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10.P — Worked Problems: Bonding, Reactions and Matter
Twelve problems across Part 10. Every solution shows the arithmetic. Attempt each before opening it.
Problem 1 — Lattice energy from Born–Landé
Compute the lattice energy of magnesium oxide. r_0 = 212 pm, Madelung constant 1.7476 (rock salt), Born exponent n = 7, both ions doubly charged.
Solution
U = -\frac{N_A M z^+z^-e^2}{4\pi\varepsilon_0 r_0}\left(1-\frac{1}{n}\right)
With z^+z^- = 2\times2 = 4:
Numerator:
(6.022\times10^{23})(1.7476)(4)(1.602\times10^{-19})^2 = (6.022\times10^{23})(1.7476)(4)(2.566\times10^{-38})
= 1.080\times10^{-13}
Denominator:
4\pi(8.854\times10^{-12})(2.12\times10^{-10}) = (1.113\times10^{-10})(2.12\times10^{-10}) = 2.360\times10^{-20}
U = -\frac{1.080\times10^{-13}}{2.360\times10^{-20}}\times\left(1-\frac{1}{7}\right) = -(4.576\times10^{6})(0.857)
U = -3.92\times10^{6}\ \text{J/mol} = -3920\ \text{kJ/mol}
Experimental: -3795 kJ/mol. Within 3 %.
What to notice. Compare with NaCl's -787 kJ/mol from Chapter 10.1. MgO is five times larger, and the two factors are the charge product (4 instead of 1, giving a factor of 4) and the shorter separation (212 vs 282 pm, giving 1.33).
And the melting points follow directly: MgO 2852 °C, NaCl 801 °C. This is why magnesium oxide lines furnaces and why the general rule "higher charge means higher melting point" works for ionic solids.
Problem 2 — Predicting shape and polarity
For each molecule, give the electron pair arrangement, the shape, and whether it is polar: (a) SO₂, (b) CCl₄, (c) NH₃, (d) XeF₄, (e) CHCl₃.
Solution
(a) SO₂. Sulphur has 6 valence electrons; two double bonds to oxygen use 4, leaving one lone pair. Three regions: trigonal planar arrangement, bent shape, angle about 119°. Polar — the two bond dipoles do not cancel.
(b) CCl₄. Four bonding pairs, no lone pairs. Tetrahedral, 109.5°. Non-polar — four identical bonds pointing to tetrahedron corners cancel exactly.
(c) NH₃. Three bonds, one lone pair. Tetrahedral arrangement, trigonal pyramidal shape, 107°. Polar — the three N–H dipoles add, and the lone pair contributes too.
(d) XeF₄. Xenon has 8 valence electrons; four bonds use 4, leaving two lone pairs. Six regions: octahedral arrangement, lone pairs opposite each other, giving a square planar shape. Non-polar — the four Xe–F dipoles cancel in the plane.
(e) CHCl₃. Four bonding pairs. Tetrahedral, and the shape is tetrahedral. Polar, because one bond is C–H and three are C–Cl, so the dipoles do not cancel.
What to notice. CCl₄ and CHCl₃ have the same shape and opposite polarity, because polarity depends on the bonds as well as the geometry. CCl₄ boils at 77 °C and CHCl₃ at 61 °C — the lighter one boils higher, because its dipole–dipole attraction outweighs the mass difference.
And XeF₄ being non-polar is worth pausing on: two lone pairs and four bonds, arranged so everything cancels. A noble gas compound whose polarity you can predict from a 1957 rule.
Problem 3 — MO diagram and bond order
Use molecular orbital theory to find the bond order and magnetic behaviour of (a) N₂, (b) O₂, (c) NO, (d) F₂.
Solution
Filling order for second-row diatomics (with the ordering for O₂ and F₂, where \sigma_{2p} lies below \pi_{2p}):
\sigma_{2s}\ \sigma^*_{2s}\ \sigma_{2p}\ \pi_{2p}\ \pi^*_{2p}\ \sigma^*_{2p}
(a) N₂, 10 valence electrons. (For N₂ the \pi_{2p} actually lies below \sigma_{2p}, which does not change the count.)
\sigma_{2s}^2\sigma_{2s}^{*2}\pi_{2p}^4\sigma_{2p}^2
\text{Bond order} = \frac{8-2}{2} = 3
Triple bond, all paired — diamagnetic. Consistent with the 945 kJ/mol bond strength.
(b) O₂, 12 valence electrons.
\sigma_{2s}^2\sigma_{2s}^{*2}\sigma_{2p}^2\pi_{2p}^4\pi_{2p}^{*2}
\text{Bond order} = \frac{8-4}{2} = 2
The last two electrons are in two degenerate \pi^* orbitals, so by Hund's rule they are unpaired. Paramagnetic.
