JEE Advanced 2023 Paper 1, Chemistry Section 3 Q3: Law of Chemical Equilibrium And Equilibrium Constant

JEE Advanced2023Paper 1Chemistry Section 3

The plot of $lo g k_{f}$ versus $1/ T$ for a reversible reaction $A (g) ⇌ P (g)$ is shown.

Pre-exponential factors for the forward and backward reactions are $1 0^{15} s^{- 1}$ and $1 0^{11} s^{- 1}$ , respectively. If the value of $lo g K$ for the reaction at $500$ K is $6$ , the value of $∣ lo g k_{b} ∣$ at $250$ K is ______.

[ $K$ = equilibrium constant of the reaction, $k_{f}$ = rate constant of forward reaction, $k_{b}$ = rate constant of backward reaction]

Your answer

Solution

Forward activation energy from the plot. Arrhenius: $lo g k_{f} = lo g A_{f} - \frac{E _{f}}{2.303 R T}$ . At $1/ T = 0.002$ , $lo g k_{f} = 9$ , with $A_{f} = 1 0^{15}$ : $9 = 15 - \frac{E _{f}}{2.303 R} (0.002) \Rightarrow \frac{E _{f}}{2.303 R} = \frac{6}{0.002} = 3000 K .$

Use the equilibrium relation at 500 K. $K = \frac{k _{f}}{k _{b}} = \frac{A _{f}}{A _{b}} exp [- \frac{E _{f} - E _{b}}{R T}]$ , so $lo g K = lo g (\frac{A _{f}}{A _{b}}) - \frac{E _{f} - E _{b}}{2.303 R T} .$ With $lo g (A_{f} / A_{b}) = 4$ and $lo g K = 6$ at $T = 500$ K: $6 = 4 - \frac{E _{f} - E _{b}}{2.303 R \cdot 500} \Rightarrow \frac{E _{b} - E _{f}}{2.303 R} = 1000 K .$ Hence $\frac{E _{b}}{2.303 R} = 3000 + 1000 = 4000$ K.

Compute $lo g k_{b}$ at 250 K. $lo g k_{b} = lo g A_{b} - \frac{E _{b}}{2.303 R T} = 11 - \frac{4000}{250} = 11 - 16 = - 5.$

Therefore $∣ lo g k_{b} ∣ = 5$ .

Answer: 5.

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JEE Advanced 2023 Paper 1, Chemistry Section 3 Q3: Law of Chemical Equilibrium And Equilibrium Constant

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