Question

A certain reaction $\text{A}\to \text{B}$ 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 $3.5\times 10^{-3}M{s}^{-1}$
• 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 $\text{[A]}=0.1$
$\therefore \text{rate}=\frac{0.693}{20}$
$=3.46\times 10^{-3}{\text{Ms}}^{-1}$
$=3.5\times 10^{-3}{\text{s}}^{-1}$