(c) NO, 11 valence electrons.
\text{Bond order} = \frac{8-3}{2} = 2.5
One unpaired electron — paramagnetic, and it is a radical.
(d) F₂, 14 valence electrons.
\text{Bond order} = \frac{8-6}{2} = 1
Single bond, diamagnetic.
What to notice. Bond order 2.5 for NO is impossible in Lewis theory and it is exactly what MO gives. The measured bond length, 115 pm, sits between O₂'s 121 and N₂'s 110, as a bond order of 2.5 requires.
NO being a stable radical is unusual and biologically crucial: it is a signalling molecule that relaxes blood vessels, which is how nitroglycerine treats angina and how Viagra works. The 1998 Nobel Prize in Medicine went to that discovery, and the molecule's reactivity — and its short lifetime — trace directly to that single electron in an antibonding orbital.
Problem 4 — Boiling points from intermolecular forces
Rank these by boiling point and justify: butane (C₄H₁₀), propanone (C₃H₆O), propan-1-ol (C₃H₈O), and propane (C₃H₈). Molar masses: 58, 58, 60, 44.
Solution
Propane, 44 g/mol. Non-polar, dispersion only, and the lightest. Boils at -42 °C.
Butane, 58 g/mol. Non-polar, dispersion only, but larger so more polarisable. Boils at -0.5 °C.
Propanone (acetone), 58 g/mol. Has a C=O, so dipole–dipole (2.9 D) on top of dispersion. No O–H, so no hydrogen bonding to itself. Boils at 56 °C.
Propan-1-ol, 60 g/mol. Has O–H, so hydrogen bonding. Boils at 97 °C.
\text{propane} < \text{butane} < \text{propanone} < \text{propan-1-ol}
What to notice. Butane and propanone have identical molar masses and boil 57 degrees apart, entirely from the dipole. Propanone and propan-1-ol differ by 41 degrees, from hydrogen bonding.
The ordering of forces is clear from the numbers: dispersion sets the baseline, dipole–dipole adds tens of degrees, hydrogen bonding adds tens more.
Note that propanone can accept hydrogen bonds from water, even though it cannot donate them, which is why acetone is fully miscible with water despite not hydrogen bonding to itself. Donating and accepting are separate abilities.
Problem 5 — Limiting reagent and percent yield
Ammonia is made from 28.0 kg of nitrogen and 5.00 kg of hydrogen. Find the theoretical yield, the limiting reagent, and the percent yield if 25.0 kg of ammonia is obtained.
Solution
\text{N}_2+3\text{H}_2 \to 2\text{NH}_3
Moles:
n_{\text{N}_2} = \frac{28000}{28.02} = 999.3\ \text{mol}
n_{\text{H}_2} = \frac{5000}{2.016} = 2480\ \text{mol}
Required ratio 1:3. With 999.3 mol of N₂ you need 2998 mol of H₂ and you have 2480. Hydrogen is limiting.
Theoretical yield from hydrogen:
n_{\text{NH}_3} = 2480\times\frac{2}{3} = 1653\ \text{mol}
m = 1653\times17.03 = 28{,}150\ \text{g} = 28.2\ \text{kg}
Percent yield:
\frac{25.0}{28.2}\times100 = 88.7\ \%
What to notice. Nitrogen left over: 999.3-826.7 = 172.6 mol, or 4.84 kg. In a real plant that is recycled, along with the unreacted hydrogen, which is why the overall conversion approaches 98 % despite the single-pass equilibrium yield being only about 30 % (Chapter 10.4).
The recycling is what makes the process economic, and it is a direct application of Le Chatelier: removing the ammonia and returning the reactants keeps Q permanently below K.
Problem 6 — Equilibrium calculation
At 500 K, K_c = 0.500 for \text{PCl}_5 \rightleftharpoons \text{PCl}_3+\text{Cl}_2. Starting with 2.00 M PCl₅, find the equilibrium concentrations.
Solution
Let x be the amount that dissociates:
| PCl₅ | PCl₃ | Cl₂ | |
|---|---|---|---|
| Initial | 2.00 | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium | 2.00-x | x | x |
K_c = \frac{x^2}{2.00-x} = 0.500
x^2 = 1.00-0.500x
x^2+0.500x-1.00 = 0
x = \frac{-0.500\pm\sqrt{0.250+4.00}}{2} = \frac{-0.500\pm2.062}{2}
Taking the positive root:
x = 0.781
[\text{PCl}_5] = 1.22\ \text{M}, \qquad [\text{PCl}_3] = [\text{Cl}_2] = 0.781\ \text{M}
Check:
\frac{(0.781)^2}{1.22} = \frac{0.610}{1.22} = 0.500\ \checkmark
What to notice. The small-x approximation would give x = \sqrt{1.00} = 1.00, which is 28 % off — and 50 % of the initial concentration, far above the 5 % threshold where the approximation is valid.
