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(A) The reaction is $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$.At equilibrium,the total pressure $X$ is distributed according to the

Vedclass · Accepted Answer

$-RT (3 \ln X + \ln \frac{4}{27})$

Vedclass · Answer

(A) The reaction is $NH_2COONH_{4(s)} \rightleftharpoons 2NH_{3(g)} + CO_{2(g)}$.At equilibrium,the total pressure $X$ is distributed according to the stoichiometric coefficients.$P_{NH_3} = \frac{2}{3} X$ and $P_{CO_2} = \frac{1}{3} X$.$K_p = (P_{NH_3})^2 (P_{CO_2}) = (\frac{2}{3} X)^2 (\frac{1}{3} X) = \frac{4}{9} X^2 \cdot \frac{1}{3} X = \frac{4}{27} X^3$.Using the relation $\Delta_r G^o = -RT \ln K_p$:$\Delta_r G^o = -RT \ln (\frac{4}{27} X^3)$.Using logarithmic properties $\ln(ab) = \ln a + \ln b$ and $\ln(a^n) = n \ln a$:$\Delta_r G^o = -RT (\ln X^3 + \ln \frac{4}{27}) = -RT (3 \ln X + \ln \frac{4}{27})$.

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 $X \ bar$,then $\Delta_r G^o$ would be :-

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