Write the formula relating the equilibrium constant $K$ and $\Delta G^{\circ}$.
(N/A) The relationship between the standard Gibbs free energy change $\Delta_{r}G^{\circ}$ and the equilibrium constant $K$ is given by the equation: $\Delta_{r}G^{\circ} = -RT \ln K$ or $\Delta_{r}G^{\circ} = -2.303 RT \log K$.
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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 $(K)$?
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 $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$.
The equilibrium constant of the following given reaction is $K_p = 6.022 \times 10^{-5}$ at $298 \ K$ temperature. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. Calculate the value of $\Delta_r G^o$.
Medium
View SolutionDescribe the relationship between Gibbs energy change and chemical equilibrium.
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View SolutionHydrolysis of sucrose gives,
$Sucrose + H_{2}O \rightleftharpoons Glucose + Fructose$
Equilibrium constant $K_{c}$ for the reaction is $2 \times 10^{13}$ at $300 \ K$. Calculate $\Delta G^{\ominus}$ at $300 \ K$.
$Sucrose + H_{2}O \rightleftharpoons Glucose + Fructose$
Equilibrium constant $K_{c}$ for the reaction is $2 \times 10^{13}$ at $300 \ K$. Calculate $\Delta G^{\ominus}$ at $300 \ K$.
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