In a one-litre flask, 6 moles of A undergoes the reaction A(~g) leftharpoons P(~g) . The progress of product formation at two temperatures (in Kelvin), T_1 and T_2 , is shown in the figure: If T_1=2 ~T_2 and (\Delta G_2^\Theta-\Delta G_1^\Theta)=RT_2 \ln x , then the value of x is \underline\hspace2cm . [\Delta G_1^\Theta. and \Delta G_2^\Theta are standard Gibb's free energy change for the reaction at temperatures T_1 and T_2 , respectively.]

Thermodynamics PYQ — JEE Advanced 2023 · Shift 1 · Q12 with Solution

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Correct answer: 1/8

For the reaction $A(g) \rightleftharpoons P(g)$ in a $1$ L flask, initially $6$ moles of $A$ are present.

From the graph, the equilibrium amount of product formed is:

So equilibrium compositions are:

Since volume is $1$ L, concentrations equal mole numbers.

For $A(g) \rightleftharpoons P(g)$ we have $K=\frac{[P]}{[A]}$

Thus,

We know: $\Delta G^\circ=-RT\ln K$

Hence,

$\Delta G_1^\circ=-RT_1\ln K_1$ $\Delta G_2^\circ=-RT_2\ln K_2$

Substitute $K_1=\tfrac12$ and $K_2=2$ :

$\Delta G_1^\circ=-RT_1\ln\left(\frac12\right)=RT_1\ln 2$ $\Delta G_2^\circ=-RT_2\ln 2$

Given $T_1=2T_2$ ,

$\Delta G_2^\circ-\Delta G_1^\circ = (-RT_2\ln 2) - (RT_1\ln 2)$

$= -R\ln 2\,(T_2+T_1)$

Using $T_1=2T_2$ :

$\Delta G_2^\circ-\Delta G_1^\circ = -R\ln 2\,(T_2+2T_2)$ $= -3RT_2\ln 2$

Now, $-3\ln 2 = \ln(2^{-3}) = \ln\left(\frac{1}{8}\right)$

Therefore, $\Delta G_2^\circ-\Delta G_1^\circ = RT_2\ln\left(\frac{1}{8}\right)$

Comparing with $\Delta G_2^\circ-\Delta G_1^\circ = RT_2\ln x$ we get $x=\frac{1}{8}$

The mathematically derived value is $x=\frac{1}{8}$ not $8$ .

So the stored answer appears to have omitted the reciprocal.