Topic 13

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Everything from this topic in one place, for the last read before an exam. If anything here is unfamiliar, follow the link back to the full page.

Constants and conversions

Quantity	Value
gas constant, R	8.31 J mol⁻¹ K⁻¹
temperature	K = °C + 273
pressure for $K_\text{p}$	atm
entropy	J K⁻¹ mol⁻¹
enthalpy	kJ mol⁻¹, converted to J before dividing by T

The expressions

To find	Use	Page
$K_\text{c}$	products over reactants, each [X] to the power of its coefficient, equilibrium concentrations in mol dm⁻³	Writing Kc
mole fraction	$x_\text{A}$ = moles of A ÷ total moles of gas	Partial pressures and Kp
partial pressure	$p_\text{A} = x_\text{A} \times P_\text{total}$ , in atm	Partial pressures and Kp
$K_\text{p}$	the same shape as $K_\text{c}$ , with equilibrium partial pressures in atm	Partial pressures and Kp
heterogeneous	leave out solids and pure liquids; $K_\text{p}$ has gases only	Heterogeneous systems
units	power = (sum of coefficients on top) − (sum on the bottom); zero means no units	Units of K
reverse reaction	$K$ becomes $1/K$	Writing Kc
ln K	$\Delta S_\text{total} = R \ln K$ , so $K = e^{\Delta S_\text{total}/R}$	Entropy and K
total entropy	$\Delta S_\text{total} = \Delta S_\text{system} - \Delta H/T$	Entropy and K

Changing conditions

Change	Position of equilibrium	Value of K
add a reactant	moves right	unchanged
remove a product	moves right	unchanged
raise the pressure	towards the side with fewer moles of gas; no effect if equal	unchanged
add a catalyst	no effect; equilibrium is reached sooner	unchanged
add an inert gas at constant volume	no effect	unchanged
raise the temperature, exothermic	moves left	decreases
raise the temperature, endothermic	moves right	increases

The reasons are on What changes K, Pressure and catalysts and Temperature changes K.

Definitions and statements to learn

Homogeneous: every species is in the same phase.
Heterogeneous: species are in more than one phase.
Mole fraction: moles of the gas divided by the total moles of gas.
Partial pressure: mole fraction multiplied by the total pressure.
The value of $K$ is not affected by a change in concentration, a change in pressure or a catalyst. It changes only with temperature.
A catalyst speeds up the forward and backward reactions equally, so the composition and $K$ are unchanged.
Temperature changes the value of $K$ , and the position moves to match the new value.
Large $K$ : products favoured. Small $K$ : reactants favoured. $K$ says how far, not how fast.

Methods

$K_\text{c}$ from amounts: table of start, change, equilibrium (moles) → divide by volume → substitute.
$K_\text{p}$ from amounts: table → total moles of gas → mole fractions → × total pressure → substitute.
Titration: moles of NaOH → scale to the whole mixture → subtract the blank → ratio from the equation → substitute.
Temperature question: exothermic or endothermic → $K$ rises or falls → position moves to match.
$K$ from entropy: $\Delta S_\text{total}$ in J → ÷ 8.31 → ln K → $e^x$ .

The ten traps

Adding the terms instead of multiplying, or writing [2NH₃] instead of [NH₃]².
Including a solid or a pure liquid in an expression, or [ ] in $K_\text{p}$ .
Using the starting amounts instead of the equilibrium ones.
Forgetting to divide by the volume when the volume does not cancel.
Writing 0.20 mol of NO₂ when 0.20 mol of N₂O₄ dissociates: the ratio is 1 : 2.
Mole fractions that do not add up to 1, or a partial pressure bigger than the total pressure.
Forgetting to subtract the blank in the titration.
Saying " $K$ changes" for a pressure, concentration or catalyst change.
Saying "a catalyst increases the yield", or "a large $K$ means a fast reaction".
Using kJ in $R \ln K$ without converting to J.

Mixed test

Try these with no notes, then check the answers.

