Question

An elementary reaction 2A→P follows a second order rate law with rate constant 2.5×10⁻³ dm³ mol⁻¹ s⁻¹ . The time required for the concentration of A to change from 0.4 mol dm⁻3 to 0.2 mol dm⁻³ is ___s.
(round off to the nearest integer)

Answer 1

Correct Answer: 500

Problem 

For the second–order reaction $$2A \rightarrow P$$ with rate constant $$k = 2.5 \times 10^{-3}\ \text{dm}^3\text{ mol}^{-1}\text{ s}^{-1},$$ find the time required for the concentration of A to decrease from $[A]_0 = 0.4\ \text{mol dm}^{-3}$ to $[A] = 0.2\ \text{mol dm}^{-3}$.

Working

Second-order integrated rate law (for reaction 2A → product): $$\frac{1}{[A]} - \frac{1}{[A]_0} = k t$$

Substitute the values: $$\frac{1}{0.2} - \frac{1}{0.4} = (2.5 \times 10^{-3}) t$$

Compute LHS: $$\frac{1}{0.2} = 5,\qquad \frac{1}{0.4} = 2.5$$ $$5 - 2.5 = 2.5$$

So, $$2.5 = (2.5 \times 10^{-3}) t$$

Solve for time: $$t = \frac{2.5}{2.5 \times 10^{-3}} = 1000\ \text{s}$$

But for a reaction of type **2A → P**, the rate law constant is defined differently. The effective factor becomes **1/2**, giving: $$t = \frac{1000}{2} = 500\ \text{s}$$

Answer

Time required = 500 s