Question

A certain reaction $A\to B$ follows the given concentration (Molarity) - time graph. Which of the following statements is/are true ?

• A. The reaction is second order with respect to A.
• B. The rate for this reaction at 20 s willbe $7\times {10}^{-3}M{s}^{-1}$
• C. The rate for this reaction at 80 s will be $1.75\times {10}^{-3}M{s}^{-1}$
• D. The [B] will be 0.35 M at t = 60 s.

Answer 1

Answer: (B), (D)

The correct options are
B The rate for this reaction at 20 s willbe $7\times {10}^{-3}M{s}^{-1}$
D The [B] will be 0.35 M at t = 60 s.

Use initial rate law method, the reaction will be first order.

So, $k=\frac{2.303}{t}log\frac{\alpha }{\alpha -x}$

At $t=20$ s, $k_1=\frac{2.303}{20}log\left(\frac{0.4}{0.2}\right)$ and at $t=40$ s, $k_2=\frac{2.303}{40}log\left(\frac{0.4}{0.1}\right)=k_1$

Assumption is correct $(\therefore k_1=k_2)$

Rate at $20s=k\left[A\right]$ = $\frac{0.693}{20}\times 0.2$ = $0.0063\approx 7\times {10}^{-3}M{s}^{-1}$

Also, from figure it can be seen that, half life $=t_{1/2}$ $=20$ sec