In a one-litre flask, 6 moles of A undergoes the reaction A (g) ⇌P (g). The progress of product formation at two temperatures (in Kelvin), $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$, is shown in the figure: (see figure) If $\mathrm{T}_{1}=2 \mathrm{~T}_{2}$ and $(\Delta \mathrm{G}_{2}^{\Theta}-\Delta \mathrm{G}_{1}^{\Theta})=\mathrm{RT}_{2} \ln \mathrm{x}$, then the value of x is ____. [$\Delta \mathrm{G}_{1}^{\Theta}$ and $\Delta \mathrm{G}_{2}^{\Theta}$ are standard Gibb's free energy change for the reaction at temperatures $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$, respectively.]

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Chemical Thermodynamics

In a one-litre flask, 6 moles of A undergoes the reaction A (g) ⇌P (g). The progress of product formation at two temperatures (in Kelvin),

T_{1}

and

T_{2}

, is shown in the figure:

(see figure)

If

T_{1} = 2 T_{2}

and

(Δ G_{2}^{Θ} - Δ G_{1}^{Θ}) = RT_{2} ln x

, then the value of x is ____.

[

Δ G_{1}^{Θ}

and

Δ G_{2}^{Θ}

are standard Gibb's free energy change for the reaction at temperatures

T_{1}

and

T_{2}

, respectively.]

Answer: 8

From the figure:

At $T_{1}$ : equilibrium has 4 moles of P, so 2 moles of A remain. $K_{P_{1}} = 4/2 = 2$
At $T_{2}$ : equilibrium has 2 moles of P, so 4 moles of A remain. $K_{P_{2}} = 2/4 = 1/2$

Standard Gibbs free energies:

Δ G_{1}^{\circ} = - R T_{1} ln K_{P_{1}} = - R T_{1} ln 2 = - 2 R T_{2} ln 2

Δ G_{2}^{\circ} = - R T_{2} ln K_{P_{2}} = - R T_{2} ln (1/2) = R T_{2} ln 2

Difference:

Δ G_{2}^{\circ} - Δ G_{1}^{\circ} = R T_{2} ln 2 + 2 R T_{2} ln 2 = 3 R T_{2} ln 2 = R T_{2} ln 2^{3} = R T_{2} ln 8

Therefore:

x = 8

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Integer answer