Question

\text{For a certain thermochemical reaction } \text{M} \rightarrow \text{N at } T = 400 \, \text{K}, \, \Delta H^\circ = 77.2 \, \text{kJ mol}^{-1}, \, \Delta S = 122 \, \text{JK}^{-1}, \\\ \text{log equilibrium constant (log K) is } - \\_\\_\\_\\_\\_ \times 10^{-1}.

Answer 1

The standard Gibbs free energy change ($\Delta G^\circ$) is related to the enthalpy change ($\Delta H^\circ$) and entropy change ($\Delta S^\circ$) by the equation:

$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ.$

Substitute the given values:
$\Delta H^\circ = 77.2 \times 10^3 \, \text{J}, \, T = 400 \, \text{K}, \, \Delta S^\circ = 122 \, \text{J/K}.$

$\Delta G^\circ = 77.2 \times 10^3 - 400 \times 122 = 28400 \, \text{J}.$

The relationship between $\Delta G^\circ$ and the equilibrium constant ($K$) is:

$\Delta G^\circ = -2.303 RT \log K.$

Substitute $\Delta G^\circ = 28400 \, \text{J}, R = 8.314 \, \text{J K}, T = 400 \, \text{K}$:
$28400 = -2.303 \times 8.314 \times 400 \log K.$

Simplify:
$\log K = \frac{-28400}{2.303 \times 8.314 \times 400}.$

Calculate:
$\log K = \frac{-28400}{7668.8} = -3.708.$

Thus: $K = 10^{\log K} = 10^{-3.708}.$

The Correct answer is: 37