Chemical EquilibriumHard

Question

The value of equilibrium constant for the following reaction at 300 K and constant pressure is

A(g) + B(g) $\rightleftharpoons$C(g) + D(g) + E(g); ΔEo = 30 kcal and ΔSo = 100 cal/K.

Options

A.e
B.$\frac{1}{e}$
C.e2
D.$\frac{1}{e^{2}}$

Solution

$\Delta H^{o} = \Delta E^{o} + \Delta n_{g}.RT = ( + 30) + (3 - 2) \times \frac{2}{1000} \times 300 = + 30.6\text{ cal}$

Now, ΔG° – RT ⋅ ln Keq = ΔH° – T ⋅ ΔS°

or, – 2 × 300 × ln Keq = 30.6 × 103 – 300 × 100

$\therefore K_{eq} = \frac{1}{e}$

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