(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

Medium

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 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

Similar Questions

Calculate $\Delta G^\circ$ for the conversion of oxygen to ozone $\frac{3}{2} O_{2(g)} \to O_{3(g)}$ at $298 \ K$,if $K_p$ for this conversion is $2.47 \times 10^{-29}$.

Derive the relation between the equilibrium constant $K$,reaction quotient $Q_{c}$,and Gibbs energy change $\Delta G$.

Which condition among the following holds true at the state of half-completion for the reaction $A \rightleftharpoons B$?

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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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