The equilibrium constant for a reaction is 20 | vedclass

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Question

(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K$.

Vedclass · Accepted Answer

$-7.19 \ kJ \ mol^{-1}$

Vedclass · Answer

(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by the equation: $\Delta G^{\circ} = -RT \ln K$. Given values: $K = 20$,$T = 300 \ K$,$R = 8 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$. Substituting the values into the formula: $\Delta G^{\circ} = -(8 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) 	imes (300 \ K) 	imes \ln(20)$. Since $\ln(20) \approx 2.996$,we have: $\Delta G^{\circ} = -2.4 	imes 2.996 \approx -7.19 \ kJ \ mol^{-1}$. Therefore,the correct value is $-7.19 \ kJ \ mol^{-1}$.

The equilibrium constant for a reaction is $20$. What is the value of $\Delta G^{\circ}$ at $300 \ K$? (Given: $R = 8 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,$\ln(20) \approx 2.996$)

Similar Questions

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At $298 \ K$, if the standard Gibbs energy change $\Delta_r G^{\ominus}$ of a reaction is $-115 \ kJ \ mol^{-1}$, the value of $\log_{10} K_{p}$ will be $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$.

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