energy curves have minima at (127 pm, −431 kJ/mol), (141 pm, −366 kJ/mol), and (161 pm, −298 kJ/mol). They belong to HCl, HBr, and HI. Match each and explain the trend.

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   HCl: 127 pm, 431 kJ/mol; HBr: 141 pm, 366 kJ/mol; HI: 161 pm, 298 kJ/mol. Down the group the halogen is larger, with its bonding electrons in a higher shell and more core electrons in between, so the nuclei sit farther apart at the minimum (longer bond) and the attraction between them and the shared pair is weaker at that distance (shallower well). Longer bond, weaker bond.

3. Which compound has the largest lattice energy?

   NaCl KCl MgO CaO

   NaCl has 1+ and 1− ions, so its charge product is a quarter that of MgO or CaO, and its lattice energy (787 kJ/mol) is roughly a quarter of theirs. Small ions help, but in \(E \propto q_1 q_2 / r\) the charges dominate.

   KCl has the same 1+/1− charges as NaCl but a larger cation, so the ion centers are farther apart and its lattice energy (715 kJ/mol) is the smallest of the four.

   Lattice energy follows \(E \propto q_1 q_2 / r\). MgO and CaO have 2+/2− ions, four times the charge product of the chlorides; between them Mg²⁺ is smaller than Ca²⁺, so the ion centers are closer in MgO (about 3800 vs. 3400 kJ/mol).

   CaO has the right 2+/2− charges, but Ca²⁺ has one more electron shell than Mg²⁺, so the distance \(r\) is larger and the attraction weaker than in MgO. When the charges tie, the smaller ions win.

4. Solid KBr does not conduct electricity, but molten KBr does; copper conducts as a solid and bends without breaking. Explain all three observations at the particle level. Then classify steel (C in Fe) and brass (Zn in Cu) as interstitial or substitutional alloys.

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   In solid KBr the K⁺ and Br⁻ ions are locked in a lattice; no charged particle can move. Melting frees the ions to migrate, so the liquid conducts. Copper's valence electrons are delocalized in an electron sea that carries current, and because cations are attracted to that uniform sea rather than to specific partners, planes of atoms slide past each other without shattering. Steel is interstitial (small C atoms fill gaps between Fe atoms); brass is substitutional (similar-sized Zn replaces Cu). Both are harder than the pure metal because the foreign atoms pin the layers.

5. Describe the Lewis diagram of the sulfite ion, SO₃²⁻: total valence electrons, bonds, lone pairs, and the formal charge on each atom. Then give its molecular geometry and approximate bond angle.

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   Electrons: \(6 + 3(6) + 2 = 26\). S central with three S–O single bonds (6 e⁻), three lone pairs on each O (18), and the remaining 2 as a lone pair on S. Every atom has an octet. Formal charges: S \(= 6 - 2 - \tfrac{1}{2}(6) = +1\); each O \(= 6 - 6 - \tfrac{1}{2}(2) = -1\); sum \(+1 - 3 = -2\). ✓ Four electron domains with one lone pair: trigonal pyramidal, about 107°, sp³ sulfur.

6. Count the valence electrons in XeF₂ and describe its Lewis diagram. Why is this arrangement allowed for xenon? Separately, how many valence electrons does ClO₂ have, and what does that number tell you?

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   XeF₂: \(8 + 2(7) = 22\). Two Xe–F bonds (4), three lone pairs on each F (12), and the remaining 6 as three lone pairs on Xe: ten electrons around Xe, an expanded octet, allowed because a period-5 atom is large enough to hold more than four pairs. Five domains with three lone pairs make it linear. ClO₂: \(7 + 2(6) = 19\), an odd number, so one electron must be unpaired: ClO₂ is a radical, which is why it is so reactive.

7. Draw (in words) the resonance structures of the nitrite ion, NO₂⁻. Give the formal charge on each atom in one structure, the N–O bond order, and what experiment shows about the two bond lengths.

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   Electrons: \(5 + 2(6) + 1 = 18\). N central with one N=O and one N–O, a lone pair on N, two lone pairs on the doubly bonded O and three on the singly bonded O. The double bond can be on either oxygen: two equivalent structures. Formal charges: N \(= 5 - 2 - 3 = 0\); double-bonded O \(= 6 - 4 - 2 = 0\); single-bonded O \(= 6 - 6 - 1 = -1\); sum −1. ✓ Bond order \(= 3/2\); both N–O bonds are the same length, between a typical N–O single and N=O double bond.

8. Dinitrogen monoxide, N₂O, has the skeleton N–N–O and 16 valence electrons. Three Lewis structures are possible: N≡N–O, N=N=O, and N–N≡O. Compute the formal charges and identify the dominant structure.

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   Formal charges (terminal N, central N, O): N≡N–O: 0, +1, −1. N=N=O: −1, +1, 0. N–N≡O: −2, +1, +1. The third has large, separated charges and contributes little. The first two each have charges of one unit; N≡N–O puts the negative charge on oxygen, the most electronegative atom, so it is the largest contributor, with N=N=O a close second. (Check each: the charges sum to 0, the molecule's charge.)

9. For each molecule give the number of electron domains on the central atom, the molecular geometry, the approximate bond angle, and whether it is polar: (a) PCl₃, (b) SF₄, (c) CS₂.

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   (a) PCl₃: 4 domains (3 bonds + 1 lone pair); trigonal pyramidal; ~107°; sp³; polar: the three P–Cl dipoles don't cancel. (b) SF₄: 5 domains (4 bonds + 1 lone pair); seesaw; 90°/120° ideal, compressed; polar: the lone pair breaks the symmetry. (c) CS₂: 2 domains (two double bonds); linear; 180°; sp carbon; nonpolar: equal and opposite C=S dipoles cancel.

10. Propene is CH₃–CH=CH₂. How many σ and π bonds does it contain? State the hybridization of each carbon, and which carbon–carbon bond is shorter.

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    Six C–H σ bonds, one C–C σ bond, and one C=C (σ + π): 8 σ and 1 π. The CH₃ carbon has 4 domains → sp³ (~109.5°); the two alkene carbons have 3 domains each → sp² (~120°). The C=C bond is shorter: two shared pairs put more electron density between the nuclei and pull them closer than the C–C single bond does.

Lesson 3.1 · Unit 3 · CED topic 3.1

Intermolecular forces

Bonds hold a molecule together; intermolecular forces (IMFs) hold molecules near each other. Far weaker than bonds, they still decide whether a substance is a gas, liquid, or solid, and they are the most-tested explanation in the course.

Lesson 3.2 · Unit 3 · CED topics 3.2–3.3

Properties of solids and liquids

Once you can name the forces between particles, you can predict how a liquid behaves and what kind of solid forms. The question is always: what must be overcome for the particles to move apart?

Exam tip: "it has strong bonds" is wrong for a molecular solid. The covalent bonds stay intact when it melts: only IMFs break.

Lesson 3.3 · Unit 3 · CED topics 3.4–3.5

Ideal gases

Gases are the easiest state to calculate with, because in an ideal gas the particles don't attract each other at all. One equation connects pressure, volume, moles, and temperature, and almost every gas problem is that equation plus a mole conversion.

Lesson 3.4 · Unit 3 · CED topics 3.5–3.6

Kinetic molecular theory and real gases

The ideal gas law works because of a picture: tiny particles in constant random motion, colliding elastically, with no attraction between them. That picture also tells you exactly when the law breaks down.

Lesson 3.5 · Unit 3 · CED topics 3.7–3.8

Solutions and mixtures

Most reactions you'll run happen in solution, so you need a way to count particles by volume. Molarity does that, and one rule, moles of solute don't change when you add water, handles every dilution.

Lesson 3.6 · Unit 3 · CED topics 3.9–3.10

Separations and solubility

Every separation technique and solubility rule comes back to Lesson 3.1: which attractions are strongest? A substance goes where its IMFs are best satisfied.

Lesson 3.7 · Unit 3 · CED topics 3.11–3.13

Spectroscopy and the Beer-Lambert law

Light is a probe. Different energies of light make molecules do different things, and how much light a solution absorbs tells you how much of the absorbing substance is there. That second idea is the basis of the most common lab technique on the exam.

Unit 3 practice · 10 problems

Unit 3 practice: Properties of Substances and Mixtures

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer. Calculator allowed; \(R = 0.08206\) L·atm/(mol·K).

1. Propane (CH₃CH₂CH₃), dimethyl ether (CH₃OCH₃), and ethanol (CH₃CH₂OH) all have 26 electrons. Rank them by boiling point and justify each placement with the intermolecular forces present.

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   Propane (−42 °C) < dimethyl ether (−24 °C) < ethanol (78 °C). Equal electron counts make the London dispersion forces comparable, so the difference is in the other forces. Propane is nonpolar: LDF only. Dimethyl ether is polar (bent C–O–C) and adds dipole–dipole attractions. Ethanol has an O–H bond, so it hydrogen-bonds to the lone pairs on neighboring oxygens: the strongest of the three, needing the most energy to separate.

