Question

Hydrogenation of vegetable ghee at $25^{\circ} C$ reduces pressure of $H _{2}$ from 2 atm to 1.2 atm in 50 minute. The rate of reaction in terms of molarity per second is -

• A. $1.09 \times 10^{-6}$
• B. $1.09 \times 10^{-5}$
• C. $1.09 \times 10^{-7}$
• D. $1.09 \times 10^{-9}$

Answer 1

Answer: (B)

The change in molarity $=\frac{ n }{ V }=\frac{\Delta P }{ R T }$

$=\frac{0.8}{0.0821 \times 298}=0.0327$

rate of reaction $=\frac{0.0327}{50 \times 60}$

$=1.09 \times 10^{-5}\, mol\,$ litre $^{-1}\, \sec ^{-1}$