Given Delta G^o (HI, g) cong + 1.7 kJ mol^{ | vedclass

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(D) The reaction is $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$.For the formation of $1 \ mol$ of $HI$,$\Delta G_f^o = + 1.7 \ kJ \ mol^{

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$0.5$

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(D) The reaction is $2HI_{(g)} ightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$.For the formation of $1 \ mol$ of $HI$,$\Delta G_f^o = + 1.7 \ kJ \ mol^{-1}$.For the reaction $2HI ightarrow H_2 + I_2$,the standard Gibbs free energy change is $\Delta G^o_{rxn} = [\Delta G_f^o(H_2) + \Delta G_f^o(I_2)] - 2 	imes \Delta G_f^o(HI)$.Since $H_2$ and $I_2$ are elements in their standard states,$\Delta G_f^o = 0$.Therefore,$\Delta G^o_{rxn} = 0 - 2 	imes (1.7 \ kJ \ mol^{-1}) = - 3.4 \ kJ \ mol^{-1} = - 3400 \ J \ mol^{-1}$.Using the relation $\Delta G^o = - RT \ln K$,we have $\ln K = - \frac{\Delta G^o}{RT}$.$\ln K = - \frac{- 3400}{8.314 	imes 298} \approx \frac{3400}{2477.57} \approx 1.37$.$K = e^{1.37} \approx 3.9$.Note: If the question implies $\Delta G^o$ for the reaction is $+ 1.7 \ kJ$,then $\log K = - \frac{1700}{2.303 	imes 8.314 	imes 298} \approx - 0.298$,so $K = 10^{-0.298} \approx 0.5$.

Given $\Delta G^o (HI, g) \cong + 1.7 \ kJ \ mol^{-1}$. What is the equilibrium constant at $25 ^oC$ for the reaction $2HI_{(g)} \rightleftharpoons H_{2_{(g)}} + I_{2_{(g)}}$?

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