Edexcel IAL Chemistry Topic 13 Chemical equilibria

Start from the balanced equation, construct the correct concentration or partial-pressure expression and build an equilibrium table before explaining what can change K.

Exam points

Write Kc or Kp expressions with stoichiometric powers and omitted pure solids or liquids.
Calculate equilibrium concentrations or partial pressures and constants with appropriate units.
Distinguish changes to composition from the temperature-dependent value of the equilibrium constant.

Covered syllabus points

13.1—Deduce an expression for Kc , for homogeneous and heterogeneous systems, in terms of equilibrium concentrations 13.2—Deduce an expression for Kp for homogeneous and heterogeneous systems, in terms of equilibrium partial pressures in atm 13.3—A value, with units where appropriate, for the equilibrium constants (Kc and Kp) for homogeneous and heterogeneous 13.4—How, if at all, a change in temperature, pressure or the presence of a catalyst affects the equilibrium composition in 13.5—The value of the equilibrium constant is not affected by changes in concentration or pressure or by the addition of 13.6—The effect of changing the temperature on the equilibrium constant (Kc and Kp) for both exothermic and endothermic reactions 13.7—The effect of temperature on the position of equilibrium is explained using a change in the value of the equilibrium 13.8—The effect of a change in temperature on: i the value of ∆Stotal ii the magnitude of the equilibrium constant, since ∆Stotal 13.9—Apply knowledge of the value of equilibrium constants to predict the extent to which a reaction takes place

[Maximum number: 2]

The equation for the reaction between iron and steam is shown.

3 \mathrm{Fe}(\mathrm{~s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{~s})+4 \mathrm{H}_{2}(\mathrm{~g}) \quad \Delta H=-151.2 \mathrm{~kJ} \mathrm{~mol}^{-1}

(a)

What is the equilibrium constant expression for this reaction?

$K_{\mathrm{c}}=\frac{\left[\mathrm{Fe}_{3} \mathrm{O}_{4}\right] \times\left[\mathrm{H}_{2}\right]^{4}}{[\mathrm{Fe}]^{3} \times\left[\mathrm{H}_{2} \mathrm{O}\right]^{4}}$

$K_{\mathrm{c}}=\frac{\left[\mathrm{Fe}_{3} \mathrm{O}_{4}\right] \times\left[\mathrm{H}_{2}\right]^{4}}{\left[\mathrm{H}_{2} \mathrm{O}\right]^{4}}$ □

$K_{\mathrm{c}}=\frac{\left[\mathrm{H}_{2}\right]^{4}}{[\mathrm{Fe}]^{3} \times\left[\mathrm{H}_{2} \mathrm{O}\right]^{4}}$ □

$K_{\mathrm{c}}=\frac{\left[\mathrm{H}_{2}\right]^{4}}{\left[\mathrm{H}_{2} \mathrm{O}\right]^{4}}$

[ 1 ]

(b)

What is the effect on the equilibrium constant for this reaction if small pieces of iron are replaced by iron powder and if the temperature is increased?
□ A
□ B
□ C
□ D

Table for Question (b) — Edexcel A-Level Chemistry A2

[ 1 ]

The only correct answer is D ( $K_{\mathrm{c}}$ unchanged; $K_{\mathrm{c}}$ decreases)

A is incorrect because $K_{\mathrm{c}}$ is not affected by the state of division of the solid and $K_{c}$ decreases for an exothermic reaction

B is incorrect because $K_{c}$ decreases for an exothermic reaction

C is incorrect because $K_{\mathrm{c}}$ is not affected by the state of division of the solid

[Maximum number: 8]

This question is about sulfuric acid and its salts.

(a)

The manufacture of sulfuric acid involves the equilibrium

2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H=-197 \mathrm{~kJ} \mathrm{~mol}^{-1}

[ 8 ]

(i)

A catalyst of vanadium(V) oxide is used in this reaction.

State the effect, if any, of the catalyst on the value of the equilibrium constant, $K_{p}$ .

[ 1 ]

(ii)

The temperature used for this reaction in industry is 700 K .

Explain, in terms of the equilibrium constant and the equilibrium position, the effect of an increase in temperature on the equilibrium yield of sulfur trioxide.
(2)

[ 2 ]

An explanation that makes reference to the following points:

- (equilibrium) yield (of sulfur trioxide / $\mathrm{SO}_{3}$ / product) decreases

- the equilibrium constant $/ K_{\mathrm{p}} / K_{c} / K$ decreases (as temperature increases)

and because the (forward / right) reaction is exothermic / releases heat (energy) / $\Delta H$ is negative

(1)

Allow less sulfur trioxide / $\ce{SO3}$ / product forms Ignore equilibrium position shifts to the left Ignore more reactants formed

Allow the equilibrium constant $/ K_{\mathrm{p}} / K_{c} / K$ decreases (as temperature increases) and because the reverse / backward / left reaction is endothermic / absorbs heat (energy)

Allow K decreases because $\Delta S_{\text {surroundings }} / \Delta S_{\text {total }}$ decreases / becomes less positive (as temperature increases and assuming $\Delta S_{\text {system }}$ is constant)

Ignore reference to rate

(2)

(iii)

Write the expression for the equilibrium constant, $K_{p}$ , for this equilibrium. State symbols are not required.

