Edexcel IAL Chemistry Unit 4 Rates and Equilibria

[Maximum number: 16]

A group of students investigated the kinetics of a 'clock' reaction.
The reaction investigated was that between hydrogen peroxide and iodide ions in the presence of acid.

$Reaction $1 \quad \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{I}_{2}(\mathrm{aq})$$

In this 'clock' reaction, a fixed volume of aqueous sodium thiosulfate, $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$ , and a small amount of starch were added to the reaction mixture.

The added thiosulfate ions react with the iodine produced in Reaction 1.

$Reaction $22 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{aq}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})$$

When all the thiosulfate ions have reacted, the presence of iodine is detected by the formation of a starch-iodine complex. The students recorded the time taken for this complex to form.

(a)

State the final colour of the mixture containing the starch-iodine complex.

[ 1 ]

(b)

Under appropriate conditions, the reciprocal of time can be used as an approximate measure of the initial rate of the reaction.

Explain why the concentration of the sodium thiosulfate must be low compared with the initial concentrations of the other reagents.

[ 2 ]

(c)

Four reaction mixtures, with different initial concentrations of hydrogen peroxide, hydrogen ions and iodide ions, were prepared.

Each mixture had the same volume and contained the same amount of sodium thiosulfate and starch.

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

[ 6 ]

(i)

Use the results in the table to deduce the order of Reaction 1 with respect to hydrogen peroxide, hydrogen ions and iodide ions. Justify each answer by referring to relevant data from the table.

Hydrogen peroxide

Hydrogen ions
lodide ions

[ 3 ]

(ii)

Write the overall rate equation for Reaction 1 using your answers to (b)(i).

[ 1 ]

An answer that makes reference to the following point:

- rate $=k\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\left[I^{-}\right]$

Allow rate $=k\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]\left[\left[^{-}\right]\left[\mathrm{H}^{+}\right]^{\circ}\right.$

Allow species in any order TE from (b)(i)

(iii)

Calculate a value for the rate constant of Reaction 1 using data from Mixture 1 and your answers to (b)(ii) and (b)(iv).
Include the units of the rate constant.

[ 2 ]

rearrangement of the rate equation and calculation of the rate constant

Example of calculation

$k=\operatorname{rate} \div\left(\left[\mathrm{H}_{2} \mathrm{O}_{2}\right] \times\left[\mathrm{I}^{-}\right]\right)$ $k=4.359 \times 10^{-6} \div\left(5.4 \times 10^{-2} \times 8.2 \times 10^{-3}\right)$ $=4.359 \times 10^{-6} \div 4.428 \times 10^{-4}$ $=9.844 \times 10^{-3}$

Ignore SF except 1SF

(d)

The activation energy for Reaction 1 may be found by repeating the experiment at different temperatures.
Each student carried out an experiment at a different temperature.
One of the students misread the thermometer in their experiment.

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

The activation energy, $E_{\mathrm{a}}$ , for a reaction may be found by plotting a graph of In rate against 1 / T.
The gradient of the resulting line of best fit can be used in the Arrhenius equation to determine a value for $E_{\mathrm{a}}$ , in $\mathrm{kJ} \mathrm{mol}^{-1}$ .

[ 7 ]

(i)

Determine the value for $E_{\mathrm{a}}$ for Reaction 1 by plotting a graph using the axes provided.
You should take into account the error made by one of the students.

\text { In rate }=-\frac{E_{\mathrm{a}}}{R} \times \frac{1}{T}+\text { constant } \quad R=8.31 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}

Figure for Question (i) — Edexcel A-Level Chemistry A2

$E_{a}$ $\mathrm{kJ} \mathrm{mol}^{-1}$

[ 5 ]

axes correct with scales allowing plot to cover at least half the grid and $\ln$ rate values becoming more negative down the axis

(1)

- both axes labelled with units $\mathrm{K}^{-1}$ on x axis

(1)

- points correctly plotted and best fit line drawn

(1)

- gradient in the range -7200 to -6800

(1)

- $\mathrm{E}_{\mathrm{a}}=-(-7000 \times 8.31) \div 1000$ $=+58.2\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)$

range 56.5 to 59.9 TE on incorrect gradient value but Ea

(1)

Example of calculation

Allow horizontal axis at bottom of graph Ignore extrapolation

Figure for Answer to Question (i) — Edexcel A-Level Chemistry A2

Allow plotting of all points except the incorrect one

TE on incorrect gradient

Ignore SF

(ii)

The students all used thermometers capable of reading to the same precision. Use your graph to deduce the temperature that the student who made the error should have read on their thermometer.

