Question

Given the following balanced equation, determine the rate of reaction with respect to [Cl2]. If the rate of disappearance of Cl2 is 4.24 × 10–2 M/s, what is the rate of formation of NO? 2 NO(g) + Cl2(g) → 2 NOCl(g)

Answer 1

Answer : The rate of formation of $NOCl$ is, $8.48\times 10^{-2}M/s$

Explanation : Given,

Rate of disappearance of $Cl_2$ = $4.24\times 10^{-2}M/s$

The given rate of reaction is,

$2NO(g)+Cl_2(g)\rightarrow 2NOCl$

The expression for rate of reaction :

$\text{Rate of disappearance}=-\frac{1}{2}\frac{d[NO]}{dt}=-\frac{d[Cl_2]}{dt}$

$\text{Rate of formation}=\frac{1}{2}\frac{d[NOCl]}{dt}$

From this we conclude that,

$\frac{1}{2}\frac{d[NOCl]}{dt}=-\frac{d[Cl_2]}{dt}$

$\frac{1}{2}\frac{d[NOCl]}{dt}=-\frac{d[Cl_2]}{dt}$

$\frac{d[NOCl]}{dt}=2\times \frac{d[Cl_2]}{dt}$

Now put the value of rate of disappearance of $Cl_2$, we get:

$\frac{d[NOCl]}{dt}=2\times (4.24\times 10^{-2}M/s)=8.48\times 10^{-2}M/s$

Therefore, the rate of formation of $NOCl$ is, $8.48\times 10^{-2}M/s$