Always check the approximation. The rule of thumb is that it holds when [\text{initial}]/K > 400; here the ratio is 4, so the quadratic is mandatory.
Problem 7 — Buffer preparation
How many grams of sodium acetate must be added to 500 mL of 0.200 M acetic acid to make a buffer of pH 5.00? (K_a = 1.8\times10^{-5}, M_{\text{NaOAc}} = 82.03)
Solution
\text{p}K_a = -\log(1.8\times10^{-5}) = 4.74
5.00 = 4.74+\log\frac{[\text{A}^-]}{[\text{HA}]}
\log\frac{[\text{A}^-]}{[\text{HA}]} = 0.26 \quad\Longrightarrow\quad \frac{[\text{A}^-]}{[\text{HA}]} = 1.82
Moles of acid:
n_{\text{HA}} = (0.500)(0.200) = 0.100\ \text{mol}
Moles of acetate needed:
n_{\text{A}^-} = 1.82\times0.100 = 0.182\ \text{mol}
m = 0.182\times82.03 = 14.9\ \text{g}
What to notice. The Henderson–Hasselbalch equation uses the ratio, so the volume cancels and you can work in moles directly. This is why diluting a buffer does not change its pH — both concentrations fall by the same factor.
What dilution does change is the buffer capacity. A buffer made from 0.100 mol of each resists ten times more added acid than one made from 0.010 mol of each, at the same pH. Capacity depends on absolute amounts; pH depends on the ratio.
Problem 8 — Cell voltage and free energy
For the cell \text{Zn}|\text{Zn}^{2+}(0.10\ \text{M})||\text{Cu}^{2+}(1.0\ \text{M})|\text{Cu}, find the cell voltage and \Delta G. Standard potentials: Cu²⁺/Cu +0.34 V, Zn²⁺/Zn -0.76 V.
Solution
E^\circ = 0.34-(-0.76) = 1.10\ \text{V}
The cell reaction:
\text{Zn}+\text{Cu}^{2+} \to \text{Zn}^{2+}+\text{Cu}, \qquad n = 2
Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{0.10}{1.0} = 0.10
Nernst:
E = 1.10-\frac{0.0592}{2}\log(0.10) = 1.10-(0.0296)(-1) = 1.13\ \text{V}
Free energy:
\Delta G = -nFE = -(2)(96485)(1.13) = -218{,}000\ \text{J/mol} = -218\ \text{kJ/mol}
What to notice. Lowering the product concentration by a factor of ten raised the voltage by only 30 mV. The Nernst correction is logarithmic and therefore small unless the concentrations differ by many orders of magnitude.
This is why batteries hold a nearly constant voltage through most of their discharge, and then fall off sharply at the end when a reactant runs out and Q changes by orders of magnitude.
And the equilibrium constant follows from E^\circ:
\log K = \frac{nE^\circ}{0.0592} = \frac{2\times1.10}{0.0592} = 37.2 \quad\Longrightarrow\quad K = 1.6\times10^{37}
Effectively complete. Zinc metal and copper ions cannot coexist.
Problem 9 — Gibbs free energy and spontaneity
For \text{CaCO}_3(s) \to \text{CaO}(s)+\text{CO}_2(g), \Delta H^\circ = +178.3 kJ/mol and \Delta S^\circ = +160.6 J/mol/K. Find the temperature above which it becomes spontaneous, and K at 1200 K.
Solution
At the crossover, \Delta G = 0:
T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{178300}{160.6} = 1110\ \text{K} = 837\ ^\circ\text{C}
At 1200 K:
\Delta G^\circ = 178300-(1200)(160.6) = 178300-192720 = -14{,}420\ \text{J/mol}
\ln K = -\frac{\Delta G^\circ}{RT} = \frac{14420}{(8.314)(1200)} = \frac{14420}{9977} = 1.445
K = e^{1.445} = 4.24
Since K_p = P_{\text{CO}_2} for this reaction (solids omitted), the equilibrium CO₂ pressure is 4.24 bar.
What to notice. This is lime burning, one of the oldest chemical processes there is, and it has been done in kilns since antiquity at exactly the temperature this calculation gives — around 900 °C, comfortably above the 837 °C threshold.
The reaction is entropy-driven. It is strongly endothermic and happens only because a gas is produced from a solid, and T\Delta S eventually beats \Delta H.
And the CO₂ released is a serious climate problem. Cement production accounts for about 8 % of global CO₂ emissions, and roughly half of that is this reaction rather than the fuel used to heat the kiln. It is chemically unavoidable as long as cement is made from limestone, which is why cement decarbonisation is so difficult.
