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

To find the log equilibrium constant (log K) for the reaction at 400 K, use the Gibbs free energy change equation: \(\Delta G^\circ = \Delta H^\circ - T\Delta S\).

Given: \(\Delta H^\circ = 77.2 \, \text{kJ mol}^{-1}\), \(\Delta S = 122 \, \text{J K}^{-1}\), \(T = 400 \, \text{K}\).

Convert \(\Delta H^\circ\) to J: \(\Delta H^\circ = 77.2 \times 10^3 \, \text{J/mol}\).

Calculate \(\Delta G^\circ\): 

\[\Delta G^\circ = 77.2 \times 10^3 - (400 \times 122)\]

\[\Delta G^\circ = 77,200 - 48,800 = 28,400 \, \text{J mol}^{-1}\]

Use the relation: \(\Delta G^\circ = -RT \ln K\).

Convert to \(\ln K\):

\[\ln K = -\frac{\Delta G^\circ}{RT}\]

With \(R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}\):

\[\ln K = -\frac{28,400}{8.314 \times 400}\]

\[\ln K = -8.544\] 

Convert \(\ln K\) to \(\log K\) using \(\log K = \ln K / \ln 10\):

\[ = \frac{-8.544}{2.302}\]

\[ \log K = -3.71\]

The equilibrium constant is expressed as: \(-0.37 \times 10^{1}\).

Check if the solution fits within the given range:

The value \(-0.37 \times 10^{1} = -3.7\) is within the expected range of 37,37.

Answer 2

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