Calculate Standard Gibbs Energy Difference Constant from Equilibrium Progress

For the given reversible reaction: $\text{A (g)} \rightleftharpoons \text{P (g)}$

The reaction takes place in a $1\text{ L}$ flask, and the initial amount of reactant $\text{A}$ is $6\text{ moles}$. Therefore, the initial concentration of $\text{A}$ is: $[\text{A}]_0 = \frac{6\text{ moles}}{1\text{ L}} = 6\text{ mol L}^{-1}$

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Step 1: Calculate equilibrium constants at temperatures $\text{T}_1$ and $\text{T}_2$

From the graph showing the progress of product formation:

1. At temperature $\text{T}_1$:

   • The equilibrium concentration of product $\text{P}$ is $[\text{P}]_{eq, 1} = 4\text{ mol L}^{-1}$.
   • The equilibrium concentration of reactant $\text{A}$ is $[\text{A}]_{eq, 1} = [\text{A}]_0 - [\text{P}]_{eq, 1} = 6 - 4 = 2\text{ mol L}^{-1}$.
   • The equilibrium constant $K_1$ is given by: $K_1 = \frac{[\text{P}]_{eq, 1}}{[\text{A}]_{eq, 1}} = \frac{4}{2} = 2$

2. At temperature $\text{T}_2$:

   • The equilibrium concentration of product $\text{P}$ is $[\text{P}]_{eq, 2} = 2\text{ mol L}^{-1}$.
   • The equilibrium concentration of reactant $\text{A}$ is $[\text{A}]_{eq, 2} = [\text{A}]_0 - [\text{P}]_{eq, 2} = 6 - 2 = 4\text{ mol L}^{-1}$.
   • The equilibrium constant $K_2$ is given by: $K_2 = \frac{[\text{P}]_{eq, 2}}{[\text{A}]_{eq, 2}} = \frac{2}{4} = \frac{1}{2}$

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Step 2: Relate Standard Gibbs Free Energy Change ($\Delta G^\ominus$) with Equilibrium Constants

The standard Gibbs free energy change for a reaction at temperature $T$ is given by: $\Delta G^\ominus = -\text{RT} \ln K$

• At temperature $\text{T}_1$: $\Delta G_1^\ominus = -\text{RT}_1 \ln K_1 = -\text{RT}_1 \ln 2$

• At temperature $\text{T}_2$: $\Delta G_2^\ominus = -\text{RT}_2 \ln K_2 = -\text{RT}_2 \ln \left(\frac{1}{2}\right) = \text{RT}_2 \ln 2$

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Step 3: Calculate $(\Delta G_2^\ominus - \Delta G_1^\ominus)$

Subtracting $\Delta G_1^\ominus$ from $\Delta G_2^\ominus$: $\Delta G_2^\ominus - \Delta G_1^\ominus = \text{RT}_2 \ln 2 - (-\text{RT}_1 \ln 2) = \text{RT}_2 \ln 2 + \text{RT}_1 \ln 2$

Given that $\text{T}_1 = 2\text{T}_2$, substitute $\text{T}_1$ into the equation: $\Delta G_2^\ominus - \Delta G_1^\ominus = \text{RT}_2 \ln 2 + \text{R}(2\text{T}_2) \ln 2$ $\Delta G_2^\ominus - \Delta G_1^\ominus = 3 \text{RT}_2 \ln 2$ $\Delta G_2^\ominus - \Delta G_1^\ominus = \text{RT}_2 \ln(2^3)$ $\Delta G_2^\ominus - \Delta G_1^\ominus = \text{RT}_2 \ln 8$

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Step 4: Find the value of $x$

We are given that: $\Delta G_2^\ominus - \Delta G_1^\ominus = \text{RT}_2 \ln x$

Comparing the two expressions: $\text{RT}_2 \ln x = \text{RT}_2 \ln 8 \implies x = 8$

Final Answer: The value of $x$ is $8$.