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 is 10 min. The rate of the reaction is $3\times {10}^{-x}Mmi{n}^{-1}$. Find the value of x.

[Given R = 0.08 L atm $K^{-1}mo{l}^{-1}$]

Answer 1

$Rate=\frac{(10-2)atm}{10min}=0.8atmmi{n}^{-1}$
Rate in $Mmi{n}^{-1}$ is given by:

Rate $=\frac{\Delta \left(n/v\right)}{\Delta t}Mmi{n}^{-1}=\frac{\Delta P}{\Delta t}\times \frac{1}{RT}$
$\frac{0.8}{0.08\times 300}=0.03Mmi{n}^{-1}$

Rate of the reaction $3\times {10}^{-x}=3\times {10}^{-2}$

we get, x=2

The correct answer is 2.