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:
$\Delta G^{\ominus} = -8.314 \ J \ mol^{-1} \ K^{-1} \times 300 \ K \times \ln(2 \times 10^{13})$
$\Delta G^{\ominus} = -2494.2 \times (\ln 2 + \ln 10^{13})$
$\Delta G^{\ominus} = -2494.2 \times (0.693 + 13 \times 2.303)$
$\Delta G^{\ominus} = -2494.2 \times (0.693 + 29.939)$
$\Delta G^{\ominus} = -2494.2 \times 30.632$
$\Delta G^{\ominus} \approx -76400 \ J \ mol^{-1} = -76.4 \ kJ \ mol^{-1}$
Explore More
Similar Questions
$A$ reaction attains equilibrium when the free energy change accompanying it is
Easy
View SolutionWrite the formula relating the equilibrium constant $K$ and the standard Gibbs free energy change $\Delta G^{\circ}$.
Easy
View SolutionFor a certain reaction at $300 \ K$,$K=10$,then $\Delta G^{\circ}$ for the same reaction is . . . . . . $\times 10^{-1} \ kJ \ mol^{-1}$. (Given $R=8.314 \ J \ K^{-1} \ mol^{-1}$)
DifficultJEE MAIN 2024
View SolutionAssertion $(A)$: For every chemical reaction at equilibrium,standard Gibbs energy change of the reaction is zero.
Reason $(R)$: At constant temperature and pressure,chemical reactions are spontaneous in the direction of decreasing Gibbs energy.
Reason $(R)$: At constant temperature and pressure,chemical reactions are spontaneous in the direction of decreasing Gibbs energy.
For the reaction $CH_{4(g)} + H_{2(g)} \longrightarrow C_2H_{6(g)}$,$K_p = 3.356 \times 10^{17}$,calculate $\Delta G^{\circ}$ for the reaction at $298 \ K$.
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo