Question

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), T₁ and T₂, is shown in the figure: 

If T₁ = 2T₂ and (∆G₂^Θ − ∆G₁^Θ) = RT₂ ln x, then the value of x is ___ .
[∆G₁^Θ and ∆G₂^Θ are standard Gibb’s free energy change for the reaction at temperatures T₁ and T₂, respectively.]

Answer 1

Correct Answer: 8

Gibbs Free Energy Calculation 

Given: The chemical equilibrium for the reaction:

\(A(g) \rightleftharpoons P(g)\)

Initially: 6 moles of A and 0 moles of P.

At equilibrium: \(6 - a\) moles of A and \(a\) moles of P.

Solution:

At temperature \( T_1 \), we are given \( a = 4 \). Therefore, the equilibrium constant \( K_{eq1} \) is:

\(K_{eq1} = \frac{4}{2} = 2\)

At temperature \( T_2 \), we are given \( a = 2 \). Therefore, the equilibrium constant \( K_{eq2} \) is:

\(K_{eq2} = \frac{2}{4} = \frac{1}{2}\)

Using the equation for Gibbs free energy change:

\(\Delta G^\circ_1 = -RT \ln(K_{eq1})\)

\(\Delta G^\circ_1 = -2RT \ln(K_{eq1}) \quad \text{[Given: } T_1 = 2T_2 \text{]}\)

\(\Delta G^\circ_2 = -RT \ln(K_{eq2})\)

Now, we subtract the two equations to find the difference in Gibbs free energy:

\(\Delta G^\circ_2 - \Delta G^\circ_1 = RT_2 \ln \left( \frac{(K_{eq1})^2}{K_{eq2}} \right)\)

\(= RT_2 \ln \left( \frac{2^2}{\frac{1}{2}} \right)\)

\(= RT_2 \ln 8\)

Therefore, \( \Delta G^\circ = RT \ln 8 \), and since \( x = 8 \), we conclude that:

The value of \( \Delta G^\circ \) is given by \( RT \ln x \) where \( x = 8 \).

The correct answer is \( x = 8 \).