2. Acetone (CH₃COCH₃) and 2-propanol (CH₃CHOHCH₃) have similar molar masses. Which has the higher vapor pressure at 25 °C, and which would distill out of a mixture first? Explain.

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   Acetone on both counts. Both are polar with similar LDF, but 2-propanol has an O–H group and hydrogen-bonds; acetone has no H on O and relies on dipole–dipole forces. At the same temperature the molecules share the same kinetic-energy distribution, but a larger fraction of acetone molecules have enough energy to escape their weaker attractions, so its vapor pressure is higher (about 230 vs. 44 torr) and its boiling point lower (56 vs. 82 °C): it boils off first.

3. Classify each solid as molecular, covalent network, ionic, or metallic: diamond, I₂(s), CaF₂, Fe. Which has the lowest melting point and which the highest? Explain in terms of what must be overcome.

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   Diamond: covalent network; I₂: molecular; CaF₂: ionic; Fe: metallic. Lowest: I₂ (114 °C): melting only disrupts London dispersion forces between intact I₂ molecules; the I–I bonds stay put. Highest: diamond (above 3500 °C), every atom is covalently bonded to four neighbors, so melting means breaking C–C bonds throughout. CaF₂ (Coulombic attraction between Ca²⁺ and F⁻) and Fe (cations in an electron sea) fall in between.

4. A 2.00 L flask at 300 K and 1.00 atm holds 3.58 g of an unknown gas. Find its molar mass and identify it as CO₂, N₂, or Ar.

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   \(n = \dfrac{PV}{RT} = \dfrac{(1.00)(2.00)}{(0.08206)(300)} = 0.0812\) mol. \(M = \dfrac{3.58\ \mathrm{g}}{0.0812\ \mathrm{mol}} = 44.1\) g/mol → CO₂ (44.01). N₂ would be 28.0 and Ar 39.9.

5. Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g). What volume of hydrogen gas at 1.00 atm and 298 K is produced when 2.00 g of zinc reacts with excess acid?

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   \(2.00 / 65.38 = 0.0306\) mol Zn; the ratio is 1 : 1, so 0.0306 mol H₂. \(V = \dfrac{nRT}{P} = \dfrac{(0.0306)(0.08206)(298)}{1.00} = 0.748\) L. Grams → moles → mole ratio → gas law.

6. 0.250 mol N₂ and 0.750 mol He share a 10.0 L container at 298 K. Find the total pressure and the partial pressure of N₂.

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   \(P_{\text{total}} = \dfrac{(1.000)(0.08206)(298)}{10.0} = 2.45\) atm. Mole fraction \(X_{\mathrm{N_2}} = 0.250/1.000 = 0.250\), so \(P_{\mathrm{N_2}} = 0.250 \times 2.45 = 0.611\) atm (and He contributes 1.83 atm). Each gas acts as if it were alone in the container.

7. Two rigid 1.0 L flasks at the same temperature hold 0.10 mol He and 0.10 mol Ar respectively. Which statement is true?

   The He atoms have the greater average kinetic energy. The Ar atoms have the greater average speed. The two gases exert the same pressure. Ar exerts more pressure because its atoms are heavier.

   Average kinetic energy depends only on temperature (\(KE_{\text{avg}} = \tfrac{3}{2}k_BT\)), so at the same \(T\) the He and Ar atoms have exactly the same average kinetic energy. He atoms are faster, not more energetic.

   Since \(KE = \tfrac{1}{2}mv^2\) is the same for both, the lighter He atoms must move faster; Ar atoms, ten times heavier, are the slower ones by a factor of \(\sqrt{10}\). This choice reverses the mass–speed relationship.

   Equal \(n\), \(V\), and \(T\) give equal \(P\) by \(PV = nRT\); heavier atoms hit the wall harder but less often, and the two effects cancel exactly for an ideal gas. (Compressed to very high pressure, Ar would deviate more from ideal behavior: its atoms are larger and more polarizable, so their own volume and their London dispersion forces matter more.)

   Pressure comes from the momentum transferred per collision times the collision frequency. Ar atoms carry more momentum per hit but hit less often because they are slower, so the pressure is the same: mass drops out of the ideal gas law.

8. What volume of 2.50 M Na₂SO₄ stock is needed to prepare 500.0 mL of 0.150 M Na₂SO₄? What is [Na⁺] in the final solution?

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   \(V_1 = \dfrac{M_2 V_2}{M_1} = \dfrac{(0.150)(500.0\ \mathrm{mL})}{2.50} = 30.0\) mL, then dilute to the 500.0 mL mark. Each formula unit releases two Na⁺: \([\mathrm{Na^+}] = 2 \times 0.150 = 0.300\) M (and \([\mathrm{SO_4^{2-}}] = 0.150\) M).

9. Two compounds are spotted on polar paper and run with a nonpolar solvent. Spot A moves 4.5 cm and spot B moves 1.5 cm while the solvent front moves 7.5 cm. Compute each \(R_f\) and decide which compound is more polar.

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   \(R_f(\mathrm{A}) = 4.5/7.5 = 0.60\); \(R_f(\mathrm{B}) = 1.5/7.5 = 0.20\). B is more polar: it is held back by strong dipole–dipole or hydrogen-bonding attractions to the polar stationary phase and has little affinity for the nonpolar mobile phase. A interacts weakly with the paper and travels with the solvent.

10. A standard solution of a dye at \(1.50 \times 10^{-4}\) M has absorbance 0.210 in a 1.00 cm cuvette. (a) Find the molar absorptivity. (b) An unknown reads 0.455 in the same cuvette; find its concentration. (c) What absorbance would the standard give in a 2.00 cm cuvette?

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    (a) \(\varepsilon = \dfrac{A}{bc} = \dfrac{0.210}{(1.00\ \mathrm{cm})(1.50 \times 10^{-4}\ \mathrm{M})} = 1400\) M⁻¹cm⁻¹. (b) \(c = \dfrac{0.455}{(1400)(1.00)} = 3.25 \times 10^{-4}\) M. (c) Absorbance is proportional to path length: \(0.210 \times 2 = 0.420\).

Lesson 4.1 · Unit 4 · CED topics 4.1–4.2

Chemical and physical change; balancing equations and net ionic equations

A chemical change rearranges atoms into new substances; a physical change leaves the substances intact. Its equation must respect one law, atoms are neither created nor destroyed, and the most useful form shows only the particles that actually change.

Exam tip: a net ionic equation must be balanced for atoms and charge, and every species needs a state symbol. Points are lost for splitting a solid or a weak acid into ions.

Lesson 4.2 · Unit 4 · CED topics 4.3–4.4, 4.7

Representations of reactions and reaction types

An equation is a summary. The exam also wants you to see the reaction as particles, boxes of circles before and after, and to classify it from the formulas alone.

Lesson 4.3 · Unit 4 · CED topic 4.5

Stoichiometry: mole ratios, limiting reactant, and percent yield

Coefficients count molecules, and therefore moles, never grams. So every stoichiometry problem has the same spine: grams → moles, apply the mole ratio, moles → grams. When two reactant amounts are given, one of them runs out first and decides everything.

Lesson 4.4 · Unit 4 · CED topic 4.6

Introduction to titration

A titration is stoichiometry done with a buret. You add a solution of known concentration to a measured volume of unknown until the reaction is exactly complete, and the volume you used tells you how much unknown was there.

Exam tip: error-analysis questions ask what happens if the buret is rinsed with water instead of titrant. The titrant is diluted, so more volume is needed to reach the endpoint; the student computes more moles of base than were really delivered and reports an acid concentration that is too high.

Lesson 4.5 · Unit 4 · CED topics 4.8–4.9

Acid–base and redox reactions

Two of the exam's big reaction categories are defined by what gets transferred: a proton in acid–base, electrons in redox. Oxidation numbers are the bookkeeping for following the electrons.

Unit 4 practice · 10 problems

Unit 4 practice: Chemical Reactions

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer. Calculator allowed for the stoichiometry.

1. Balance the combustion of butane: C₄H₁₀ + O₂ → CO₂ + H₂O. Classify the reaction in two ways.

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   Balance C (4 CO₂), then H (5 H₂O), then O: the right side has 8 + 5 = 13 O, so 6½ O₂; double everything: 2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O. Check: 8 C, 20 H, 26 O each side. It is a combustion reaction and also redox: O goes 0 → −2 and C is oxidized.

2. Aqueous iron(III) nitrate is mixed with aqueous sodium hydroxide and a rust-colored solid forms. Write the molecular, complete ionic, and net ionic equations, and name the spectator ions.

