The rate constant for the forward reaction A (g) \underset{}{\rightleftharpoons} 2B(g) is 1.5 × 10–3 s–1 at 100 K. If 10–5 mol of A and 100 mol of B are present in a 10 L vessel at equilibrium then rate constant for the backward reaction at this temperature is

Engineering

Chemistry

Opposed or Reversible Elementary Reaction

Question

The rate constant for the forward reaction

A (g) $\underset{}{⇌}$ 2B(g)

is 1.5 × 10^–3 s^–1 at 100 K. If 10^–5 mol of A and 100 mol of B are present in a 10 L vessel at equilibrium then rate constant for the backward reaction at this temperature is

1.50× 10⁴ L mol^–1 s^–1

1.5 × 10¹¹ L mol^–1 s^–1

None

1.5 × 10¹⁰ L mol^–1 s^–1

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

For a reversible reaction A ⇌ 2B at equilibrium, the rate of the forward reaction equals the rate of the backward reaction. Therefore, k_f[A]_eq = k_b[B]_eq², where k_f and k_b are the rate constants.

Given: k_f = $1.5 \times 10^{- 3} s^{- 1}$

First, calculate the equilibrium concentrations. Volume is 10 L.

[A]_eq = (10^-5 mol) / (10 L) = $10^{- 6} mol L$ ⁻¹

[B]_eq = (100 mol) / (10 L) = $10 mol L$ ⁻¹

Now solve for k_b: k_b = k_f [A]_eq / [B]_eq²

$k_{b} = \frac{(1.5 \times 10^{- 3}) (10^{- 6})}{{(10)}^{2}} = \frac{1.5 \times 10^{- 9}}{100} = 1.5 \times 10^{- 11}$

The units for k_b must be L mol^–1 s^–1 for a second-order reverse reaction.

Final Answer: 1.5 × 10¹¹ L mol^–1 s^–1