Question

For the reaction: $2NO+C{l}_2\to 2NOCl$, the following mechanism was proposed on the basis of experimental observation.
$NO+C{l}_2\stackrel{Fast}{\rightleftharpoons }NOC{l}_2$
$NOC{l}_2+NO\stackrel{Slow}{\to }2NOBr$
The order of reaction is:

• A. 2nd order
• B. pseudo 2nd order
• C. 3rd order
• D. first order

Answer 1

Answer: (C)

3rd order

From the slow step, the rate law for the reaction is rate $=k^{\prime }\left[NOC{l}_2\right]\left[NO\right]$.....(1)

$NOC{l}_2$ is an intermediate and its concentration is obtained from fast step (equilibrium step)

$\left[NOC{l}_2\right]=K\left[NO\right]\left[C{l}_2\right]$......(2)

Substitute (2) in (1)
$rate=k^{\prime }K\left[NO\right]\left[C{l}_2\right]\left[NO\right]$

Hence, rate $=k\left[NO{]}^2\right[Cl]$

The reaction is of second order in NO and first order in $C{l}_2$. The overall reaction order is $2+1=3$