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   Molecular: Fe(NO₃)₃(aq) + 3 NaOH(aq) → Fe(OH)₃(s) + 3 NaNO₃(aq). Complete ionic: Fe³⁺ + 3 NO₃⁻ + 3 Na⁺ + 3 OH⁻ → Fe(OH)₃(s) + 3 Na⁺ + 3 NO₃⁻. Net ionic: Fe³⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s); charge +3 − 3 = 0 on both sides. Spectators: Na⁺ and NO₃⁻.

3. A particle diagram shows a box containing 4 CO molecules and 3 O₂ molecules. Describe the box after 2 CO + O₂ → 2 CO₂ runs to completion, and identify the limiting reactant.

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   4 CO need only 2 O₂, so CO is limiting. After: 4 CO₂ and 1 O₂, no CO. Atom check: C 4 → 4; O \(4 + 6 = 10\) → \(8 + 2 = 10\). The leftover O₂ must appear in the "after" box.

4. Which reaction is not an oxidation–reduction reaction?

   2 Na + Cl₂ → 2 NaCl CH₄ + 2 O₂ → CO₂ + 2 H₂O AgNO₃(aq) + KBr(aq) → AgBr(s) + KNO₃(aq) Zn + CuSO₄ → ZnSO₄ + Cu

   Free elements appear on the left, so oxidation numbers must change: Na goes from 0 to +1 and Cl from 0 to −1. Any reaction that consumes or produces a free element is redox.

   Combustion is always redox: C goes from −4 in CH₄ to +4 in CO₂, and O goes from 0 in O₂ to −2. It involves no metals, which is why students sometimes miss the electron transfer.

   This is a precipitation reaction: ions swap partners and every oxidation number is unchanged (Ag +1, Br −1, K +1, N +5, O −2 before and after). No electrons are transferred.

   This is a single-replacement redox reaction: Zn goes from 0 to +2 (oxidized) and Cu from +2 to 0 (reduced). A free metal on each side is the giveaway.

5. N₂ + 3 H₂ → 2 NH₃. 28.0 g of N₂ reacts with 8.00 g of H₂. Find the limiting reactant, the theoretical yield of NH₃, the mass of excess reactant left over, and the percent yield if 28.5 g of NH₃ is collected.

   Show answer

   Moles: \(28.0/28.02 = 0.999\) mol N₂; \(8.00/2.016 = 3.97\) mol H₂. N₂ needs \(3 \times 0.999 = 3.00\) mol H₂; 3.97 is available, so N₂ is limiting. Theoretical yield: \(0.999 \times 2 = 2.00\) mol NH₃ \(\times 17.03 = 34.0\) g. Excess: \(3.97 - 3.00 = 0.97\) mol H₂ \(\times 2.016 = 1.96\) g. Percent yield: \(28.5/34.0 \times 100\% = 83.8\%\).

6. C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O. What mass of oxygen is required to burn 10.0 g of ethanol completely, and what mass of CO₂ forms?

   Show answer

   \(10.0 / 46.07 = 0.217\) mol ethanol. O₂: \(0.217 \times 3 = 0.651\) mol \(\times 32.00 = 20.8\) g. CO₂: \(0.217 \times 2 = 0.434\) mol \(\times 44.01 = 19.1\) g. Both chains: g → mol → mole ratio → g.

7. 20.00 mL of barium hydroxide solution requires 27.30 mL of 0.1000 M HCl to reach the endpoint. Write the balanced equation and find [Ba(OH)₂].

   Show answer

   Ba(OH)₂ + 2 HCl → BaCl₂ + 2 H₂O. Moles HCl: \(0.02730 \times 0.1000 = 2.730 \times 10^{-3}\) mol. Each Ba(OH)₂ neutralizes two H⁺, so moles base \(= 1.365 \times 10^{-3}\). \([\mathrm{Ba(OH)_2}] = \dfrac{1.365 \times 10^{-3}}{0.02000} = 0.06825\) M: four sig figs, and don't skip the factor of 2.

8. In a titration of an acid with standardized NaOH, describe the effect on the calculated acid concentration (too high, too low, or unchanged) if (a) the flask of acid is diluted with extra distilled water before titrating; (b) an air bubble in the buret tip escapes during the titration; (c) the buret is rinsed with water instead of NaOH before filling.

   Show answer

   (a) Unchanged: dilution adds no moles of acid, and the calculation uses moles of base and the original acid volume. (b) Too high: the bubble's volume is read as delivered titrant, so the student overstates the moles of NaOH and therefore the moles (and concentration) of acid. (c) Too high: the diluted NaOH is weaker than its label, more volume is needed, and the student again computes too many moles of base.

9. (a) In HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻, identify the Brønsted–Lowry acid, base, and the two conjugate pairs. (b) Assign the oxidation number of P in H₃PO₄, Mn in MnO₂, N in NO₃⁻, and S in S₂O₃²⁻.

   Show answer

   (a) H₂O donates a proton: acid, conjugate base OH⁻. HCO₃⁻ accepts it: base, conjugate acid H₂CO₃. Pairs: H₂O/OH⁻ and H₂CO₃/HCO₃⁻. (b) H₃PO₄: \(3(+1) + \mathrm{P} + 4(-2) = 0\) → P = +5. MnO₂: Mn = +4. NO₃⁻: \(\mathrm{N} - 6 = -1\) → N = +5. S₂O₃²⁻: \(2\mathrm{S} - 6 = -2\) → each S = +2.

10. Magnesium metal placed in a solution of Fe³⁺ produces iron metal. Identify what is oxidized and reduced, name the oxidizing and reducing agents, and write the balanced net ionic equation.

    Show answer

    Mg goes 0 → +2: oxidized, so Mg is the reducing agent. Fe goes +3 → 0: reduced, so Fe³⁺ is the oxidizing agent. Half-reactions: Mg → Mg²⁺ + 2 e⁻ (×3) and Fe³⁺ + 3 e⁻ → Fe (×2), six electrons each: 3 Mg(s) + 2 Fe³⁺(aq) → 3 Mg²⁺(aq) + 2 Fe(s). Charge check: +6 on each side.

Lesson 5.1 · Unit 5 · CED topics 5.1–5.2

Reaction rates and the factors that affect them

Thermodynamics tells you whether a reaction can happen; kinetics tells you how fast it does. A rate is a change in concentration per unit time, and because coefficients tie every species together, one measured rate gives you all of them.

Lesson 5.2 · Unit 5 · CED topics 5.3–5.5

Rate laws, integrated rate laws, and half-life

A rate law is the experimental fact that connects rate to concentration. You can't read it off the balanced equation: you have to measure it. Two kinds of data give it to you: initial rates from several trials, or one concentration–time run that you test for linearity.

Lesson 5.3 · Unit 5 · CED topics 5.6–5.7

Elementary reactions, collision theory, and energy profiles

Why does a reaction have a rate at all, instead of happening the instant reactants meet? Because most collisions fail. Bonds must stretch and partially break before new ones form, and that costs energy up front.

Lesson 5.4 · Unit 5 · CED topics 5.8–5.10

Reaction mechanisms and the rate-determining step

Most reactions don't happen in one collision. A mechanism is the proposed sequence of elementary steps, and it must pass two tests: the steps add up to the overall equation, and the predicted rate law matches the measured one.

Lesson 5.5 · Unit 5 · CED topic 5.11

Catalysis

A catalyst makes a reaction faster without being used up. It doesn't push molecules over the same hill harder: it opens a route with a lower hill. Because it changes only the path, not the start or end, \(\Delta H\) and the equilibrium constant are untouched.

Unit 5 practice · 10 problems

Unit 5 practice: Kinetics

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer. Calculator allowed.

1. 4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g). If NH₃ is consumed at 0.240 M/s, at what rates is O₂ consumed and H₂O produced?

   Show answer

   Rates scale with coefficients: O₂ \(= 0.240 \times \tfrac{5}{4} = 0.300\) M/s consumed; H₂O \(= 0.240 \times \tfrac{6}{4} = 0.360\) M/s produced. (NO forms at 0.240 M/s.)

2. For A + B → products, initial-rate data at one temperature are shown. Find the rate law and the value and units of \(k\).

   Trial	[A]₀ (M)	[B]₀ (M)	Initial rate (M/s)
   1	0.100	0.100	4.0 × 10⁻³
   2	0.200	0.100	8.0 × 10⁻³
   3	0.200	0.200	3.2 × 10⁻²

   Show answer

   Trials 1 → 2: [A] doubles, rate doubles → first order in A. Trials 2 → 3: [B] doubles, rate ×4 → second order in B. \(\text{rate} = k[\mathrm{A}][\mathrm{B}]^2\), third order overall. From trial 1: \(k = \dfrac{4.0 \times 10^{-3}}{(0.100)(0.100)^2} = 4.0\ \mathrm{M^{-2}\,s^{-1}}\).

