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(D) The standard Gibbs free energy change is given by $\Delta G^{\circ} = - RT \ln K_p$.For the reaction $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g

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$- RT \ln 4 - 3 RT \ln X$

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(D) The standard Gibbs free energy change is given by $\Delta G^{\circ} = - RT \ln K_p$.For the reaction $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$,the total equilibrium pressure is $P_{total} = 3X \ bar$.Let the partial pressure of $CO_2$ be $p$ and the partial pressure of $NH_3$ be $2p$ based on the stoichiometry.Thus,$p + 2p = 3X$,which implies $3p = 3X$,so $p = X$.Therefore,$p_{CO_2} = X$ and $p_{NH_3} = 2X$.The equilibrium constant $K_p$ is given by $K_p = (p_{NH_3})^2 (p_{CO_2}) = (2X)^2 (X) = 4X^3$.Substituting this into the Gibbs free energy equation:$\Delta G^{\circ} = - RT \ln(4X^3) = - RT (\ln 4 + 3 \ln X) = - RT \ln 4 - 3 RT \ln X$.

For the reaction taking place at a certain temperature $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$. If the equilibrium pressure is $3X \ bar$,then $\Delta _r G^o$ would be:

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