If Delta G and Delta S for the reaction A_{(g | vedclass

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(A) Given: $\Delta G = -40 \ kJ \ mol^{-1}$,$\Delta S = 0.22 \ kJ \ K^{-1} \ mol^{-1}$,$T = 2000 \ K$. Using the relation $\Delta G = \Delta H - T \De

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$366.7$

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(A) Given: $\Delta G = -40 \ kJ \ mol^{-1}$,$\Delta S = 0.22 \ kJ \ K^{-1} \ mol^{-1}$,$T = 2000 \ K$. Using the relation $\Delta G = \Delta H - T \Delta S$,we find $\Delta H = \Delta G + T \Delta S$. $\Delta H = -40 + (2000 	imes 0.22) = -40 + 440 = 400 \ kJ \ mol^{-1}$. For the reaction $A_{(g)} ightarrow B_{(g)} + 2C_{(g)}$,the change in moles of gas is $\Delta n_g = (1 + 2) - 1 = 2$. Using the relation $\Delta H = \Delta U + \Delta n_g RT$,we have $\Delta U = \Delta H - \Delta n_g RT$. $\Delta U = 400 - 2 	imes (8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) 	imes 2000 \ K$. $\Delta U = 400 - 33.256 = 366.744 \approx 366.7 \ kJ \ mol^{-1}$.

If $\Delta G$ and $\Delta S$ for the reaction $A_{(g)} \rightarrow B_{(g)} + 2C_{(g)}$ at $2000 \ K$ are $-40 \ kJ \ mol^{-1}$ and $0.22 \ kJ \ K^{-1} \ mol^{-1}$ respectively,the change in internal energy for the same reaction approximately (in $kJ \ mol^{-1}$) is

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