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Last updated date: 29th Nov 2023
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# For the chemical reaction $X\rightleftharpoons Y$, the standard reaction Gibbs energy depends on temperature $T({\rm{in K}})$ as: ${\Delta _r}{G^ \circ }(in\,kJmo{l^{ - 1}}) = 120 - \dfrac{3}{8}T$. The major component of the reaction mixture at $T$ is:(This question has multiple correct options)A. $X\,{\rm{if}}\,T = 315\,K$B. $Y\,{\rm{if}}\,T = 350K$C. $Y\,{\rm{if}}\,T = 300K$D. $Y\,{\rm{if}}\,T = 280\,K$

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Hint: Gibbs free energy: It is a thermodynamic potential which is used to determine the maximum work done by the system at constant pressure and temperature. It is also used to determine the spontaneity of the reaction i.e.; it determines the extent of formation of product during the chemical reaction.

For a reaction at equilibrium, the rate of forward reaction is equivalent to the rate of backward reaction. So, if the change in Gibbs free energy for forward reaction is $G$. Then the change in Gibbs free energy for backward reaction will be $- G$. Therefore, total change in Gibbs free energy for overall reaction will be as follows:
${\Delta _r}{G^ \circ }_{eq.} = G + ( - G)$
$\therefore {\Delta _r}{G^ \circ }_{eq.} = 0$
Hence, the change in Gibbs free energy at equilibrium is zero.
According to the given conditions, change in Gibbs free energy depends on temperature as follows:
${\Delta _r}{G^ \circ }(in\,kJmo{l^{ - 1}}) = 120 - \dfrac{3}{8}T$
At equilibrium,
$120 - \dfrac{3}{8}T = 0$
$\Rightarrow 3T = 960$
$\Rightarrow T = 320\,K$
Now, as per options given:
If $T = 315K$, then the change in Gibbs free energy is as follows:
${\Delta _r}{G^ \circ }(in\,kJmo{l^{ - 1}}) = 120 - \dfrac{3}{8}T$
$\Rightarrow \,120 - \dfrac{3}{8} \times 315$
$\Rightarrow 120 - 118.13$
$\Rightarrow 1.87\,kJmo{l^{ - 1}}$
As the change in Gibbs free energy is positive which indicates that the reaction is nonspontaneous. Therefore, the major component of the reaction will be $X$ at temperature $T = 315K$.
If $T = 350K$, then the change in Gibbs free energy is as follows:
${\Delta _r}{G^ \circ }(in\,kJmo{l^{ - 1}}) = 120 - \dfrac{3}{8}T$
$\Rightarrow \,120 - \dfrac{3}{8} \times 350$
$\Rightarrow 120 - 131.25$
$\Rightarrow - 11.25\,kJmo{l^{ - 1}}$
As the change in Gibbs free energy is negative which indicates that the reaction is spontaneous. Therefore, the major component of the reaction will be $Y$ at temperature $T = 350K$.
If $T = 300K$, then the change in Gibbs free energy is as follows:
${\Delta _r}{G^ \circ }(in\,kJmo{l^{ - 1}}) = 120 - \dfrac{3}{8}T$
$\Rightarrow \,120 - \dfrac{3}{8} \times 300$
$\Rightarrow 120 - 112.5$
$\Rightarrow 7.5\,kJmo{l^{ - 1}}$
As the change in Gibbs free energy is positive which indicates that the reaction is nonspontaneous. Therefore, the major component of the reaction will be $X$ at temperature $T = 300K$.
If $T = 280K$, then the change in Gibbs free energy is as follows:
${\Delta _r}{G^ \circ }(in\,kJmo{l^{ - 1}}) = 120 - \dfrac{3}{8}T$
$\Rightarrow \,120 - \dfrac{3}{8} \times 280$
$\Rightarrow 120 - 105$
$\Rightarrow 15\,kJmo{l^{ - 1}}$
As the change in Gibbs free energy is positive which indicates that the reaction is nonspontaneous. Therefore, the major component of the reaction will be $X$ at temperature $T = 280K$.

Hence, options (a) and (b) are correct.

Note:
Spontaneous reaction: It is a chemical reaction which favours the formation of products at a particular temperature. Change in Gibbs free energy i.e., $\Delta G < 0$ is the mandatory condition for a reaction to be spontaneous.
Non-spontaneous reaction: It is a chemical reaction which does not favour the formation of products for specific conditions. Hence, for these reactions the reactants are the major components.