[ 2 ]

[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?
□ 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: 16]

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)

At 298 K the standard entropy change of the system, $\Delta S_{\text {system }}^{\ominus}=-98 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ .

Calculate the standard entropy of one mole of ammonia.
Use the value of $\Delta S_{\text {system }}^{\ominus}$ and the data in the table.

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

[ 2 ]

states or uses equation

- calculate $S^{\circ}$ products

(1)

(1)

$\begin{aligned} \Delta S_{\text {system }}^{\ominus} & =S_{\text {products }}^{\ominus}-S_{\text {reactants }}^{\ominus} -98.0 & =S_{\text {products }}^{\ominus}-((0.5 \times 192)+(1.5 \times 131)) S_{\text {products }}^{\ominus} & =292.5-98 S_{\text {products }}^{\ominus} & =(+) 194.5 / 195\left(\mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \end{aligned}$

If units are given they must be correct

Allow TE on incorrect $S^{\ominus}$ reactants

Comment

Correct answer with no working scores (2)
$\begin{aligned} S_{\text {products }}^{\ominus} & =63.5 \text { scores max (1) } S_{\text {products }}^{\ominus} & =225 \text { scores max (1) } \end{aligned}$

(b)

The relationship between $\Delta S_{\text {total }}$ and 1 / T can be found by combining the two equations:

\Delta S_{\text {total }}=\Delta S_{\text {surroundings }}+\Delta S_{\text {system }}

and $\Delta S_{\text {surroundings }}=-\Delta H / T$
to give

\Delta S_{\text {total }}=-\Delta H / T+\Delta S_{\text {system }}

[ 2 ]

(i)

Identify the thermodynamic quantity that can be obtained from this gradient.

[ 1 ]

An answer that makes reference to the following points:

- enthalpy change of reaction / $\Delta_{\mathrm{r}} H$ (of the Haber process)

Allow $-_{\mathrm{r}} H$

Allow enthalpy change / $\Delta H /-\Delta H$

(ii)

Determine the temperature at which the reaction ceases to be thermodynamically feasible at a pressure of 100 kPa .

[ 1 ]

An answer that makes reference to the following points:

- value of T found either by reading from the graph the value of T when $\Delta S_{\text {total }}=0$

or

by calculation

460 (K)

Allow an answer between 440-480

$=\frac{\text { answer to (c)(i) }}{98}$ $=\frac{45150}{98}=460.71 / 460(\mathrm{~K})$

Or

$=\frac{- \text { answer to (b) }}{-98}$ $=\frac{-45150}{-98}=460.71 / 460(\mathrm{~K})$

ALLOW TE on graph or on answer to (c)(i)

(c)

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).

[ 6 ]

(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

(iii)

Explain why $\Delta S_{\text {total }}$ decreases with an increase in temperature.

[ 3 ]

An answer that makes reference to the following points:

Either

- ( $\Delta S_{\text {total }}$ decreases because) $\Delta S_{\text {system }}$ (and $\Delta H$ ) do not change with temperature (significantly)

- therefore $\Delta S_{\text {surroundings }}$ must decrease (so that ( $\Delta S_{\text {total }}$ decreases)

- this is because $\Delta S_{\text {surroundings }}=-\Delta H / T$ (so as T increases $-\Delta H / T$ becomes less positive because $\Delta H$ is exothermic)

Or

- the reaction is exothermic and

so increasing temperature shits the equilibrium to the left / towards the reactants

- the value of K decreases

- because $\Delta S_{\text {total }}$ is proportional to K /

$S_{\text {total }}=R \ln K$ the value of $\Delta S_{\text {total }}$ decreases

(d)

Ammonia from the Haber Process reacts with acids.

With phosphoric acid, $\mathrm{H}_{3} \mathrm{PO}_{4}$ , a number of products are formed in solution. One of these is the fertiliser diammonium hydrogenphosphate.

[ 6 ]

(i)

Write an equation for the production of this fertiliser.

State symbols are not required.

