Question

A certain reaction $\text{A} \rightarrow \text{B}$ follows the given concentration (molarity)-time graph. Which of the following statement is true?

• A. The reaction is of second order with respect to A
• B. The rate for this reaction at 40 second will be approximately $3.5\times 10^{- 3}Ms^{- 1}$
• C. The rate for this reaction at 80 second will be $1.75\times 10^{- 3}M \, s^{- 1}$
• D. The [B] will be 0.25 M at 60 second

Answer 1

Answer: (B)

Time required for concentration to change from $0.4$ to $0.2$ is same as the time required for the concentration to change from $0.2$ to $0.1$ , it means $\text{t}_{\text{1/2}}$ is independent of initial concentration, i.e. it is a first order reaction.
And from the graph $\text{t}_{\text{1/2}} = 20 \, \text{s}$
$\therefore \, \text{K} = \frac{0.693}{20} \text{s}^{- 1}$
$\therefore \text{Rate} = \text{k[A]}$
At $40$ second $\text{[A]} = 0.1$
$\therefore \, \text{rate} = \frac{0.693}{20}$
$= 3.46 \times 10^{- 3} \, \text{Ms}^{- 1}$
$= 3.5 \times 10^{- 3} \text{s}^{- 1}$