Problem 10 — Activation energy from rate data
A reaction has rate constants k = 2.5\times10^{-4} s⁻¹ at 300 K and 3.8\times10^{-3} s⁻¹ at 340 K. Find E_a and the pre-exponential factor.
Solution
\ln\frac{k_2}{k_1} = -\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)
\ln\frac{3.8\times10^{-3}}{2.5\times10^{-4}} = \ln(15.2) = 2.721
\frac{1}{340}-\frac{1}{300} = 2.9412\times10^{-3}-3.3333\times10^{-3} = -3.922\times10^{-4}
2.721 = -\frac{E_a}{8.314}\times(-3.922\times10^{-4})
E_a = \frac{2.721\times8.314}{3.922\times10^{-4}} = \frac{22.62}{3.922\times10^{-4}} = 5.77\times10^{4}\ \text{J/mol} = 57.7\ \text{kJ/mol}
Pre-exponential factor, from the 300 K point:
A = k\,e^{E_a/RT} = (2.5\times10^{-4})e^{57700/(8.314\times300)} = (2.5\times10^{-4})e^{23.13}
= (2.5\times10^{-4})(1.11\times10^{10}) = 2.8\times10^{6}\ \text{s}^{-1}
What to notice. E_a = 57.7 kJ/mol is close to the typical value that makes the "doubles every 10 °C" rule work (Chapter 10.6). Check it: over 10 K from 300 to 310,
\frac{k_{310}}{k_{300}} = e^{-\frac{57700}{8.314}\left(\frac{1}{310}-\frac{1}{300}\right)} = e^{6940\times1.075\times10^{-4}} = e^{0.746} = 2.11
Just over double. ✔
Two data points are the minimum, and in practice you measure k at five or six temperatures and take the slope of \ln k against 1/T, which averages out experimental scatter.
Problem 11 — Enthalpy from bond energies
Estimate \Delta H for the combustion of methane using bond energies: C–H 413, O=O 498, C=O 799, O–H 463 kJ/mol. Compare with the true value of -802 kJ/mol (gaseous water).
Solution
\text{CH}_4+2\text{O}_2 \to \text{CO}_2+2\text{H}_2\text{O}(g)
Bonds broken: 4 C–H, 2 O=O
4(413)+2(498) = 1652+996 = 2648\ \text{kJ}
Bonds formed: 2 C=O, 4 O–H
2(799)+4(463) = 1598+1852 = 3450\ \text{kJ}
\Delta H = 2648-3450 = -802\ \text{kJ/mol}
Exactly the accepted value.
What to notice. The agreement here is better than the method usually gives, because bond energies are averages over many compounds and methane is close to typical.
The comparison with liquid water matters. Chapter 10.6 gave -890 kJ/mol for combustion producing liquid water. The difference, 88 kJ/mol, is the condensation of two moles of steam at 44 kJ/mol each. This is exactly the difference between a boiler's "gross" and "net" calorific value, and a condensing boiler recovers it by cooling the exhaust below the dew point — which is why condensing boilers are about 10 % more efficient and why they produce liquid water that must be drained.
Bond energies also explain why hydrocarbons are good fuels. The bonds broken (C–H, O=O) are weaker than those formed (C=O, O–H), and the C=O bond in CO₂ at 799 kJ/mol is one of the strongest known. Carbon dioxide is the ash of an efficient fire, which is precisely why turning it back into fuel is so energetically expensive.
Problem 12 — ATP yield and efficiency
Glucose oxidation releases 2870 kJ/mol. If 32 ATP are made and ATP hydrolysis under cellular conditions releases 50 kJ/mol, find the efficiency. Then compute how much ATP a person turns over daily at 100 W.
Solution
Efficiency:
\eta = \frac{32\times50}{2870} = \frac{1600}{2870} = 55.7\ \%
Daily turnover. 100 W for 86,400 s:
E = 100\times86400 = 8.64\times10^{6}\ \text{J} = 8640\ \text{kJ}
Moles of ATP, at 50 kJ/mol:
n = \frac{8640}{50} = 173\ \text{mol}
Mass, with M_{\text{ATP}} = 507 g/mol:
m = 173\times507 = 87{,}700\ \text{g} = 88\ \text{kg}
What to notice. You make and consume about 88 kg of ATP a day, which is more than your body weight — and your body contains only about 250 g of it at any moment.
\text{Recycling rate} = \frac{88000}{250} = 350\ \text{times per day}
Each ATP molecule is used and remade roughly every four minutes.
And the 56 % efficiency is remarkable. A car engine manages 25–35 % and a power station 40 % (Chapter 3.4), and both are limited by Carnot. Biology is not a heat engine — the energy goes from chemical bonds to a proton gradient to ATP without ever passing through a thermal step — so the Carnot limit does not apply.
The rest of the energy does become heat, and that is why you produce roughly 100 W of it and why a room full of people warms up.