3. A reaction is first order in A with \(k = 1.20 \times 10^{-2}\ \mathrm{s^{-1}}\). Starting from [A] = 0.800 M, what is [A] after 120 s? What is the half-life?

   Show answer

   \(\ln[\mathrm{A}]_t = \ln(0.800) - (0.0120)(120) = -0.223 - 1.44 = -1.663\), so \([\mathrm{A}]_t = e^{-1.663} = 0.190\) M. \(t_{1/2} = 0.693 / 0.0120 = 57.8\) s. Check: 120 s is about two half-lives, and \(0.800/4 = 0.200\). ✓

4. Concentration data for the decomposition of A: \(t = 0, 10, 20, 30\) s; \([\mathrm{A}] = 1.000, 0.500, 0.333, 0.250\) M. Determine the order and \(k\), and explain how you know.

   Show answer

   Test the plots. \(1/[\mathrm{A}] = 1.00, 2.00, 3.00, 4.00\ \mathrm{M^{-1}}\): a straight line versus \(t\), so the reaction is second order, with \(k = \text{slope} = 0.100\ \mathrm{M^{-1}\,s^{-1}}\). It is not first order: the successive half-lives (10 s from 1.000 to 0.500, then 20 s from 0.500 to 0.250) double as the concentration falls, whereas a first-order half-life is constant.

5. A first-order reaction has a half-life of 20.0 min. What fraction of the reactant remains after 60.0 min?

   1/3 1/4 1/8 1/6

   1/3 treats the decay as linear, “60 minutes is three 20-minute periods, so a third is left”, but first-order decay halves the remaining amount each half-life rather than removing a fixed amount each interval.

   1/4 is what remains after two half-lives (40.0 min), not three. Stopping one half-life early, or counting \(60/20 - 1\) half-lives, is the usual slip.

   60.0 min is three half-lives, and each half-life halves whatever is present regardless of concentration: \((\tfrac{1}{2})^3 = \tfrac{1}{8}\). A concentration-independent half-life is the fingerprint of first-order kinetics.

   1/6 comes from multiplying 2 by 3, the “half” times three half-lives, instead of raising \(\tfrac{1}{2}\) to the third power. Half-lives compound multiplicatively, not additively.

6. On an energy profile the reactants sit at 30 kJ/mol, the transition state at 145 kJ/mol, and the products at 75 kJ/mol. Find the forward activation energy, \(\Delta H\), and the reverse activation energy. Is the reaction endothermic or exothermic?

   Show answer

   \(E_{a,\text{fwd}} = 145 - 30 = 115\) kJ/mol. \(\Delta H = 75 - 30 = +45\) kJ/mol, endothermic. \(E_{a,\text{rev}} = 145 - 75 = 70\) kJ/mol. Check: \(E_{a,\text{fwd}} - E_{a,\text{rev}} = 45 = \Delta H\). ✓ The reverse reaction has the smaller barrier.

7. Raising the temperature of a reaction from 20 °C to 30 °C roughly doubles its rate, yet the average kinetic energy rises only about 3%. Explain, using the Maxwell–Boltzmann distribution, in the language graders expect.

   Show answer

   Collision frequency rises only slightly with a 3% increase in average kinetic energy. The dominant effect is on the distribution: at higher T the curve flattens and its high-energy tail extends further, so a much larger fraction of collisions have energy greater than or equal to the activation energy. Because that fraction is exponential in \(-E_a/RT\), a small temperature change produces a large change in the number of successful collisions.

8. Proposed mechanism for 2 NO₂ + F₂ → 2 NO₂F: Step 1 (slow): NO₂ + F₂ → NO₂F + F. Step 2 (fast): NO₂ + F → NO₂F. Show the steps sum to the overall equation, identify any intermediate, and predict the rate law.

   Show answer

   Adding: 2 NO₂ + F₂ + F → 2 NO₂F + F; the F atom cancels, leaving 2 NO₂ + F₂ → 2 NO₂F. ✓ The F atom is an intermediate (made in step 1, consumed in step 2). The slow step is bimolecular in NO₂ and F₂: rate = k[NO₂][F₂]: first order in NO₂ even though its overall coefficient is 2.

9. The experimental rate law for H₂ + I₂ → 2 HI is rate = k[H₂][I₂]. Proposed mechanism: Step 1 (fast, reversible): I₂ ⇌ 2 I. Step 2 (slow): H₂ + 2 I → 2 HI. Show that this mechanism is consistent with the rate law.

   Show answer

   The slow step gives rate \(= k_2[\mathrm{H_2}][\mathrm{I}]^2\), but I is an intermediate. Step 1 is at equilibrium, so \(k_1[\mathrm{I_2}] = k_{-1}[\mathrm{I}]^2\), giving \([\mathrm{I}]^2 = \dfrac{k_1}{k_{-1}}[\mathrm{I_2}]\). Substituting: \(\text{rate} = \dfrac{k_2 k_1}{k_{-1}}[\mathrm{H_2}][\mathrm{I_2}] = k[\mathrm{H_2}][\mathrm{I_2}]\). ✓ The steps also sum to H₂ + I₂ → 2 HI.

10. Nitrogen dioxide speeds the oxidation of SO₂: Step 1: SO₂ + NO₂ → SO₃ + NO (×2). Step 2: 2 NO + O₂ → 2 NO₂. Write the overall reaction, identify the catalyst and the intermediate, and state what the catalyst does and does not change.

    Show answer

    Overall: 2 SO₂ + O₂ → 2 SO₃ (the NO₂ and NO cancel). NO₂ is consumed first and regenerated later: the catalyst. NO is produced first and consumed later: the intermediate. The catalyst provides a pathway with a lower activation energy, so a larger fraction of collisions succeed and \(k\) increases: for both directions. It does not change \(\Delta H\), the equilibrium constant, or the equilibrium composition; those depend only on reactants and products.

Lesson 6.1 · Unit 6 · CED topics 6.1–6.2

Endothermic and exothermic processes; energy diagrams

Every chemical or physical change moves energy between the thing you're studying and everything around it. Thermochemistry is bookkeeping for that energy, and the first rule is deciding whose account you're tracking.

Lesson 6.2 · Unit 6 · CED topics 6.3–6.4

Heat capacity and calorimetry

You can't measure heat directly. You measure a temperature change in something whose heat capacity you know, usually water, and work backward. That's calorimetry.

Lesson 6.3 · Unit 6 · CED topic 6.5

Energy of phase changes

Heat a block of ice steadily and its temperature rises, then stalls at 0 °C while it melts, then rises again, then stalls at 100 °C while it boils. The flat parts of that heating curve are where the energy goes into breaking intermolecular attractions instead of speeding molecules up.

Lesson 6.4 · Unit 6 · CED topics 6.6–6.7

Enthalpy of reaction and bond enthalpies

Breaking a bond always costs energy; forming one always releases it. A reaction's enthalpy is the net of those two, so if you know roughly how strong each bond is, you can estimate \(\Delta H\) without a calorimeter.

Lesson 6.5 · Unit 6 · CED topics 6.8–6.9

Enthalpy of formation and Hess's law

Enthalpy is a state function: the change depends only on where you start and end, not on the route. That one fact lets you compute \(\Delta H\) for reactions nobody has run in a calorimeter, by adding up reactions somebody has.

Unit 6 practice · 10 problems

Unit 6 practice: Thermochemistry

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer. Calculator allowed; \(c_{\text{water}} = 4.18\) J/(g·°C), and for water \(\Delta H_{\text{fus}} = 6.01\) kJ/mol, \(\Delta H_{\text{vap}} = 40.7\) kJ/mol.

1. An ice cube melts in your hand. Taking the ice as the system, give the sign of \(q\) and \(\Delta H\), classify the process, and say where the products sit on an energy diagram. Explain at the particle level why your hand feels cold.

   Show answer

   The ice absorbs heat: \(q \gt 0\), \(\Delta H \gt 0\), endothermic; liquid water sits above ice on the diagram. Fast-moving particles in your skin collide with the slower water molecules in the ice and transfer kinetic energy to them; that energy goes into pulling molecules apart against their hydrogen bonds. Your skin's particles are left with less kinetic energy, a lower temperature, so it feels cold.

2. How much heat is needed to warm 250.0 g of water from 20.0 °C to 85.0 °C?

   Show answer

   \(q = mc\,\Delta T = (250.0\ \mathrm{g})(4.18\ \mathrm{J/(g·°C)})(65.0\ \mathrm{°C}) = 6.79 \times 10^{4}\ \mathrm{J} = 67.9\) kJ. Positive: heat flows into the water.

3. A 75.0 g metal sample at 95.0 °C is placed in 150.0 g of water at 20.0 °C. The final temperature is 24.0 °C. Find the specific heat of the metal.

