Question

For a gaseous reaction, the rate is expressed interms of $\frac{dP}{dt}$ in place of $\frac{dC}{dt}$ or $\frac{dn}{dt}$, where $C$ is concentration, $n$ is number of moles and ${}^{\prime }{P}^{\prime }$ ispressure of reactant. The three are related as:

• A. $\left[\frac{dP}{dt}\right]=\frac{RT}{V}\left[\frac{dn}{dt}\right]=\left[\frac{dC}{dt}\right]$
• B. $\frac{1}{RT}\left[\frac{dP}{dt}\right]=\frac{1}{V}\left[\frac{dn}{dt}\right]=\left[\frac{dC}{dt}\right]$
• C. $\left[\frac{dP}{dt}\right]=\left[\frac{dn}{dt}\right]=\left[\frac{dC}{dt}\right]$
• D. None of these

Answer 1

The correct option is B $\frac{1}{RT}\left[\frac{dP}{dt}\right]=\frac{1}{V}\left[\frac{dn}{dt}\right]=\left[\frac{dC}{dt}\right]$

Rate expression $=\frac{dP}{dt}$
From ideal gas equation, $P=\frac{n}{V}RT$
$\therefore C=\frac{n}{V}$ and $C=\frac{P}{RT}$
$\frac{dC}{dt}=\frac{d}{dt}\left(\frac{n}{V}\right)=\frac{1}{V}\left(\frac{dn}{dt}\right)$
$\frac{d}{dt}\left(\frac{P}{RT}\right)=\frac{1}{RT}\left(\frac{dP}{dt}\right)$