the equilibrium constant $k_p$ for the reaction $co(g)+cl_2(g)\rightleftharpoons cocl_2(g)$ is $5.62\times…

the equilibrium constant $k_p$ for the reaction $co(g)+cl_2(g)\rightleftharpoons cocl_2(g)$ is $5.62\times 10^{35}$ at $25^{circ}c$. calculate $delta g_{f}^{circ}$ for $cocl_2$ at $25^{circ}c$. round your answer to 4 significant digits.

the equilibrium constant $k_p$ for the reaction $co(g)+cl_2(g)\rightleftharpoons cocl_2(g)$ is $5.62\times 10^{35}$ at $25^{circ}c$. calculate $delta g_{f}^{circ}$ for $cocl_2$ at $25^{circ}c$. round your answer to 4 significant digits.

Answer

Explanation:

Step1: Recall the relationship between $\Delta G^{\circ}$ and $K_p$

The formula is $\Delta G^{\circ}=-RT\ln K_p$. Here, $R = 8.314\ J/(mol\cdot K)$ and $T=(25 + 273.15)\ K= 298.15\ K$, $K_p = 5.62\times 10^{35}$.

Step2: Calculate $\Delta G^{\circ}$

First, calculate $\ln K_p=\ln(5.62\times 10^{35})=\ln5.62+35\ln10\approx1.726+35\times2.303 = 1.726 + 80.605=82.331$. Then, $\Delta G^{\circ}=-8.314\ J/(mol\cdot K)\times298.15\ K\times82.331$. $\Delta G^{\circ}=-8.314\times298.15\times82.331\ J/mol\approx - 2.047\times10^{5}\ J/mol=-204.7\ kJ/mol$.

Answer:

$-204.7\ kJ/mol$