Question

In hydrogenation reaction at ${25}^{o}C$, it is observed that hydrogen gas pressure falls from $2$ $atm$ to $1.2$ $atm$ in $50$ $min$. Calculate the rate of reaction in molarity per $sec$. $(R=0.0821litre\cdot atm{K}^{-1}{mol}^{-1})$

Answer 1

Rate $=\frac{dP}{dt}=\frac{2-1.2}{50\times 60}$
$=2.666\times {10}^{-4}$ $atm$ $s^{-1}$
$PV=nRT$
$\frac{P}{sec}=\left(\frac{n}{V}\right)\frac{RT}{sec}$
$\left[\frac{n}{V}\right]\frac{1}{sec}=\left[\frac{P}{sec}\right]\frac{1}{RT}$
Rate (molarity/sec) $=\frac{2.666\times {10}^{-4}}{0.0821\times 298}$
$=1.09\times {10}^{-5}mol{litre}^{-1}{s}^{-1}$