[ 2 ]

formula of diammonium hydrogenphosphate

(1)

- balanced equation

(1)

$\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}$ $2 \mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}$

Allow multiples

Allow ions for the product

Allow for M2
$\mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow\left(\mathrm{NH}_{4}\right) \mathrm{H}_{2} \mathrm{PO}_{4}$

Allow ions for the product

No other TE

Ignore state symbols even if incorrect

(ii)

Write an ionic equation to show that ammonium ions are acidic in aqueous solution.
State symbols are not required.

[ 1 ]

$\mathrm{NH}_{4}^{+} \rightleftharpoons \mathrm{NH}_{3}+\mathrm{H}^{+}$

OR

- $\mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+}$

Allow → instead of ⇌

Do not award reactions reversed

Allow
$\begin{aligned} & \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} & \text { Allow } \rightleftharpoons \text { instead of } \rightarrow \end{aligned}$

Ignore state symbols even if incorrect

(iii)

A solution containing both ammonia and ammonium ions acts as a buffer. Explain, using a relevant ionic equation, the effect of adding a small amount of acid to this buffer.

[ 3 ]

An answer that makes reference to the following points:

- the mixture contains a large amount/ (large) reservoir of both ammonium ions and ammonia (1) / of $\mathrm{NH}_{4}^{+}$ and $\mathrm{NH}_{3}$

Either

- added $\mathrm{H}^{+}$ reacts with ammonia to form ammonium ions $/ \mathrm{H}^{+}+\mathrm{NH}_{3} \rightleftharpoons \mathrm{NH}_{4}^{+}$

Or

- added $\mathrm{H}^{+}$ combines with $\mathrm{OH}^{-}$ ions in water to form water $/ \mathrm{H}^{+}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}$

And ammonia reacts with water to produce $\mathrm{OH}^{-} \quad$ (1) ions $/ \mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}$

- ratio of ammonium ions to ammonia hardly (1) changes

Do not award incorrect formulae such as $\mathrm{NH}_{3}{ }^{-}$ in M1 and M2 but allow TE in M3 Ignore comments about acid / base in relation to $\mathrm{NH}_{4}^{+}$ / $\mathrm{NH}_{3}$ unless defined

Allow → instead of ⇌ Allow $\mathrm{H}_{3} \mathrm{O}^{+}$

Allow → instead of ⇌

This marking point must include at least one ionic equation

Allow remains constant

Allow pH is unchanged / changes very little because added $\mathrm{H}^{+}$ removed and change in concentration of $\mathrm{NH}_{3}$ and $\mathrm{NH}_{4}^{+}$ is small

[Maximum number: 4]

Ethyl butanoate reacts with water.

\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}

A student investigated the kinetics of this reaction.

Procedure

Step 150 cm^3 of ethyl butanoate was heated under reflux with a very large excess of water.

Step 2 After 30 minutes a sample of known volume was removed from the reaction mixture and placed in a conical flask.

Step 3 This sample was titrated using aqueous sodium hydroxide solution.
Step 4 Further samples were removed from the refluxing mixture at 30-minute intervals. Each sample was titrated until the titres of three consecutive samples were the same.

(a)

A graph of the results obtained by the student is shown.

The student plotted V_final -V_t on the y -axis against time on the x -axis.
V_final is the volume of the titre at 4.5 hours.
V_t is the volume of the titre at time t .

Figure for Question (a) — Edexcel A-Level Chemistry A2

[ 4 ]

(i)

Explain how the data collected indicates first order kinetics.

Show your working on the graph.

[ 2 ]

(ii)

The student concluded that the reaction is first order overall.

Explain whether or not this statement is valid.

[ 2 ]

Edexcel IAL Chemistry Unit 4 Rates, equilibria and further organic chemistry

Covered syllabus points

Covered syllabus points

Unit 4: Rates, Equilibria and Further Organic Chemistry question 1

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (iii)

Question (d)

Question (i)

Question (ii)

Unit 4: Rates, Equilibria and Further Organic Chemistry question 2

Question (a)

Question (b)

Unit 4: Rates, Equilibria and Further Organic Chemistry question 3

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question (i)

Question (ii)

Question (iii)

Question (d)

Question (i)

Question (ii)

Question (iii)

Unit 4: Rates, Equilibria and Further Organic Chemistry question 4

Question (a)

Question (i)

Question (ii)