(D) The equilibrium condition is given by ${\Delta_r}{G^o} = 0$.Setting the equation to zero: $120 - \frac{3}{8} \ T = 0$.Solving for $T$: $T = \frac{

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(D) The equilibrium condition is given by ${\Delta_r}{G^o} = 0$.Setting the equation to zero: $120 - \frac{3}{8} \ T = 0$.Solving for $T$: $T = \frac{120 \times 8}{3} = 320 \ K$.For $T 0$,which implies the reaction is non-spontaneous in the forward direction and $X$ is the major component.For $T > 320 \ K$,${\Delta_r}{G^o} Checking the options:$A$: $T = 300 \ K $B$: $T = 280 \ K $C$: $T = 350 \ K > 320 \ K$,so $Y$ is major.$D$: $T = 315 \ K Thus,option $D$ is correct.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Standard free energy

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For the chemical reaction $X \rightleftharpoons Y$,the standard reaction Gibbs energy depends on temperature $T$ (in $K$) as
${\Delta_r}{G^o}$ (in $kJ \ mol^{-1}$) $= 120 - \frac{3}{8} \ T$
The major component of the reaction mixture at $T$ is

JEE MAIN 2019 All JEE MAIN

A
$Y$ if $T = 300 \ K$
B
$Y$ if $T = 280 \ K$
C
$X$ if $T = 350 \ K$
D
$X$ if $T = 315 \ K$

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6-1.Equilibrium (Chemical Equilibrium)

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Standard free energy

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For a certain reaction at $300 \ K$,$K=10$,then $\Delta G^{\circ}$ for the same reaction is . . . . . . $\times 10^{-1} \ kJ \ mol^{-1}$. (Given $R=8.314 \ J \ K^{-1} \ mol^{-1}$)

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

EasyMHT CET 2020

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At $60^{\circ} C$,dinitrogen tetroxide is $50 \%$ dissociated. Find its standard free energy change at this temperature and $1 \ atm$. [ Given: $\log 1.33 = 0.1239 ]$

MediumAP EAMCET 2021

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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.

EasyAP EAMCET 2017

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Calculate $\Delta G^\circ$ for the reaction, $CH_4(g) + H_2(g) \rightarrow C_2H_6(g)$ at $298 \text{ K}$, given $K_p = 2 \times 10^{17}$ and $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

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For the chemical reaction $X \rightleftharpoons Y$,the standard reaction Gibbs energy depends on temperature $T$ (in $K$) as${\Delta_r}{G^o}$ (in $kJ \ mol^{-1}$) $= 120 - \frac{3}{8} \ T$The major component of the reaction mixture at $T$ is

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For a certain reaction at $300 \ K$,$K=10$,then $\Delta G^{\circ}$ for the same reaction is . . . . . . $\times 10^{-1} \ kJ \ mol^{-1}$. (Given $R=8.314 \ J \ K^{-1} \ mol^{-1}$)

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

At $60^{\circ} C$,dinitrogen tetroxide is $50 \%$ dissociated. Find its standard free energy change at this temperature and $1 \ atm$. [ Given: $\log 1.33 = 0.1239 ]$

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.

Calculate $\Delta G^\circ$ for the reaction, $CH_4(g) + H_2(g) \rightarrow C_2H_6(g)$ at $298 \text{ K}$, given $K_p = 2 \times 10^{17}$ and $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

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For the chemical reaction $X \rightleftharpoons Y$,the standard reaction Gibbs energy depends on temperature $T$ (in $K$) as
${\Delta_r}{G^o}$ (in $kJ \ mol^{-1}$) $= 120 - \frac{3}{8} \ T$
The major component of the reaction mixture at $T$ is

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.