The value of K for the reaction A B at 298 ~K is ___________. (Nearest integer) Given: H^=-54.07 ~kJ ~mol^-1 S^=10 ~J ~K^-1 ~mol^-1 (Take 2.303 8.314 298=5705 )

Given: align H^0 & = -54.07 \, kJ/mol = -54070 \, J/mol \\\\ S^0 & = 10 \, J/K mol \\\\ T & = 298 \, K align We find the change in Gibbs free energy G^0: align G^0 & = H^0 - T S^0 \\\\ & = -54070 - (10 298) \\\\ & = -54070 - 2980 \\\\ & = -57050 \, J/mol align Now, we'll use the

The value of log K for the reaction A leftharpoons B at 298 : null (Chemistry)

Given:

\begin{align} \Delta H^0 & = -54.07 \, \text{kJ/mol} = -54070 \, \text{J/mol} \\\\ \Delta S^0 & = 10 \, \text{J/K}\cdot \text{mol} \\\\ T & = 298 \, \text{K} \end{align}

We find the change in Gibbs free energy $\Delta G^0$ :

\begin{align} \Delta G^0 & = \Delta H^0 - T \Delta S^0 \\\\ & = -54070 - (10 \times 298) \\\\ & = -54070 - 2980 \\\\ & = -57050 \, \text{J/mol} \end{align}

Now, we'll use the given expression with the correct constant for $R$ as 8.314 J/(mol·K):

\begin{align} \Delta G^0 & = -2.303 \times 8.314 \times 298 \times \log K \\\\ \log K & = \frac{-57050}{2.303 \times 8.314 \times 298} \\\\ & \approx 10 \end{align}

So, the answer is $\log K = 10$ .

Explanation

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