Question

For a gaseous reaction, $A\left(g\right)⟶$ Product, which one of the following is correct relation among $\frac{dP}{dt},\frac{dn}{dt}$ and $\frac{dc}{dt}$?
($\frac{dP}{dt}=$ Rate of reaction in $atm$ ${sec}^{-1}$; $\frac{dc}{dt}=$ Rate of reaction in molarity ${sec}^{-1}$; $\frac{dn}{dt}=$ Rate of reaction in $mol$ ${sec}^{-1}$)

• A. $\frac{dc}{dt}=\frac{dn}{dt}=-\frac{dP}{dt}$
• B. $-\frac{dc}{dt}=-\frac{1}{V}\frac{dn}{dt}=-\frac{1}{RT}\frac{dP}{dt}$
• C. $\frac{dc}{dt}=\frac{V}{RT}\frac{dn}{dt}=\frac{dP}{dt}$
• D. None of the above

Answer 1

Answer: (B)

$-\frac{dc}{dt}=-\frac{1}{V}\frac{dn}{dt}=-\frac{1}{RT}\frac{dP}{dt}$

$PV=nRT$

$\frac{dP}{dt}.V=RT.\frac{dn}{dt}$=> $\frac{1}{V}\frac{dn}{dt}=\frac{1}{RT}\frac{dP}{dt}$-(1)

Now, PV=nRT or $P=\frac{n}{V}RT$

Concentration $c=\frac{n}{V}$

P=cRT

$\frac{dP}{dt}=\frac{dc}{dt}.RT$=> \frac{dc}{dt}=\frac{1}{RT}\frac{dP}{dt}$

$-\frac{dc}{dt}=-\frac{1}{V}\frac{dn}{dt}=-\frac{1}{RT}\frac{dP}{dt}$