The value of log _{10} K for a reaction A rig | vedclass

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(B) The standard Gibbs free energy change is given by $\Delta _{r} G^{\circ} = \Delta _{r} H^{\circ} - T \Delta _{r} S^{\circ}$.Substituting the value

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$10$

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(B) The standard Gibbs free energy change is given by $\Delta _{r} G^{\circ} = \Delta _{r} H^{\circ} - T \Delta _{r} S^{\circ}$.Substituting the values: $\Delta _{r} G^{\circ} = (-54.07 	imes 1000 \ J \ mol^{-1}) - (298 \ K 	imes 10 \ J \ K^{-1} \ mol^{-1}) = -54070 - 2980 = -57050 \ J \ mol^{-1}$.We know that $\Delta _{r} G^{\circ} = -2.303 \ RT \log _{10} K$.Substituting the values: $-57050 = - (2.303 	imes 8.314 	imes 298) \log _{10} K$.Given $2.303 	imes 8.314 	imes 298 = 5705$,we have: $-57050 = -5705 \log _{10} K$.Therefore,$\log _{10} K = \frac{57050}{5705} = 10$.Hence,the correct option is $(B)$.

The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is
(Given : $\Delta _{r} H_{298 K}^{\circ} = -54.07 \ kJ \ mol^{-1}$,$\Delta _{r} S_{298 K}^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)

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The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is(Given : $\Delta _{r} H_{298 K}^{\circ} = -54.07 \ kJ \ mol^{-1}$,$\Delta _{r} S_{298 K}^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)