Question

Two second-order reactions are given below having identical frequency factors:
(i) $A\to$ Product
(ii) $B\to$ Product
The $E_a$ for first reaction is $10.46$ kJ mol${}^{-1}$ more than that of $B$ at ${100}^{o}C$. How many time it will take to reach 70% completion at the same temperature if the initial concentration of $B$ is 0.05 mol

litre${}^{-1}$?

• A. $22.22$ min
• B. $11.11$ min
• C. $2.2$ min
• D. $4.4$ min

Answer 1

The correct option is A $22.22$ min

For I: $K=\frac{1}{r}-\frac{x}{a(a-x)}$ (IInd order equation)
Given, $a=0.1M,x=\frac{0.1\times 30}{100}=0.03,t=60min$
$\therefore K_1=\frac{1}{60\times 0.1}\times \frac{0.03}{(0.1-0.03)}=0.07mi{n}^{-1}$
Now, For II: $K_2=\frac{1}{t}-\frac{x}{(a-x)};a=0.05;x=\frac{0.05\times 70}{100}=0.035$
$\therefore t=\frac{1}{2.10\times 0.05}\times \frac{0.035}{(0.05-0.035)}=22.22min$