# What is the answer? With an explanation, please.

##### 1 Answer
Mar 14, 2018

The essence of these problems is determining the rate determining step: which the rate law is derived from.

Consider,

$\text{Rate} = k {\left[N O C l\right]}^{2}$

In the first mechanism, the slow step is the RDS, and there is only one mole of $N O C l$. So, that is wrong.

In the second mechanism, the slow step is the RDS, and there are two moles of $N O C l$. This may be a mechanism for this reaction.

In the third mechanism, we'll need to do some analysis,

$N O C l r i g h t \le f t h a r p \infty n s N O + C l$ fast, eq.

This isn't the RDS, but it makes the intermediate in the RDS (the slow step after). So let's substitute,

${k}_{1} \left[N O C l\right] = {k}_{- 1} \left[N O\right] \left[C l\right]$
$\implies \left[C l\right] = \frac{{k}_{1} \left[N O C l\right]}{{k}_{- 1} \left[N O\right]}$

for the intermediate in the RDS and see what we come up with.

Consider,

$N O C l + C l \to N O + C {l}_{2}$ slow (RDS)

where,

$\text{Rate} = \frac{{k}_{1} \left[N O C l\right]}{{k}_{- 1} \left[N O\right]} \cdot {k}_{2} \left[N O C l\right] = \frac{{k}_{1} {k}_{2} {\left[N O C l\right]}^{2}}{{k}_{- 1} \left[N O\right]}$

Hence, the only reliable mechanism we have that matches that rate law is $d .$ or the second mechanism.