Edexcel IAL Chemistry 13.8 Total entropy and equilibrium constants

Preserve natural logarithms and compatible units, infer K above or below one from the entropy sign and track temperature through ΔSsurroundings when required.

Exam points

Apply ΔStotal = R ln K and relate the sign of total entropy to whether K exceeds one.
Calculate K or total entropy with natural logarithms, R and consistent units.
Explain temperature effects on total entropy and therefore on equilibrium-constant magnitude.

[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

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13.8—The effect of a change in temperature on: i the value of ∆Stotal ii the magnitude of the equilibrium constant, since ∆Stotal question 1

Question (a)

Question (i)

Question (ii)