• # question_answer 12) Find out the value of equilibrium constant for the following reaction at 298 K: $2N{{H}_{3}}(g)+C{{O}_{2}}(g)\rightleftharpoons N{{H}_{2}}CON{{H}_{2}}(aq)+{{H}_{2}}O(l)$ Standard Gibbs free energy change$\Delta {{G}^{{}^\circ }}$at the given temperature is$-13.6kJ\,mo{{l}^{-1}}$ .

Answer:

$\Delta {{G}^{{}^\circ }}=-2.303RT\,{{\log }_{10}}{{K}_{p}}$ $\log {{K}_{p}}=\frac{-\Delta {{G}^{{}^\circ }}}{2.303RT}$ $=-\frac{(-13.6\times {{10}^{3}})}{2.303\times 8.314\times 298}$ $=2.38$ ${{K}_{p}}=Anti\log (2.38)=2.4\times {{10}^{2}}$

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