Question

At ${27}^{\circ }$C it was observed in the hydrogenation of a reaction, the pressure of $H_2\left(g\right)$ decreases from $10$ atm to $2$ atm in $10$ min. Calculate the rate of reaction in M $mi{n}^{-1}$ (Given R = $0.08$ L atm $K^{-1}mo{l}^{-1}$) :

• A. $0.02$
• B. $0.03$
• C. $0.04$
• D. $0.05$

Answer 1

Answer: (B)

$0.03$

$Rate=\frac{(10-2)atm}{10min}$

$=0.8atmmi{n}^{-1}$

Rate in M $mi{n}^{-1}$ is given by:

Rate $=\frac{\Delta \left(n/v\right)}{\Delta t}Mmi{n}^{-1}$

$=\frac{\Delta P}{\Delta t}\times \frac{1}{RT}$

$\frac{0.8}{0.08\times 300}$

$=0.03Mmi{n}^{-1}$
Option B is correct.