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 (∆G2Θ − ∆G1Θ) = RT2 ln x, then the value of x is ___ . [∆G1Θ and ∆G2Θ are standard Gibb’s free energy change for the reaction at temperatures T1 and T2, respectively.]
If T1 = 2T2 and (∆G2Θ − ∆G1Θ) = RT2 ln x, then the value of x is ___ . [∆G1Θ and ∆G2Θ are standard Gibb’s free energy change for the reaction at temperatures T1 and T2, respectively.]
Solution and Explanation
Correct answer is 8.
The given graph depicts the progress of product formation for the reaction A(g)⇌P(g) at two different temperatures, 1T1 and 2T2. The correct answer is 88, which refers to the ratio of 2T2 to 1T1.
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 2T2 favors the product formation (P) more than at temperature 1T1. This suggests that increasing the temperature from 1T1 to 2T2 has shifted the equilibrium towards the products. This is in line with Le Chatelier's principle because the forward reaction (A→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:
1=8KT1KT2=T1T2=8
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Concepts Used:
Law of Chemical Equilibrium
Law of Chemical Equilibrium states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to a power equal to the corresponding stoichiometric coefficients as represented by the balanced chemical equation.
Let us consider a general reversible reaction;
A+B ↔ C+D
After some time, there is a reduction in reactants A and B and an accumulation of the products C and D. As a result, the rate of the forward reaction decreases and that of backward reaction increases.
Eventually, the two reactions occur at the same rate and a state of equilibrium is attained.
By applying the Law of Mass Action;
The rate of forward reaction;
Rf = Kf [A]a [B]b
The rate of backward reaction;
Rb = Kb [C]c [D]d
Where,
[A], [B], [C] and [D] are the concentrations of A, B, C and D at equilibrium respectively.
a, b, c, and d are the stoichiometric coefficients of A, B, C and D respectively.
Kf and Kb are the rate constants of forward and backward reactions.
However, at equilibrium,
Rate of forward reaction = Rate of backward reaction.
Kc is called the equilibrium constant expressed in terms of molar concentrations.
The above equation is known as the equation of Law of Chemical Equilibrium.
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