The reaction 2NO(g) + Cl2(g) --> 2NOCl(g) obeys the rate law rate = k [NO]2 [Cl2]....

The reaction 2NO(g) + Cl₂(g) --> 2NOCl(g) obeys the rate law rate = k [NO]² [Cl₂]. The following mechanism is proposed:

NO (g) + Cl₂ (g) --> NOCl₂(g)

NOCl₂(g) + NO (g) -->2NOCl (g)

a) What would the rate law be if the first step was rate determining?

b) Based on the observed rate law, what can be concluded about the relative rates of the 2 reactions?

Solutions

Expert Solution

Rate laws must be determined experimentally, however if the first step is a bimolecular elementary step, then the rate law would be: rate = k[NO]^2 [Cl2].

The observed rate law appears to be consistent with the first step being fast and the second step being slow. This can be proven as follows:

Equation 1) NO + Cl2 <---> NOCl2 (fast)

Equation 2) NOCl2 + NO --> 2NOCl (slow)
2NO + Cl2 --> 2NOCl (overall equation)

rate = k2 [NOCl2] [NO] The slow step determines the rate law, however intermediates cannot be in the rate law, so we have to substitute for it.

The fast step is at equilibrium, so the rate in the forward reaction is equal to the rate in the reverse rxn.
k1 [NO] [Cl2] = k-1 [NOCl2]

Solve for the intermediate concentration.

[NOCl2] = k1/k-1 [NO] [Cl2] (substitute into the rate law)
rate = k2k1/k-1 [NO]^2 [Cl2]

The assumption that the fast and slow steps are what they are is because it matches the experimental rate law.

rate = k2 [NOCl2] [NO]
rate = k[NO]^2 [Cl2]

queen_honey_blossom answered 3 months ago

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