Edexcel IAL Chemistry 12.9 Entropy and reaction feasibility

Use ΔStotal > 0 as the criterion, set total entropy to zero at the threshold and rearrange with compatible ΔH and ΔSsystem units before interpreting above or below it.

Exam points

State feasibility at a given temperature from the sign of total entropy change.
Calculate a threshold temperature by setting ΔStotal to zero and rearranging H/T.
Interpret how temperature changes the surroundings contribution and feasibility range.

[Maximum number: 4]

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 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 }}

[ 1 ]

(i)

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)

(b)

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)

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

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12.9—The feasibility of a reaction depends on: i the balance between ∆Ssystem and ∆Ssurroundings, so that even endothermic question 1

Question (a)

Question (i)

Question (b)

Question (i)