Write $K_\text{c}$ and $K_\text{p}$ , with units, for $\ce{C(s) + H2O(g) <=> CO(g) + H2(g)}$ .
0.500 mol of PCl₅ is heated in a 2.50 dm³ vessel. At equilibrium 0.100 mol of PCl₅ remains. Calculate $K_\text{c}$ for $\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$ , with units.
1.00 mol of PCl₅ is 25.0% dissociated at a total pressure of 2.00 atm. Calculate $K_\text{p}$ , with units.
0.0400 mol of ethyl ethanoate and 0.200 mol of water are left to reach equilibrium with an acid catalyst. Titration shows that the whole mixture needs 0.0340 mol of NaOH, and the catalyst alone needs 0.0040 mol. Calculate $K_\text{c}$ for $\ce{CH3COOC2H5(l) + H2O(l) <=> CH3COOH(l) + C2H5OH(l)}$ .
$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$ has $\Delta H$ = −196 kJ mol⁻¹. State the effect on $K_\text{p}$ and on the yield of SO₃ of (a) raising the pressure, (b) raising the temperature, (c) adding a catalyst.
Use $\Delta S_\text{total} = R \ln K$ to explain why $K_\text{p}$ decreases when the temperature rises for an exothermic reaction.
A reaction has $\Delta S_\text{total}$ = +80 J K⁻¹ mol⁻¹ at 298 K. Calculate K and predict the extent.
A student says "K = 1 × 10¹² means the reaction is fast". Comment.

Answers cover these until you have tried the questions

$K_\text{c} = \dfrac{[\ce{CO}][\ce{H2}]}{[\ce{H2O}]}$ , units mol dm⁻³. $K_\text{p} = \dfrac{p_{\ce{CO}}\,p_{\ce{H2}}}{p_{\ce{H2O}}}$ , units atm. Carbon is a solid, so it is left out. (Top 2, bottom 1, power +1.)
PCl₅ used = 0.500 − 0.100 = 0.400 mol, so PCl₃ = Cl₂ = 0.400 mol. Concentrations: PCl₅ 0.100 ÷ 2.50 = 0.0400; PCl₃ and Cl₂ 0.400 ÷ 2.50 = 0.160 mol dm⁻³. $K_\text{c}$ = 0.160² ÷ 0.0400 = 0.640 mol dm⁻³
Dissociated = 0.250 mol. Equilibrium: PCl₅ = 0.750 mol, PCl₃ = Cl₂ = 0.250 mol, total = 1.250 mol. $p_{\ce{PCl5}}$ = (0.750 ÷ 1.250) × 2.00 = 1.20 atm. $p_{\ce{PCl3}}$ = $p_{\ce{Cl2}}$ = (0.250 ÷ 1.250) × 2.00 = 0.400 atm. $K_\text{p}$ = 0.400² ÷ 1.20 = 0.133 atm
Ethanoic acid = 0.0340 − 0.0040 = 0.0300 mol = ethanol. Ester = 0.0400 − 0.0300 = 0.0100 mol. Water = 0.200 − 0.0300 = 0.1700 mol. $K_\text{c}$ = (0.0300 × 0.0300) ÷ (0.0100 × 0.1700) = 0.529, no units.
(a) $K_\text{p}$ unchanged; the yield of SO₃ increases, because the mixture moves towards fewer moles of gas (3 on the left, 2 on the right). (b) Exothermic, so $K_\text{p}$ decreases; the mixture moves left and the yield falls. (c) $K_\text{p}$ unchanged and the yield unchanged; equilibrium is reached sooner.
For an exothermic reaction $\Delta S_\text{surroundings}$ = −ΔH/T is positive. As T rises it gets smaller. $\Delta S_\text{system}$ is constant, so $\Delta S_\text{total}$ decreases. Since $\Delta S_\text{total}$ = R ln K, ln K decreases, so $K_\text{p}$ decreases.
ln K = 80 ÷ 8.31 = 9.63, so K = e^9.63 = 1.5 × 10⁴. K is much greater than 1, so products are strongly favoured. It is far below 10¹⁰, so a measurable amount of reactant remains: the reaction is not complete.
A large K shows that the equilibrium lies far to the right. It says nothing about the rate, which depends on the activation energy. A reaction with K = 10¹² can be too slow to see.