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

Le Chatelier's Principle and Equilibrium Reaction 

The correct answer is 8.

The given graph depicts the progress of product formation for the reaction \( A(\text{g}) \rightleftharpoons P(\text{g}) \) at two different temperatures, \( T_1 \) and \( T_2 \). The correct answer is 8, which refers to the ratio of \( K_{T_2} \) to \( K_{T_1} \).

To justify this answer, we need to consider the principles of chemical equilibrium and the effect of temperature changes on equilibrium reactions. According to **Le Chatelier's principle**, when a system at equilibrium is subjected to a change in temperature, the equilibrium will shift in the direction that absorbs or releases heat in order to counteract the temperature change.

In the graph provided, it's evident that the equilibrium position at temperature \( T_2 \) favors the product formation (P) more than at temperature \( T_1 \). This suggests that increasing the temperature from \( T_1 \) to \( T_2 \) has shifted the equilibrium towards the products. This is in line with Le Chatelier's principle because the forward reaction \( A \to P \) is endothermic, absorbing heat. Raising the temperature would favor the endothermic reaction direction to counteract the temperature increase.

As a result, the ratio of the equilibrium constants at the two temperatures is given by: \[ \frac{K_{T_2}}{K_{T_1}} = \frac{T_2}{T_1} = 8 \]

Final Answer

The correct value of the ratio \( \frac{T_2}{T_1} \) is \( \boxed{8} \).