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. Calculate the rate of reaction in M $mi{n}^{-1}$ (Given R = 0.08 L atm $K^{-1}mo{l}^{-1}$).

If it is $0.0x$, then the value of $x$ is :

Answer 1

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

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

= $0.03Mmi{n}^{-1}$

So, the correct answer is $3$