Question on Chemistry Destroyer (problem 70)

[MolarBear541]

NOCl is produced by the following reaction: 2NO + Cl2 --> 2NOCL
A scientist proposed the following mechanism
NO + NO --> N2O2 (FAST)
N2O2 + Cl2 --> 2NOCl (slow)

Which rate law is consistent with this reaction mechanism?
Rate=k[NO]^2
Rate= k[NO]^2 [CL2]
Rate= k[N2O2][Cl2]
Rate= k[N2O2]
Rate=k[Cl2]
The answer is Rate= k[NO]^2 [CL2]. i read the explanation but i just cant make sense of it. Could anyone enlighten me please

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[PBPL]

you have to notice that n2o2 is not in the original formula... therefore you have to go and 'solve' for it... as you see the n202 is present in the fast reaction... (but...make sure for rate you only you SLOW rxn)... you have to solve for n2o2 in order to substituted it into the rate. if you have any questions let me know! hope it helps!

 

[dl2501]

you have to notice that n2o2 is not in the original formula... therefore you have to go and 'solve' for it... as you see the n202 is present in the fast reaction... (but...make sure for rate you only you SLOW rxn)... you have to solve for n2o2 in order to substituted it into the rate. if you have any questions let me know! hope it helps!

Hey! I see the breakdown of it, but can you explain conceptually a bit more? I just thought that if you are given a mechanism, then it must me the slow step that is used to formulate the rate law? IDK if you wanted chad's videos, but he just showed the rate law being made by the slow step. Thank you so much!

 

[dl2501]

My favorite! Took a whole pchem class lol. Okay so:
Fast rxn:
forward rate = kf[NO]2 (since there are two NOs)
reverse rate= kr[N2O2]
Slow rxn:
rate = ks[N2O2][Cl2]
The rate law would equal the slow rxn's rate law. But see, N2O2 is an intermediate. Therefore, it cannot be a part of the rate law. So, we have to use what we're given from the fast rxn.
The rate laws from the fast rxn must equal each other, they're from the same reaction. So, kf[NO]2 = kr[N2O2]. Now let's solve for [N2O2] from this reaction ---> you get: [N2O2] = (kf/kr)[NO]2.
Now plug this into the slow rxn ---> rate = ((kf*ks)/kr) [NO]2[Cl2]
All the k's can be combined ----> rate = k[NO2][Cl2]

Hope this helps!

Thank you for all of that! I just want to quickly clarify... this question relates the kinetics constant K concept with that of the equilibrium constant Keq, such that Keq=kf/kr? (and of course realizing that N2O2 is an intermediate step in the reaction).

I knew the relationship between the two, I just never learned when it would be applicable...

Thank you so much, I really appreciate it!

 

[dl2501]

I think Keq would equal ((kf*ks)/kr)

Got it, thanks so much!