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(B) The reaction is $Cl_{2(g)} \rightleftharpoons 2Cl_{(g)}$.At $t=0$,let the initial pressure be $P_0$.At equilibrium,the partial pressures are $P_{C

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$-5.99$

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(B) The reaction is $Cl_{2(g)} ightleftharpoons 2Cl_{(g)}$.At $t=0$,let the initial pressure be $P_0$.At equilibrium,the partial pressures are $P_{Cl_2} = P_0(1-\alpha)$ and $P_{Cl} = 2P_0\alpha$.The total pressure $P_T = P_0(1-\alpha) + 2P_0\alpha = P_0(1+\alpha) = 15 \ atm$.Given $\alpha = 0.5$,we have $P_0(1+0.5) = 15$,so $P_0 = 10 \ atm$.The equilibrium constant $K_p = \frac{(P_{Cl})^2}{P_{Cl_2}} = \frac{(2P_0\alpha)^2}{P_0(1-\alpha)} = \frac{(2 	imes 10 	imes 0.5)^2}{10(1-0.5)} = \frac{100}{5} = 20$.Using the relation $\Delta G^o = -RT \ ln \ K_p$,where $R = 2 \ cal \ K^{-1} \ mol^{-1} = 2 	imes 10^{-3} \ kcal \ K^{-1} \ mol^{-1}$.$\Delta G^o = - (2 	imes 10^{-3} \ kcal \ K^{-1} \ mol^{-1}) 	imes (1000 \ K) 	imes ln(20)$.$\Delta G^o = -2 	imes 2.99 = -5.98 \ kcal/mole \approx -5.99 \ kcal/mole$.

Calculate $\Delta G^o$ (in $kcal/mole$) for the decomposition of $Cl_{2(g)} \rightleftharpoons 2Cl_{(g)}$,if chlorine molecules are $50\%$ dissociated at $1000 \ K$ at a pressure of $15 \ atm$ at equilibrium $(ln \ 20 = 2.99)$.

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