   Show answer

   Water gains \(q = (150.0)(4.18)(4.0) = 2508\) J, so the metal loses 2508 J: \(q_{\text{metal}} = -2508\) J with \(\Delta T = 24.0 - 95.0 = -71.0\) °C. \(c = \dfrac{-2508}{(75.0)(-71.0)} = 0.47\) J/(g·°C). The two negatives cancel; a negative result means a sign slipped. (Two sig figs, from the 4.0 °C rise.)

4. 5.00 g of NH₄NO₃ (80.05 g/mol) dissolves in 100.0 g of water in a coffee-cup calorimeter. The temperature drops from 22.00 °C to 18.35 °C. Assuming the solution has the specific heat of water, find \(\Delta H\) of solution in kJ/mol.

   Show answer

   Solution mass 105.0 g. \(q_{\text{soln}} = (105.0)(4.18)(18.35 - 22.00) = -1602\) J, so \(q_{\text{dissolving}} = +1602\) J. Moles: \(5.00/80.05 = 0.0625\) mol. \(\Delta H = \dfrac{1602\ \mathrm{J}}{0.0625\ \mathrm{mol}} = +2.56 \times 10^{4}\ \mathrm{J/mol} = +25.6\) kJ/mol. Endothermic: the accepted value is +25.7 kJ/mol.

5. How much heat is required to convert 25.0 g of liquid water at 25.0 °C to steam at 100.0 °C?

   Show answer

   Warm the liquid: \((25.0)(4.18)(75.0) = 7838\ \mathrm{J} = 7.84\) kJ. Vaporize: \(n = 25.0/18.02 = 1.387\) mol; \((1.387)(40.7) = 56.5\) kJ. Total \(= 7.84 + 56.5 = 64.3\) kJ. Boiling takes more than seven times the energy of the 75-degree warm-up, and \(\Delta H_{\text{vap}}\) is per mole, so convert the grams first.

6. While water boils at 100 °C, its temperature does not rise even though heat is added steadily. This is because

   the water stops absorbing heat the added energy raises the potential energy of the molecules by separating them against their intermolecular forces, not their kinetic energy the molecules stop moving the specific heat of water becomes zero

   The water absorbs heat the whole time: that is what drives the boiling, at 40.7 kJ per mole vaporized. The energy is going somewhere; it just isn't showing up as a temperature rise.

   Temperature measures average kinetic energy. During a phase change the incoming heat does the work of pulling molecules out of the liquid's hydrogen-bonded network, increasing their potential energy; average kinetic energy, and therefore temperature, stays fixed until the change is complete.

   Molecules at 100 °C are moving vigorously, and the escaping vapor molecules have the same average kinetic energy as the liquid ones. Motion never stops above absolute zero; it is the average kinetic energy that stays constant.

   Specific heat is a property of a single phase and does not vanish; rather, \(q = mc\Delta T\) simply doesn't apply during a phase change, where the relevant relation is \(q = n\Delta H_{\text{vap}}\). Using the wrong equation is the error, not a constant dropping to zero.

7. Using average bond enthalpies (H–H 436, O=O 495, O–H 467 kJ/mol), estimate \(\Delta H\) for 2 H₂(g) + O₂(g) → 2 H₂O(g). How much heat is released when 0.500 mol of O₂ reacts?

   Show answer

   Broken: 2 H–H + 1 O=O \(= 2(436) + 495 = 1367\) kJ. Formed: 4 O–H \(= 4(467) = 1868\) kJ. \(\Delta H \approx 1367 - 1868 = -501\) kJ for the reaction as written (accepted −484 kJ; averages are approximate). Exothermic because the bonds formed are stronger in total. For 0.500 mol O₂, half the reaction, \(0.500 \times 501 = 251\) kJ released, about 250 kJ.

8. Using \(\Delta H^\circ_f\) values CH₄(g) −74.8, CO₂(g) −393.5, and H₂O(l) −285.8 kJ/mol, find \(\Delta H^\circ\) for CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l). Why is this more negative than the bond-enthalpy estimate of −824 kJ from Lesson 6.4?

   Show answer

   \(\Delta H^\circ = [(-393.5) + 2(-285.8)] - [(-74.8) + 2(0)] = -965.1 + 74.8 = -890.3\) kJ/mol. O₂ is an element in its standard state, so its \(\Delta H^\circ_f = 0\). It is more negative partly because here the water is liquid: condensing 2 mol of water vapor releases an additional \(2 \times 44 = 88\) kJ, and the bond-enthalpy method is only an estimate from averages.

9. Given (1) 2 C(s) + O₂(g) → 2 CO(g), \(\Delta H^\circ = -221.0\) kJ, and (2) C(s) + O₂(g) → CO₂(g), \(\Delta H^\circ = -393.5\) kJ, use Hess's law to find \(\Delta H^\circ\) for 2 CO(g) + O₂(g) → 2 CO₂(g).

   Show answer

   The target has 2 CO as a reactant, so reverse (1): 2 CO → 2 C + O₂, \(+221.0\) kJ. It has 2 CO₂ as product, so double (2): 2 C + 2 O₂ → 2 CO₂, \(2(-393.5) = -787.0\) kJ. Add: the 2 C cancel and one O₂ cancels, leaving 2 CO + O₂ → 2 CO₂ with \(\Delta H^\circ = 221.0 - 787.0 = -566.0\) kJ. (Check with formation enthalpies: \(2(-393.5) - 2(-110.5) = -566.0\). ✓)

10. 2 H₂O₂(l) → 2 H₂O(l) + O₂(g) has \(\Delta H = -196\) kJ. Is the decomposition endothermic or exothermic? How much heat is released when 10.0 g of hydrogen peroxide (34.02 g/mol) decomposes?

    Show answer

    Negative \(\Delta H\): exothermic. \(10.0/34.02 = 0.294\) mol H₂O₂. The equation releases 196 kJ per 2 mol, i.e. 98 kJ per mole: \(0.294 \times 98.0 = 28.8\) kJ released. \(\Delta H\) is extensive: always ask "per mole of what?"

Lesson 7.1 · Unit 7 · CED topics 7.1–7.3

Reversible reactions and the equilibrium constant

Most reactions don't run to completion. Products, once formed, collide and re-form reactants. When the forward and reverse rates become equal, the concentrations stop changing, not because the reaction stopped, but because both directions are running at the same speed.

Lesson 7.2 · Unit 7 · CED topics 7.4–7.6

ICE tables and equilibrium calculations

An ICE table (Initial, Change, Equilibrium) is just organized stoichiometry: you let an unknown amount \(x\) react, track every species with the coefficients, and then let \(K\) pin down \(x\).

Lesson 7.3 · Unit 7 · CED topics 7.5, 7.7

Properties and magnitude of K

\(K\) is a number with a story. A huge \(K\) says the reaction runs nearly to completion; a tiny \(K\) says it barely starts. And because \(K\) comes from the balanced equation, rewriting the equation rewrites \(K\) in predictable ways.

Lesson 7.4 · Unit 7 · CED topics 7.8–7.9

Le Châtelier's principle

Disturb a system at equilibrium and it shifts to partially undo the disturbance. That's the slogan. The mechanism underneath is always the same: the stress makes \(Q \ne K\), and the reaction runs in whichever direction brings \(Q\) back to \(K\). Argue from \(Q\) and you'll never get it backward.

Lesson 7.5 · Unit 7 · CED topics 7.10–7.14

Solubility equilibria

"Insoluble" salts aren't perfectly insoluble. A little dissolves until the dissolved ions are in equilibrium with the solid: an equilibrium with its own constant, \(K_{sp}\), and everything you know about \(Q\) vs. \(K\) applies.

Unit 7 practice · 10 problems

Unit 7 practice: Equilibrium

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer. Calculator allowed.

1. Write the \(K_c\) expression for 2 NO(g) + Br₂(g) ⇌ 2 NOBr(g) and the \(K_p\) expression for C(s) + CO₂(g) ⇌ 2 CO(g). Why does carbon not appear?

   Show answer

   \(K_c = \dfrac{[\mathrm{NOBr}]^2}{[\mathrm{NO}]^2[\mathrm{Br_2}]}\). \(K_p = \dfrac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO_2}}}\). Solid carbon is omitted because a pure solid's "concentration" is fixed; adding or removing it changes nothing about the gas-phase equilibrium.

2. H₂(g) + I₂(g) ⇌ 2 HI(g) has \(K_c = 54\) at 700 K. A flask holds [H₂] = 0.20 M, [I₂] = 0.10 M, and [HI] = 0.80 M. Is the system at equilibrium? If not, which way does it proceed, and why at the particle level?

   Show answer

   \(Q = \dfrac{(0.80)^2}{(0.20)(0.10)} = 32\). \(Q \lt K\), so not at equilibrium; the reaction proceeds forward, making HI. With more reactant than equilibrium allows, H₂–I₂ collisions that form HI outnumber HI decompositions until the ratio climbs to 54.

