For the reaction A rightleftharpoons B,find t | vedclass

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(B) The standard Gibbs free energy change is given by $\Delta_rG^o = \Delta_rH^o - T\Delta_rS^o$. Substituting the values: $\Delta_rG^o = (-54.07 	im

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$10$

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(B) The standard Gibbs free energy change is given by $\Delta_rG^o = \Delta_rH^o - T\Delta_rS^o$. Substituting the values: $\Delta_rG^o = (-54.07 	imes 10^3\, J\, mol^{-1}) - (298\, K 	imes 10\, J\, K^{-1}\, mol^{-1}) = -54070 - 2980 = -57050\, J\, mol^{-1}$. We know that $\Delta_rG^o = -2.303 RT \log_{10}K$. Substituting the values: $-57050 = -2.303 	imes 8.314 	imes 298 	imes \log_{10}K$. Given $2.303 	imes 8.314 	imes 298 = 5705$,we have $-57050 = -5705 	imes \log_{10}K$. Therefore,$\log_{10}K = \frac{57050}{5705} = 10$.

For the reaction $A \rightleftharpoons B$,find the value of $log_{10}K$. Given: $\Delta_rH^o_{298\,K} = -54.07\, kJ\, mol^{-1}$,$\Delta_rS^o_{298\,K} = 10\, J\, K^{-1}\, mol^{-1}$,$R = 8.314\, J\, K^{-1}\, mol^{-1}$,$2.303 \times 8.314 \times 298 = 5705$.

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