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(D) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative. $\Delta G = \Delta H - T \Delta S$. For the reaction to

Vedclass · Accepted Answer

$971.8$

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(D) For a reaction to be spontaneous,the Gibbs free energy change $\Delta G$ must be negative. $\Delta G = \Delta H - T \Delta S$. For the reaction to be in equilibrium,$\Delta G = 0$,so $T = \frac{\Delta H}{\Delta S}$. Given: $\Delta H = -41.2 \ kJ = -41200 \ J$ and $\Delta S = -42.4 \ J \ K^{-1}$. Calculating the equilibrium temperature: $T = \frac{-41200 \ J}{-42.4 \ J \ K^{-1}} \approx 971.7 \ K$. For the reaction to proceed in the reverse direction,the reverse reaction must be spontaneous,which means $\Delta G_{reverse}  0$. Since $\Delta H$ and $\Delta S$ are both negative,the forward reaction is spontaneous at temperatures below the equilibrium temperature $(T Therefore,the reaction will proceed in the reverse direction at temperatures above $971.7 \ K$.

The entropy and enthalpy changes for the reaction $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$ at $300 \ K$ and $1 \ atm$ are respectively $-42.4 \ J \ K^{-1}$ and $-41.2 \ kJ$. The temperature at which the reaction will go in the reverse direction is (in $K$)

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