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 determine the logarithm of the equilibrium constant (\(\log K\)) at 400 K, the Gibbs free energy change equation (\(\Delta G^\circ = \Delta H^\circ - T\Delta S\)) is utilized.

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

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

Next, calculate \(\Delta G^\circ\):

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

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

The relationship \(\Delta G^\circ = -RT \ln K\) is applied.

Rearrange to solve for \(\ln K\):

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

Using the gas constant \(R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}\):

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

\[\ln K = -8.544\]

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

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

\[ \log K = -3.71\]

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

Verify if the calculated value aligns with the specified range:

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