Hydrolysis 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 $300 \ K$.
(N/A) The standard Gibbs free energy change is given by the formula:
$\Delta G^{\ominus} = -RT \ln K_{c}$
Given:
$R = 8.314 \ J \ mol^{-1} \ K^{-1}$
$T = 300 \ K$
$K_{c} = 2 \times 10^{13}$
Substituting the values:
$\Delta G^{\ominus} = -8.314 \ J \ mol^{-1} \ K^{-1} \times 300 \ K \times \ln(2 \times 10^{13})$
$\Delta G^{\ominus} = -2494.2 \times (\ln 2 + \ln 10^{13})$
$\Delta G^{\ominus} = -2494.2 \times (0.693 + 13 \times 2.303)$
$\Delta G^{\ominus} = -2494.2 \times (0.693 + 29.939)$
$\Delta G^{\ominus} = -2494.2 \times 30.632$
$\Delta G^{\ominus} \approx -76400 \ J \ mol^{-1} = -76.4 \ kJ \ mol^{-1}$
$\Delta G^{\ominus} = -RT \ln K_{c}$
Given:
$R = 8.314 \ J \ mol^{-1} \ K^{-1}$
$T = 300 \ K$
$K_{c} = 2 \times 10^{13}$
Substituting the values:
$\Delta G^{\ominus} = -8.314 \ J \ mol^{-1} \ K^{-1} \times 300 \ K \times \ln(2 \times 10^{13})$
$\Delta G^{\ominus} = -2494.2 \times (\ln 2 + \ln 10^{13})$
$\Delta G^{\ominus} = -2494.2 \times (0.693 + 13 \times 2.303)$
$\Delta G^{\ominus} = -2494.2 \times (0.693 + 29.939)$
$\Delta G^{\ominus} = -2494.2 \times 30.632$
$\Delta G^{\ominus} \approx -76400 \ J \ mol^{-1} = -76.4 \ kJ \ mol^{-1}$
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