Question

Two second-order reactions are given below having identical frequency factors:
(i) $A\to Product$
(ii) $B\to Product$

The $E_a$ for the first reaction is $10.46kJ/mol$ more than that of B. At ${100}^{o}C$, the reaction: (i) is 30% completed after 60 minute when initial concentration of A is 0.1 mol dm${}^{-3}$. How long will it take for reaction. (ii) to reach 70% completion at the same temperature if the initial concentration of B is 0.05 mol/dm${}^3$?

• A. (i) 2.10 min${}^{-1}$ (ii) 22.22 min
• B. (i) 2.10 min${}^{-2}$ (ii) 22.22 min
• C. (i) 2.10 min${}^{-1}$ (ii) 2.22 min
• D. None of these

Answer 1

Answer: (A)

(i) 2.10 min${}^{-1}$ (ii) 22.22 min

For second-order reaction

$k=\frac{x}{ta(a-x)}$

For the reaction i, which is 30% complete after 60 minutes,

$k=\frac{0.1\times 0.3}{60\times 0.1\times (0.1\times 0.7)}=0.0714/M/s$

The Arrhenius equation for the reaction (i) is

$logk=logA-\frac{E_a}{2.303RT}$.....(i)

The Arrhenius equation for the reaction (ii) is

$log{k}^{\prime }=logA-\frac{E_a^{\prime }}{2.303RT}$.....(ii)

Substract equation (ii) from equation (i)

$logk-log{k}^{\prime }=\frac{1}{2.303RT}(E_a^{\prime }-E_a)$

$log0.0714-log{k}^{\prime }=\frac{1}{2.303\times 8.314\times 373}(-10460)$

$log{k}^{\prime }=0.3184$

$k^{\prime }=2.081/min$

The second reaction 70% complete

$t=\frac{x}{ka(a-x)}$

$t=\frac{0.05\times 0.7}{2.081\times 0.05(0.05\times 0.3)}$

$t=22.4min$