The equilibrium constant for a reaction is $100$. What will be the value of standard Gibbs energy change at $298 \text{ K}$? $(R = 8.314 \text{ J K}^{-1} \text{mol}^{-1})$
- A-$11.411$ \text{ kJ/mol}
- B-$5.744$ \text{ kJ/mol}
- C-$570.584$ \text{ kJ/mol}
- D-$57.058$ \text{ kJ/mol}
Explore More
Similar Questions
Assertion $(A)$: 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.
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 SolutionFind out $\ln K_{eq}$ for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \ K$.
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
Given:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
Given:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$
Difficult
View SolutionFor 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 $298 \ K$.
For a homogeneous gaseous reaction,the equilibrium constant $K_p$ is $10^{-8}$. The standard Gibbs free energy change for the reaction is ........... $kcal$. $(R = 2.0 \, cal \, K^{-1} \, mol^{-1}, T = 298 \, K)$
Medium
View SolutionVedclass 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