Question

During the hydrogenation of vegetable oil at ${25}^{0}C$, the pressure of $H_2$ reduces from 2 atmospheres to 1.2 atmospheres in 50 minutes. 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. $1.09\times {10}^5$

Answer 1

Answer: (B)

$1.09\times {10}^{-5}$

For hydrogenation reaction the rate of reaction is written as $r=-\frac{d\left[H_2\right]}{dt}$.

Assuming the hydrogen an ideal gas $P=CRT$

Therefore $C=\frac{P}{RT}$

Hence $\Delta C=\frac{\Delta P}{RT}$

$\Delta P=-0.8$.

So $\Delta C=\frac{-0.8}{0.0821\times 298}$

So,

$r=-\frac{\frac{-0.8}{0.0821\times 298}}{50\times 60}=1.09\times {10}^{-5}$

$r=1.09\times {10}^{-5}$

Option B is correct.