At 60^{circ} C,dinitrogen tetroxide is 50 % d | vedclass

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(C) The reaction for the dissociation is: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$At $t=0$,we have $1 \ mol$ of $N_2O_4$ and $0 \ mol$ of $NO_2$.At

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$-790 \ J \cdot mol^{-1}$

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(C) The reaction for the dissociation is: $N_2O_{4(g)} ightleftharpoons 2NO_{2(g)}$At $t=0$,we have $1 \ mol$ of $N_2O_4$ and $0 \ mol$ of $NO_2$.At equilibrium,with $50 \%$ dissociation,we have $1-0.5 = 0.5 \ mol$ of $N_2O_4$ and $2 	imes 0.5 = 1 \ mol$ of $NO_2$.The total number of moles is $0.5 + 1 = 1.5 \ mol$.The partial pressures are:$P_{N_2O_4} = \frac{0.5}{1.5} 	imes 1 \ atm = \frac{1}{3} \ atm$$P_{NO_2} = \frac{1}{1.5} 	imes 1 \ atm = \frac{2}{3} \ atm$The equilibrium constant $K_p$ is:$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(2/3)^2}{1/3} = \frac{4/9}{1/3} = \frac{4}{3} \approx 1.33 \ atm$The standard free energy change is given by:$\Delta G^{\circ} = -RT \ln K_p = -2.303 	imes RT 	imes \log_{10} K_p$$\Delta G^{\circ} = -2.303 	imes (8.314 \ J \cdot K^{-1} \cdot mol^{-1}) 	imes (333 \ K) 	imes \log_{10} 1.33$$\Delta G^{\circ} = -2.303 	imes 8.314 	imes 333 	imes 0.1239 \approx -790 \ J \cdot mol^{-1}$

At $60^{\circ} C$,dinitrogen tetroxide is $50 \%$ dissociated. Find its standard free energy change at this temperature and $1 \ atm$. [ Given: $\log 1.33 = 0.1239 ]$

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