Edexcel IAL Chemistry A2 Topic 13 Chemical Equilibria

[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?

A

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

B

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

C

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

D

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

[ 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?

Iron powder used

Temperature increased

$K_{c}$ increases

$K_{c}$ unchanged

$K_{c}$ increases

$K_{c}$ decreases

$K_{c}$ unchanged

$K_{c}$ decreases

[ 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

(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

(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}

[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:

- 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 ]

[Maximum number: 1]

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}

What is the equilibrium constant expression for this reaction?

A

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

B

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

C

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

D

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

[Maximum number: 1]

This question is about sulfuric acid and its salts.

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}

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

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

[Maximum number: 4]

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.

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.

Example of calculation:

- 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)

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

[Maximum number: 3]

Ammonia is manufactured in the Haber Process.

\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})

In this process, pressures between 100 atm and 300 atm and temperatures between 675 K and 725 K are usually used.

The pressure of the system is doubled at constant temperature.

Explain the effect of this on the yield of ammonia by considering the change in the $K_{\mathrm{p}}$ expression and hence the position of equilibrium.

[Maximum number: 1]

This question is about sulfuric acid and its salts.

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}

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, $K_{p}$ .

[Maximum number: 1]

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}

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?

Iron powder used

Temperature increased

$K_{c}$ increases

$K_{c}$ unchanged

$K_{c}$ increases

$K_{c}$ decreases

$K_{c}$ unchanged

$K_{c}$ decreases

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: 1]

At $25^{\circ} \mathrm{C}$ , the pH of pure water is 7.0 and at $75^{\circ} \mathrm{C}$ the pH is 6.4 .
Why is the pH of pure water lower at $75^{\circ} \mathrm{C}$ than at $25^{\circ} \mathrm{C}$ ?

A

there are more hydrogen ions than hydroxide ions when the water is at $75^{\circ} \mathrm{C}$

B

the dissociation of water is endothermic so the concentration of hydrogen ions is lower at $75^{\circ} \mathrm{C}$

C

the dissociation of water is endothermic so the concentration of hydrogen ions is higher at $75^{\circ} \mathrm{C}$

D

the dissociation of water is exothermic so the concentration of hydrogen ions is higher at $75^{\circ} \mathrm{C}$

[Maximum number: 3]

The equation for the formation of ammonia in the Haber Process is shown

1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+1 \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})

(a)

The industrial synthesis of ammonia

1 / 2 \mathrm{~N}_{2}(\mathrm{~g})+11 / 2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{~g})

is carried out at pressures of about 20000 kPa and temperatures between 700 K and 750 K . These temperatures are higher than the answer to (c)(iii).

[ 3 ]

(i)

State the relationship between the total entropy, $\Delta S_{\text {total }}$ , and the equilibrium constant, K.

[ 1 ]

total entropy, $\Delta S=R \ln K$
or
$\ln K=\Delta S / R$
or
$\mathrm{K}=\mathrm{e}^{\frac{\Delta S}{R}}$

(ii)

Calculate the value of the equilibrium constant K at 750 K.

[ 2 ]

calculation of $\ln K$

- evaluation of K

Example of calculation
$-37.7=8.31 \times \ln K$ $\ln K=-4.5367$ $K=0.01071 / 1.071 \times 10^{-2}$

Final answer with no working scores (2)

Allow TE on M1 to M2

No TE on incorrect expression

Ignore units Ignore SF except 1 SF

[Maximum number: 2]

This question is about the reaction between nitrogen monoxide and oxygen.

\mathrm{NO}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{~g})

The equilibrium constant, $K_{p}$ , for the reaction at 298 K is $1.55 \times 10^{6} \mathrm{~atm}^{-1 / 2}$ .

State what this value of the equilibrium constant indicates about the position of the equilibrium. Justify your answer.

Edexcel A-Level Chemistry A2 13 Chemical Equilibria Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (iv)

Question 3

Question (a)

Question (i)

Question (ii)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question (a)

Question (i)

Question (ii)

Question 12