3. N₂O₄(g) ⇌ 2 NO₂(g). A flask starts with 0.200 M N₂O₄ and no NO₂; at equilibrium [NO₂] = 0.100 M. Find \(K_c\).

   Show answer

   ICE: N₂O₄ \(0.200 - x\), NO₂ \(2x\). \(2x = 0.100\) gives \(x = 0.050\), so [N₂O₄] \(= 0.150\) M. \(K_c = \dfrac{(0.100)^2}{0.150} = 0.0667\). Watch the coefficient: the change in NO₂ is \(+2x\), not \(+x\).

4. COCl₂(g) ⇌ CO(g) + Cl₂(g) has \(K_c = 2.2 \times 10^{-10}\) at 373 K. Starting with 0.500 M COCl₂, find the equilibrium [CO] and justify any approximation.

   Show answer

   \(\dfrac{x^2}{0.500 - x} \approx \dfrac{x^2}{0.500} = 2.2 \times 10^{-10}\), so \(x^2 = 1.1 \times 10^{-10}\) and \(x = [\mathrm{CO}] = 1.0 \times 10^{-5}\) M. Check: \(x/0.500 = 0.002\%\), far under 5%, so \(0.500 - x \approx 0.500\) is valid. Tiny \(K\), tiny \(x\).

5. N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) has \(K = 6.0 \times 10^{5}\) at 298 K. Find \(K\) for NH₃(g) ⇌ ½ N₂(g) + 3⁄2 H₂(g).

   Show answer

   Reverse (\(1/K\)) and halve (square root): \(K' = \left(\dfrac{1}{6.0 \times 10^{5}}\right)^{1/2} = (1.67 \times 10^{-6})^{1/2} = 1.3 \times 10^{-3}\). Sensible: decomposing ammonia is unfavorable when forming it is favorable.

6. At 25 °C, N₂(g) + O₂(g) ⇌ 2 NO(g) has \(K_1 = 2.0 \times 10^{-31}\) and 2 NO(g) + O₂(g) ⇌ 2 NO₂(g) has \(K_2 = 5.3 \times 10^{12}\). Find \(K\) for NO₂(g) ⇌ ½ N₂(g) + O₂(g).

   Show answer

   Adding the two gives N₂ + 2 O₂ ⇌ 2 NO₂ with \(K = K_1 K_2 = 1.06 \times 10^{-18}\). The target is that reaction reversed and halved: \(K' = \left(\dfrac{1}{1.06 \times 10^{-18}}\right)^{1/2} = (9.4 \times 10^{17})^{1/2} = 9.7 \times 10^{8}\). Large: NO₂ strongly prefers to fall apart into the elements at room temperature.

7. 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), \(\Delta H = -198\) kJ. Predict the direction of shift, if any, and justify each with \(Q\) versus \(K\): (a) adding O₂; (b) halving the volume; (c) raising the temperature; (d) adding argon at constant volume; (e) adding a catalyst.

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   (a) O₂ is in the denominator, so \(Q \lt K\): forward. (b) Every concentration doubles; the denominator (3 gas moles) grows by \(2^3 = 8\), the numerator by \(2^2 = 4\), so \(Q = K/2 \lt K\): forward, toward fewer gas moles. (c) Exothermic, so heat is a product; adding heat lowers \(K\) itself, making \(Q \gt K\): reverse. (d) No partial pressure of any reactant or product changes, so \(Q = K\): no shift. (e) A catalyst speeds both directions equally: no shift, equilibrium is just reached sooner.

8. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is endothermic. Which change increases the equilibrium amount of PCl₃?

   adding Cl₂ decreasing the volume raising the temperature adding a catalyst

   Cl₂ is a product, so adding it makes \(Q \gt K\) and the system shifts in reverse, consuming PCl₃. Adding a reactant would push forward; adding a product pushes back.

   Decreasing the volume raises the total pressure, and the system shifts toward the side with fewer moles of gas: here the PCl₅ side (1 mol vs. 2 mol). That is the reverse direction, so PCl₃ decreases.

   Endothermic means heat is effectively a reactant, so raising \(T\) increases \(K\) itself and the system shifts forward, producing more PCl₃. Temperature is the only change listed that alters \(K\).

   A catalyst lowers the activation energy for the forward and reverse reactions equally, so equilibrium is reached faster but its position, and \(K\), is unchanged. Rate and extent are different questions.

9. \(K_{sp}\)(CaF₂) \(= 3.9 \times 10^{-11}\). (a) Find its molar solubility in pure water. (b) Find it in 0.10 M NaF. (c) Would CaF₂ be more or less soluble in an acidic solution? Explain.

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   (a) \([\mathrm{Ca^{2+}}] = s\), \([\mathrm{F^-}] = 2s\): \(s(2s)^2 = 4s^3 = 3.9 \times 10^{-11}\), \(s^3 = 9.75 \times 10^{-12}\), \(s = 2.1 \times 10^{-4}\) M. (b) \([\mathrm{F^-}] \approx 0.10\): \(s(0.10)^2 = 3.9 \times 10^{-11}\), \(s = 3.9 \times 10^{-9}\) M: the common ion cuts solubility about 50,000-fold. (c) More soluble: F⁻ is a weak base, so H⁺ converts it to HF, lowering [F⁻], making \(Q \lt K_{sp}\), and pulling more solid into solution.

10. 25.0 mL of \(2.0 \times 10^{-3}\) M Pb(NO₃)₂ is mixed with 25.0 mL of \(4.0 \times 10^{-3}\) M KI. \(K_{sp}\)(PbI₂) \(= 9.8 \times 10^{-9}\). Does PbI₂ precipitate?

    Show answer

    Mixing equal volumes halves each: \([\mathrm{Pb^{2+}}] = 1.0 \times 10^{-3}\) M, \([\mathrm{I^-}] = 2.0 \times 10^{-3}\) M. \(Q = [\mathrm{Pb^{2+}}][\mathrm{I^-}]^2 = (1.0 \times 10^{-3})(2.0 \times 10^{-3})^2 = 4.0 \times 10^{-9}\). \(Q \lt K_{sp}\), so no precipitate forms; the solution is unsaturated. (Don't forget the square on iodide or the dilution.)

Lesson 8.1 · Unit 8 · CED topics 8.1–8.3

pH, pOH, and strong acids and bases

Even pure water carries a trace of H₃O⁺ and OH⁻, because one water molecule occasionally pulls a proton off another. That tiny equilibrium sets the scale every acid and base is measured against.

Lesson 8.2 · Unit 8 · CED topics 8.4–8.6

Weak acids and bases

A weak acid only partly ionizes: most of its molecules are still intact at equilibrium. That makes every weak-acid pH problem an equilibrium problem, and the ICE table from Unit 7 is the tool.

Lesson 8.3 · Unit 8 · CED topics 8.7–8.8

Acid–base reactions and buffers

Your blood holds pH 7.4 within a few hundredths despite acids from metabolism pouring in constantly. It does that with a buffer. First, though: what happens when acids and bases mix.

Lesson 8.4 · Unit 8 · CED topics 8.9–8.10

Henderson–Hasselbalch and buffer capacity

Take the \(K_a\) expression, solve for [H₃O⁺], take the negative log of both sides, and you get an equation that makes buffer design almost arithmetic.

Lesson 8.5 · Unit 8 · CED topics 8.5, 8.7

Titration curves and indicators

A titration curve plots pH against titrant volume. Its shape is a fingerprint: strong or weak acid, how much, and, for a weak acid, its p\(K_a\).

Lesson 8.6 · Unit 8 · CED topic 8.6

Molecular structure and acid strength

Why is HCl strong and HF weak, when fluorine is more electronegative? Because acid strength is about how easily the proton leaves, which depends on the bond holding it and on how well the leftover anion handles its charge.

Unit 8 practice · 10 problems

Unit 8 practice: Acids and Bases

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer. Calculator allowed; \(K_w = 1.0 \times 10^{-14}\) at 25 °C.

1. Find the pH, pOH, and [OH⁻] of (a) 0.0400 M HNO₃ and (b) 0.0020 M Sr(OH)₂, both at 25 °C.

   Show answer

   (a) Strong acid: \([\mathrm{H_3O^+}] = 0.0400\) M, pH \(= -\log(0.0400) = 1.40\), pOH \(= 12.60\), \([\mathrm{OH^-}] = 10^{-12.60} = 2.5 \times 10^{-13}\) M. (b) Two OH⁻ per formula unit: \([\mathrm{OH^-}] = 0.0040\) M, pOH \(= 2.40\), pH \(= 11.60\).

