At 298 K, the equilibrium constant of the pr | vedclass

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(D) The standard free energy change $\Delta G^{\circ}$ is related to the equilibrium constant $K$ by the equation: $\Delta G^{\circ} = -RT \ln K = -2.

Vedclass · Accepted Answer

$163$

Vedclass · Answer

(D) The standard free energy change $\Delta G^{\circ}$ is related to the equilibrium constant $K$ by the equation: $\Delta G^{\circ} = -RT \ln K = -2.303 RT \log_{10} K$. Given $R = 8.314 \ J \ mol^{-1} \ K^{-1}$, $T = 298 \ K$, and $K = 3 	imes 10^{-29}$. Substituting these values: $\Delta G^{\circ} = -2.303 	imes 8.314 	imes 298 	imes \log(3 	imes 10^{-29}) \ J \ mol^{-1}$. $\Delta G^{\circ} = -5705.8 	imes (\log 3 + \log 10^{-29}) \ J \ mol^{-1}$. $\Delta G^{\circ} = -5705.8 	imes (0.47 - 29) \ J \ mol^{-1}$. $\Delta G^{\circ} = -5705.8 	imes (-28.53) \ J \ mol^{-1} \approx 162787 \ J \ mol^{-1}$. Converting to $kJ \ mol^{-1}$: $\Delta G^{\circ} \approx 163 \ kJ \ mol^{-1}$.

At $298 \ K$, the equilibrium constant of the process $1.5 O_{2(g)} \rightleftharpoons O_{3(g)}$ is $3 \times 10^{-29}$. The standard free energy change (in $kJ \ mol^{-1}$) of the process is approximately ($R = 8.314 \ J \ mol^{-1} \ K^{-1}$; $\log 3 = 0.47$)

Similar Questions

The $INCORRECT$ match in the following is

For a chemical reaction, the standard Gibbs energy change, $\Delta G^{\circ}$ is $-7.64 \times 10^4 \ J \ mol^{-1}$. What is the value of equilibrium constant $(K)$?

The correct relation is:

For a reaction,$\Delta G^{\circ} = -115 \, kJ$. What is the value of $\log \, K_p$ at $298 \, K$?

Find the value of the equilibrium constant $(K)$ of a reaction at $300 \ K$, when standard Gibbs free energy change is $-25 \ kJ \ mol^{-1}$? (Consider $R = 8.33 \ J \ mol^{-1} \ K^{-1}$)

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