Question:
What is the \(\log K\) of the following reaction, \(2\mathrm{NH_3(g)} + \mathrm{CO_2(g)} \rightleftharpoons \mathrm{NH_2CONH_2(aq)} + \mathrm{H_2O(l)}\) at \(298\,\text{K}\), if \(\Delta_r G^\circ = -11.4\,\text{kJ mol}^{-1}\) and \(2.303RT = 5.7\,\text{kJ mol}^{-1}\)?
What is the \(\log K\) of the following reaction, \(2\mathrm{NH_3(g)} + \mathrm{CO_2(g)} \rightleftharpoons \mathrm{NH_2CONH_2(aq)} + \mathrm{H_2O(l)}\) at \(298\,\text{K}\), if \(\Delta_r G^\circ = -11.4\,\text{kJ mol}^{-1}\) and \(2.303RT = 5.7\,\text{kJ mol}^{-1}\)?
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If \(2.303RT\) is given in kJ/mol, ensure \(\Delta G^\circ\) is also in kJ/mol for direct substitution.
Updated On: Apr 27, 2026
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Solution and Explanation
Step 1: Understanding the Concept:
The relation between standard Gibbs free energy and equilibrium constant: \(\Delta_r G^\circ = -2.303 RT \log K\).
Step 2: Detailed Explanation:
Given \(\Delta_r G^\circ = -11.4 \text{ kJ mol}^{-1}\) and \(2.303 RT = 5.7 \text{ kJ mol}^{-1}\).
\(\log K = -\frac{\Delta_r G^\circ}{2.303 RT} = -\frac{(-11.4)}{5.7} = \frac{11.4}{5.7} = 2\).
Step 3: Final Answer:
\(\log K = 2\).
The relation between standard Gibbs free energy and equilibrium constant: \(\Delta_r G^\circ = -2.303 RT \log K\).
Step 2: Detailed Explanation:
Given \(\Delta_r G^\circ = -11.4 \text{ kJ mol}^{-1}\) and \(2.303 RT = 5.7 \text{ kJ mol}^{-1}\).
\(\log K = -\frac{\Delta_r G^\circ}{2.303 RT} = -\frac{(-11.4)}{5.7} = \frac{11.4}{5.7} = 2\).
Step 3: Final Answer:
\(\log K = 2\).
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