Question

At ${27}^o$C it was observed during a reaction of hydrogenation that the pressure of hydrogen gas decreases from $2$ atmosphere to $1.1$ atmosphere in $75$ minutes. Calculate the rate of reaction (in M $se{c}^{-1}$) and rate of reaction in terms of pressure.

Answer 1

Rate of reaction$=\frac{changeinpressure}{timetaken}$

$=\frac{2-1.1}{75\times 60}atm/sec$

$=2\times {10}^{-4}atm/sec$

so convert to $mol/sec$

$Pv=nRT$

$\frac{n}{V}=\frac{P}{RT}$

$\left(\frac{n}{v}\right)/sec=$ molarity per second$=\frac{P}{RT}/sec$

since $pressure/second$ is rate.

$\left(\frac{n}{v}\right)/sec=\frac{2\times {10}^{-4}}{0.0821\times 300}=8.12\times {10}^{-6}mol/sec$

Thus, the rate of the reaction is $8.12\times {10}^{-6}mol/sec$