$\text{For a certain thermochemical reaction } \text{M} \rightarrow \text{N at } T = 400 \, \text{K}, \, \Delta H^\circ = 77.2 \, \text{kJ mol}^{-1}, \, \Delta S = 122 \, \text{JK}^{-1}, \\ \text{log equilibrium constant (log K) is } - \_\_\_\_\_ \times 10^{-1}.$
Correct Answer: 37
Approach Solution - 1
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.
Approach Solution -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
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