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

Correct Answer: 37

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