2. At 50 °C, \(K_w = 5.5 \times 10^{-14}\). Pure water at this temperature has pH 6.63. The water is

   slightly acidic neutral, because [H₃O⁺] = [OH⁻] slightly basic impossible to classify without [OH⁻]

   pH below 7 means acidic only at 25 °C, where \(K_w = 1.0 \times 10^{-14}\) and neutral pH is 7.00. At 50 °C the neutral point itself has moved to 6.63, so pure water there is not acidic.

   Autoionization produces H₃O⁺ and OH⁻ in equal amounts, so \([\mathrm{H_3O^+}] = [\mathrm{OH^-}] = \sqrt{5.5 \times 10^{-14}} = 2.3 \times 10^{-7}\) M and pH \(= -\log(2.3 \times 10^{-7}) = 6.63\). Neutral means equal concentrations, not pH 7; here pH + pOH = 13.26.

   A pH of 6.63 could be basic only if it were above the neutral pH for that temperature, and at 50 °C the neutral pH is 6.63. Neither ion outnumbers the other in pure water at any temperature.

   [OH⁻] is fully determined: in pure water it must equal [H₃O⁺] by the stoichiometry of autoionization, and \(K_w\) fixes both at \(\sqrt{K_w}\). No extra information is needed.

3. Benzoic acid has \(K_a = 6.3 \times 10^{-5}\). Find the pH and percent ionization of a 0.0500 M solution. Show the ICE reasoning and check your approximation.

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   HA ⇌ H₃O⁺ + A⁻; E row: \(0.0500 - x\), \(x\), \(x\). \(\dfrac{x^2}{0.0500 - x} \approx \dfrac{x^2}{0.0500} = 6.3 \times 10^{-5}\), so \(x^2 = 3.15 \times 10^{-6}\) and \(x = 1.77 \times 10^{-3}\) M. Check: \(1.77 \times 10^{-3}/0.0500 = 3.5\%\), under 5%. ✓ pH \(= -\log(1.77 \times 10^{-3}) = 2.75\); percent ionization \(= 3.5\%\). Roughly 96 of every 100 molecules stay intact.

4. Methylamine, CH₃NH₂, has \(K_b = 4.4 \times 10^{-4}\). Find the pH of a 0.200 M solution, and find \(K_a\) of its conjugate acid CH₃NH₃⁺.

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   B + H₂O ⇌ BH⁺ + OH⁻: \(\dfrac{x^2}{0.200} = 4.4 \times 10^{-4}\), \(x = [\mathrm{OH^-}] = 9.4 \times 10^{-3}\) M (4.7% of 0.200: acceptable). pOH \(= 2.03\), pH \(= 11.97\). Conjugate pair: \(K_a = \dfrac{K_w}{K_b} = \dfrac{1.0 \times 10^{-14}}{4.4 \times 10^{-4}} = 2.3 \times 10^{-11}\): a fairly strong weak base has a very weak conjugate acid.

5. 30.0 mL of 0.200 M NaOH is mixed with 50.0 mL of 0.100 M HCl. Find the pH of the resulting solution.

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   OH⁻: \((30.0)(0.200) = 6.00\) mmol. H₃O⁺: \((50.0)(0.100) = 5.00\) mmol. They react 1 : 1, leaving 1.00 mmol OH⁻ in 80.0 mL: \([\mathrm{OH^-}] = 0.0125\) M, pOH \(= 1.90\), pH = 12.10. Use the total volume.

6. Which of these solutions are buffers? (a) 0.10 M HCN + 0.10 M NaCN; (b) 0.10 M HBr + 0.10 M NaBr; (c) 0.20 M NH₃ + 0.10 M HCl; (d) 0.10 M NaOH + 0.10 M NaCl. For solution (a), write the net ionic equation that occurs when a little NaOH is added and explain why the pH barely changes.

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   (a) Yes: weak acid + conjugate base. (b) No: HBr is strong and Br⁻ has no tendency to accept a proton. (c) Yes: HCl converts half the NH₃ to NH₄⁺, leaving 0.10 M each of a weak base and its conjugate acid. (d) No: a strong base and a neutral salt. In (a), HCN + OH⁻ → CN⁻ + H₂O: the strong base is consumed by the large reservoir of HCN, becoming an equal amount of the weak base CN⁻. Since \([\mathrm{H_3O^+}] = K_a[\mathrm{HCN}]/[\mathrm{CN^-}]\) and both amounts are large, the ratio, and so the pH, shifts only slightly.

7. A buffer is 0.30 M acetic acid (p\(K_a\) = 4.74) and 0.50 M sodium acetate. (a) Find its pH. (b) 0.020 mol of HCl is added to 1.0 L of this buffer. Find the new pH.

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   (a) \(\text{pH} = 4.74 + \log\dfrac{0.50}{0.30} = 4.74 + 0.22 = 4.96\). (b) H₃O⁺ + C₂H₃O₂⁻ → HC₂H₃O₂ + H₂O converts 0.020 mol acetate to acid: A⁻ \(= 0.48\), HA \(= 0.32\) mol. \(\text{pH} = 4.74 + \log\dfrac{0.48}{0.32} = 4.74 + 0.18 = 4.92\). A drop of 0.04; the same acid in 1.0 L of pure water would give pH 1.70.

8. You need a buffer at pH 7.20. Available acids: HNO₂ (\(K_a = 4.5 \times 10^{-4}\)), acetic acid (\(K_a = 1.8 \times 10^{-5}\)), HClO (\(K_a = 3.0 \times 10^{-8}\)). Choose the acid and find the required [A⁻]/[HA] ratio. What determines the buffer's capacity?

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   p\(K_a\) values: 3.35, 4.74, 7.52. Only HClO is within one unit of 7.20. \(7.20 = 7.52 + \log\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}\), so \(\log(\text{ratio}) = -0.32\) and \(\dfrac{[\mathrm{ClO^-}]}{[\mathrm{HClO}]} = 10^{-0.32} = 0.48\). The pH is below p\(K_a\), so more acid than base: sensible. Capacity depends on the absolute amounts of HClO and ClO⁻, not the ratio: a 1.0 M / 0.48 M buffer absorbs ten times as much strong acid or base as 0.10 M / 0.048 M.

9. 25.0 mL of 0.200 M formic acid, HCOOH (\(K_a = 1.8 \times 10^{-4}\)), is titrated with 0.100 M NaOH. Find the volume at equivalence, the pH at half-equivalence, the pH at equivalence, and a suitable indicator (methyl red 4.4–6.2, phenolphthalein 8.2–10.0).

   Show answer

   Acid: \((25.0)(0.200) = 5.00\) mmol, so equivalence needs \(5.00/0.100 = 50.0\) mL. At 25.0 mL (half-equivalence) [HA] = [A⁻]: pH = p\(K_a\) \(= 3.74\). At equivalence, 5.00 mmol formate in 75.0 mL \(= 0.0667\) M, a weak base with \(K_b = 1.0 \times 10^{-14}/1.8 \times 10^{-4} = 5.6 \times 10^{-11}\): \(x = \sqrt{(0.0667)(5.6 \times 10^{-11})} = 1.9 \times 10^{-6}\) M OH⁻, pOH \(= 5.72\), pH = 8.28. Phenolphthalein changes in the steep region near 8.3; methyl red would change in the buffer region and signal too early.

10. (a) Rank H₂O, H₂S, and H₂Se from weakest to strongest acid and explain. (b) Fluoroacetic acid, CH₂FCOOH, has \(K_a = 2.6 \times 10^{-3}\), versus \(1.8 \times 10^{-5}\) for acetic acid. Explain the difference at the molecular level.

    Show answer

    (a) H₂O < H₂S < H₂Se. Down the group the central atom is larger, the H–X bond longer and weaker, so the proton is released more easily: bond strength outweighs the decreasing polarity, just as for HF < HCl < HBr < HI. (b) The electronegative F atom near the carboxyl group withdraws electron density, making the O–H bond more polar and weaker, and it stabilizes the CH₂FCOO⁻ anion by pulling negative charge toward itself. A more stable conjugate base means ionization is more favorable: about 140 times larger \(K_a\).

Lesson 9.1 · Unit 9 · CED topics 9.1–9.2

Entropy

Drop a deck of cards and it doesn't land sorted. There is one sorted arrangement and trillions of shuffled ones, so "shuffled" wins by sheer count. Entropy is that idea applied to molecules and their energy.

Lesson 9.2 · Unit 9 · CED topics 9.3–9.4

Gibbs free energy and thermodynamic favorability

Exothermic reactions tend to go; entropy-increasing reactions tend to go. When the two disagree, temperature is the referee. Gibbs free energy combines all three into one number whose sign answers the question "will this happen on its own?"

Lesson 9.3 · Unit 9 · CED topics 9.5–9.6

Free energy and equilibrium; coupled reactions

"Favorable" and "large K" are the same statement in two languages. One equation translates between them, and it also explains how cells and blast furnaces get unfavorable reactions to run anyway.

