For a second‑order reaction $2 A \to products$ , the initial concentration of A is $0.050 M$ and the rate constant is $k = 0.20 M^{- 1} s^{- 1}$ . Calculate the time (in seconds) required for the concentration of A to decrease to $0.020 M$ .

medium 3 marks Class 12 · Chemistry · Chemical Kinetics

Answer

t = 150 s

The correct answer is $t = 150 s$ — Hence the required time is $150 s$ .

Step-by-step solution

The integrated rate law for a second‑order reaction is $\frac{1}{[ A ]} - \frac{1}{[ A ] _{0}} = k t$ .
Insert the data: $\frac{1}{0.020} - \frac{1}{0.050} = k t$ .
Calculate the left side: $\frac{1}{0.020} = 50$ , $\frac{1}{0.050} = 20$ , difference $= 30 M^{- 1}$ .
Solve for $t$ : $t = \frac{30}{0.20} = 150 s$ .
Hence the required time is $150 s$ .

Common mistake: Using the first‑order expression

t = \frac{l n ([ A ] _{0} / [ A ])}{k}

instead of the second‑order form.

Concept tested: Second‑order integrated rate law