The standard free energy change (Delta G^{cir | vedclass

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(D) The dissociation reaction is: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$At $t=0$,moles of $N_2O_4 = 1$ and $NO_2 = 0$.At equilibrium,for $50\%$ disso

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$710$

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(D) The dissociation reaction is: $N_2O_4(g) ightleftharpoons 2NO_2(g)$At $t=0$,moles of $N_2O_4 = 1$ and $NO_2 = 0$.At equilibrium,for $50\%$ dissociation: $N_2O_4 = 1 - 0.5 = 0.5\,mol$ and $NO_2 = 2 	imes 0.5 = 1.0\,mol$.Total moles at equilibrium $= 0.5 + 1.0 = 1.5\,mol$.Partial pressures are: $P_{N_2O_4} = \frac{0.5}{1.5} 	imes 1\,atm = \frac{1}{3}\,atm$ and $P_{NO_2} = \frac{1.0}{1.5} 	imes 1\,atm = \frac{2}{3}\,atm$.$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(2/3)^2}{1/3} = \frac{4/9}{1/3} = \frac{4}{3} \approx 1.333$.Using the formula $\Delta G^{\circ} = -RT \ln K_p$:$\Delta G^{\circ} = -8.31 	imes 300 	imes \ln(1.333)$.Since $\ln(1.333) = 2.303 	imes \log(1.333) \approx 2.303 	imes 0.1248 \approx 0.2875$.$\Delta G^{\circ} = -8.31 	imes 300 	imes 0.2875 \approx -716.7\,J\,mol^{-1}$.Rounding to the nearest integer,$x = 717$ (Note: Given options suggest $710$ as the intended answer based on provided constants).

The standard free energy change $(\Delta G^{\circ})$ for $50\%$ dissociation of $N_2O_4$ into $NO_2$ at $27^{\circ}C$ and $1\,atm$ pressure is $-x\,J\,mol^{-1}$. The value of $x$ is $......$ (Nearest Integer)
[Given: $R = 8.31\,J\,K^{-1}\,mol^{-1}$,$\log 1.33 = 0.1239$,$\ln 10 = 2.3$]

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The standard free energy change $(\Delta G^{\circ})$ for $50\%$ dissociation of $N_2O_4$ into $NO_2$ at $27^{\circ}C$ and $1\,atm$ pressure is $-x\,J\,mol^{-1}$. The value of $x$ is $......$ (Nearest Integer)[Given: $R = 8.31\,J\,K^{-1}\,mol^{-1}$,$\log 1.33 = 0.1239$,$\ln 10 = 2.3$]