Lesson 9.4 · Unit 9 · CED topics 9.7–9.8

Galvanic and electrolytic cells

A redox reaction moves electrons from one species to another. Separate the half-reactions and force the electrons through a wire: a battery. Run it backward with a power supply: electrolysis.

Lesson 9.5 · Unit 9 · CED topics 9.9–9.10

Cell potential, free energy, and Faraday's law

Voltage is energy per charge. Multiply a cell's potential by the charge it pushes and you get the free energy it releases: tying electrochemistry back to the rest of this unit. Count the charge and you can also predict how much metal an electrolysis deposits.

Unit 9 practice · 10 problems

Unit 9 practice: Thermodynamics and Electrochemistry

Ten problems covering the whole unit, in roughly exam order. Work each one on paper before revealing the answer. Calculator allowed; \(R = 8.314\) J/(mol·K), \(F = 96{,}485\) C/mol e⁻. Standard reduction potentials: Ag⁺/Ag +0.80 V, Cu²⁺/Cu +0.34 V, Ni²⁺/Ni −0.26 V, Zn²⁺/Zn −0.76 V, Al³⁺/Al −1.66 V.

1. Predict the sign of \(\Delta S\) for each and justify: (a) 2 H₂(g) + O₂(g) → 2 H₂O(l); (b) NH₄Cl(s) → NH₃(g) + HCl(g); (c) I₂(s) → I₂(g); (d) 2 NO₂(g) → N₂O₄(g).

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   (a) Negative: three moles of gas become a liquid; far fewer positions and energy arrangements available. (b) Positive: a solid becomes two moles of gas. (c) Positive: solid → gas; molecules go from a fixed lattice to roaming the container. (d) Negative: two gas moles become one.

2. Using \(S^\circ\) values H₂(g) 130.7, O₂(g) 205.2, H₂O(l) 69.9 J/(mol·K), calculate \(\Delta S^\circ\) for 2 H₂O(l) → 2 H₂(g) + O₂(g). Does the sign match your prediction?

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   \(\Delta S^\circ = [2(130.7) + 205.2] - 2(69.9) = 466.6 - 139.8 = +326.8\) J/(mol·K). Positive, as expected for a liquid turning into three moles of gas. Note that the elements do not have \(S^\circ = 0\), unlike \(\Delta H^\circ_f\).

3. For the reaction in problem 2, \(\Delta H^\circ = +571.6\) kJ. Find \(\Delta G^\circ\) at 298 K, decide whether the decomposition of water is favorable, and estimate the temperature above which it becomes favorable.

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   Convert units: \(\Delta S^\circ = 0.3268\) kJ/K. \(\Delta G^\circ = 571.6 - (298)(0.3268) = 571.6 - 97.4 = +474.2\) kJ: unfavorable at 298 K (water doesn't fall apart). Crossover: \(T = \dfrac{\Delta H^\circ}{\Delta S^\circ} = \dfrac{571{,}600\ \mathrm{J}}{326.8\ \mathrm{J/K}} = 1750\) K. Above roughly 1750 K the \(T\Delta S\) term wins. Mixing kJ and J here is the classic error.

4. A reaction has \(\Delta H \lt 0\) and \(\Delta S \lt 0\). It is thermodynamically favorable

   at all temperatures at no temperature only at low temperatures only at high temperatures

   Favorable at all temperatures requires \(\Delta H \lt 0\) and \(\Delta S \gt 0\), so that both terms in \(\Delta G = \Delta H - T\Delta S\) are negative. Here the entropy term opposes the reaction, so there is a temperature above which it wins.

   Favorable at no temperature is the case \(\Delta H \gt 0\) and \(\Delta S \lt 0\), where both terms are positive. Here \(\Delta H\) is negative and drives the reaction whenever \(T\) is small enough.

   \(\Delta G = \Delta H - T\Delta S\): the negative \(\Delta H\) favors the reaction, the negative \(\Delta S\) opposes it, and the opposing term \(-T\Delta S\) grows with \(T\), so at low \(T\) the enthalpy wins and \(\Delta G \lt 0\) (freezing water is the example). Remember that \(\Delta G\) says favorable, not fast: diamond → graphite has \(\Delta G^\circ = -2.9\) kJ/mol, yet diamonds persist because rearranging every C–C bond has an enormous activation energy: the change is under kinetic control.

   High temperature is where the \(-T\Delta S\) term dominates, and with \(\Delta S \lt 0\) that term is positive, making \(\Delta G \gt 0\). “Only at high temperatures” belongs to reactions with \(\Delta H \gt 0\) and \(\Delta S \gt 0\), such as melting.

5. Acetic acid has \(K_a = 1.8 \times 10^{-5}\) at 298 K. Calculate \(\Delta G^\circ\) for its ionization and interpret the sign.

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   \(\Delta G^\circ = -RT\ln K = -(8.314)(298)\ln(1.8 \times 10^{-5}) = -(2478)(-10.93) = +2.71 \times 10^{4}\) J/mol \(= +27.1\) kJ/mol. Positive: \(K \lt 1\), so at equilibrium the un-ionized acid is favored over the ions; exactly what "weak acid" means.

6. A reaction has \(\Delta G^\circ = -16.4\) kJ/mol at 298 K. Find \(K\). Are products or reactants favored?

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   \(\ln K = -\dfrac{\Delta G^\circ}{RT} = \dfrac{16{,}400\ \mathrm{J/mol}}{(8.314)(298)} = 6.62\), so \(K = e^{6.62} = 7.5 \times 10^{2}\). \(K \gt 1\): products favored. Convert kJ to J before dividing.

7. CuO(s) → Cu(s) + ½ O₂(g) has \(\Delta G^\circ = +129.7\) kJ, and C(s) + ½ O₂(g) → CO(g) has \(\Delta G^\circ = -137.2\) kJ. Show how carbon can be used to extract copper from its oxide, and find \(\Delta G^\circ\) for the coupled reaction.

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   Add the two: the ½ O₂ cancels, giving CuO(s) + C(s) → Cu(s) + CO(g) with \(\Delta G^\circ = 129.7 + (-137.2) = -7.5\) kJ. Free-energy changes add, so coupling the unfavorable decomposition to the more favorable oxidation of carbon (which consumes the O₂ the first reaction produces) makes the overall process favorable.

8. A galvanic cell is built from Ni/Ni²⁺(1.0 M) and Ag/Ag⁺(1.0 M) half-cells. Identify the anode and cathode, write the half-reactions and the overall equation, find \(E^\circ_{\text{cell}}\), write the cell notation, and describe the direction of electron flow and of ion migration in the salt bridge.

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   Ag⁺ has the higher reduction potential, so it is reduced: Ag is the cathode, Ni the anode. Anode: Ni → Ni²⁺ + 2 e⁻. Cathode: Ag⁺ + e⁻ → Ag (×2). Overall: Ni(s) + 2 Ag⁺(aq) → Ni²⁺(aq) + 2 Ag(s). \(E^\circ_{\text{cell}} = 0.80 - (-0.26) = 1.06\) V (potentials are not multiplied by 2). Notation: Ni(s) | Ni²⁺(aq) || Ag⁺(aq) | Ag(s). Electrons flow Ni → Ag through the wire; in the salt bridge, anions drift toward the Ni half-cell (to balance the Ni²⁺ being produced) and cations toward the Ag half-cell.

9. For the Ni/Ag cell in problem 8, calculate \(\Delta G^\circ\). Then predict how \(E_{\text{cell}}\) changes if [Ag⁺] is increased to 2.0 M, with reasoning.

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   Two electrons are transferred: \(\Delta G^\circ = -nFE^\circ = -(2)(96{,}485\ \mathrm{C/mol})(1.06\ \mathrm{J/C}) = -2.05 \times 10^{5}\) J \(= -205\) kJ/mol. Negative, matching \(E^\circ \gt 0\). Raising [Ag⁺] (a reactant) makes \(Q = [\mathrm{Ni^{2+}}]/[\mathrm{Ag^+}]^2\) smaller, so the system is farther from equilibrium and \(E \gt 1.06\) V. Particle-level: more Ag⁺ at the cathode means more frequent reduction events.

10. A current of 4.00 A passes through molten AlCl₃ for 30.0 min. What mass of aluminum is deposited, and at which electrode?

    Show answer

    \(q = It = (4.00\ \mathrm{A})(1800\ \mathrm{s}) = 7200\) C. Moles of electrons: \(7200 / 96{,}485 = 0.0746\) mol. Al³⁺ + 3 e⁻ → Al needs three electrons per atom: \(0.0746/3 = 0.0249\) mol Al \(\times 26.98 = 0.671\) g. Reduction happens at the cathode, even in an electrolytic cell.

Unit recap

Unit recap

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