[ 1 ]

expression for $K_{\mathrm{p}}$

Example of expression for $K_{\mathrm{p}}$ :
$\begin{aligned} K_{\mathrm{p}} & =\frac{p\left(\mathrm{SO}_{3}(\mathrm{~g})\right)^{2}}{p\left(\mathrm{SO}_{2}(\mathrm{~g})\right)^{2}(\mathrm{x}) p\left(\mathrm{O}_{2}(\mathrm{~g})\right)} \end{aligned}$

Allow P / P P / p p etc for partial pressure and this can be inside the brackets

Allow e.g. $p^{2} \mathrm{SO}_{3}$

Ignore missing (g) / brackets around formulae

Do not award square brackets

(1)

(iv)

A mixture of 2.00 mol of sulfur dioxide and 1.00 mol of oxygen is allowed to reach equilibrium at 5.00 atm pressure. 1.60 mol of sulfur trioxide is formed.

Calculate the value of $K_{p}$ .
Include units and give your answer to an appropriate number of significant figures.
(4)

[ 4 ]

calculation of eqm moles

- calculation or expressions for 3 partial pressures

(1)

- substitution of values into $K_{\mathrm{p}}$ expression

(1)

- calculation of $K_{\mathrm{p}}$ and answer to 2 / 3 SF and units

Example of calculation:

& $\mathrm{SO}_{2}$ & $\mathrm{O}_{2}$ & $\mathrm{SO}_{3}$

Initial mol & 2.00 & 1.00 & -

Eqm mol &
2.00-1.60

=0.40

1.00-0.80

=0.20

1.60

Total mol at

eqm

0.40+0.20+1.60=2.20

Partial pressure

/atm

$\frac{0.40 \times 5.00}{2.20}$

=0.90909

$\frac{0.20 \times 5.00}{2.20}$

=0.45455

$\frac{1.60 \times 5.00}{2.20}$

=3.6364

\end{tabular}

TE for partial pressures on eqm moles
$\begin{aligned} K_{\mathrm{p}} =\frac{3.6364^{2}}{0.90909^{2} \times 0.45455} =35.2 / 35 \mathrm{~atm}^{-1} \end{aligned}$

TE on expression for $K_{\mathrm{p}}$ in (a)(iii)

Allow answer from previous correct rounding to 2 or more SF e.g. 0.91, 0.45 and 3.6 gives 34.8

Penalise incorrect rounding once only e.g. 0.909 to 0.9

Allow fractions in working but not in final answer

Allow $\mathrm{atm}^{-}$ for units but do not allow any other units

Allow correct units written in (a)(iii) if not written here

Correct answer to 2 or 3 SF with units and no working scores (4)
\end{tabular}

(4)

[Maximum number: 7]

(a)

Ethyl propanoate, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3}$ , smells of pineapple and is used as a flavouring. It may be hydrolysed using hydrochloric acid as a catalyst to produce propanoic acid and ethanol.

\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\mathrm{l})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}(\mathrm{l})+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(\mathrm{l})

A mixture was prepared using 0.100 mol of ethyl propanoate and 0.200 mol of water containing the catalyst.
The mixture was left to reach equilibrium at $25^{\circ} \mathrm{C}$ .
The equilibrium mixture contained 0.0440 mol of propanoic acid.

[ 7 ]

(i)

Calculate the value for $K_{\mathrm{c}}$ for this equilibrium at $25^{\circ} \mathrm{C}$ .

Give your answer to an appropriate number of significant figures.

[ 4 ]

Example of calculation:

(4)

- expression for $K_{\mathrm{c}}$

(1)

$K_{\mathrm{c}}=\frac{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right]}{\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]}$

Ignore state symbols Brackets must be square brackets

- calculation of equilibrium quantities of reactants

(1)

$\begin{aligned} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3} =(0.100-0.0440) =0.056(0)(\mathrm{mol}) \mathrm{H}_{2} \mathrm{O}=(0.2-0.0440) =0.156(\mathrm{~mol}) \end{aligned}$

- calculation of equilibrium concentrations of
products

(1)

$\begin{aligned} {\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\right]=0.044(0) / \mathrm{V}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right)} {\left[\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right]=0.044(0) / \mathrm{V}\left(\mathrm{~mol} \mathrm{dm}^{-3}\right)} \end{aligned}$ $K_{\mathrm{c}}=\frac{0.044 \times 0.044}{0.056 \times 0.156}=0.22161=0.222 / 0.22$

(ii)

Deduce the effect of increasing the temperature on the total entropy change of this reaction, $\Delta S_{\text {total }}$ , and on the value of the equilibrium constant, $K_{c}$ . Assume that $\Delta S_{\text {system }}$ does not change with temperature.

[ 3 ]

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Covered syllabus points

Topic 13: Chemical Equilibria question 1

Question (a)

Question (b)

Topic 13: Chemical Equilibria question 2

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (iv)

Topic 13: Chemical Equilibria question 3

Question (a)

Question (i)

Question (ii)