ThermodynamicsHard

Question

The following reaction is at equilibrium at 298 K 2NO (g, 0.00001 bar) + Cl2 (g, 0.01 bar) $\rightleftharpoons$­ 2NOCl (g, 0.01 bar). ΔGo for the reaction is

Options

A.−45.65 kJ
B.−28.53 kJ
C.−22.82 kJ
D.−57.06 kJ

Solution

$\Delta G^{o} = - RT.\ln K_{P}^{o} = - RT.\ln\frac{P_{NOCl}^{o}}{P_{NO}^{2}.P_{Cl_{2}}} = - 8.314 \times 298 \times \ln\frac{(0.01)^{2}}{(0.00001)^{2} \times 0.01} = + 45.65\text{ kJ}$

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