Question

consider the reaction: icl(g) + cl₂(g) → icl₃(s) the δg° of the reaction is -17.09 kj/mol. calculate the k_eq for the reaction at 298.0 k.

Explanation:

Step1: Convert $\Delta G^{\circ}$ to J/mol

$\Delta G^{\circ}=-17.09\ \text{kJ/mol}=-17090\ \text{J/mol}$

Step2: Use the formula $\Delta G^{\circ}=-RT\ln K_{eq}$

where $R = 8.314\ \text{J/(mol·K)}$ and $T = 298.0\ \text{K}$

Rearrange the formula for $K_{eq}$: $\ln K_{eq}=-\frac{\Delta G^{\circ}}{RT}$

Substitute the values: $\ln K_{eq}=-\frac{- 17090}{8.314\times298.0}$

Step3: Calculate the value of $\ln K_{eq}$

$\ln K_{eq}=\frac{17090}{8.314\times298.0}\approx6.80$

Step4: Find $K_{eq}$

Since $\ln K_{eq} = 6.80$, then $K_{eq}=e^{6.80}\approx40.1$

Answer:

$K_{eq} = 40.1$