rate law

Baljaj89

I have a question about how to determine the rate law

In the EK Chem 101, for those of you that have it, the question is #270

given:

Step 1: NO(g) + Br2(g) <--> NOBr2(g) (fast)
Step 2: NOBr2(g) + NO(g) --> 2NOBr (Slow)

Which of the following expressions gives the rate of the reaction?

The answer is k[NOBr2][NO]

On Kaplan Full Length 8, #49 on PS was a similar question that went

given:

B + B --> B2 (fast)
A + B2 --> AB2 (slow)
AB2 + A --> 2AB (fast)

and the answer is k[A]^2

Both rates are determined by the slow step in the mechanism, however Kaplan says that a species that is not in the overall reaction cannot be in the rate law (B2 is an intermediate and is dependant on B + B, therefore ^2 is in the rate law and not [B2]. However in the EK question, NOBr2 is not the in the overall reaction but is in the rate law for the correct answer. Which one is right?

ezsanche

10+ Year Member
5+ Year Member
I have a question about how to determine the rate law

In the EK Chem 101, for those of you that have it, the question is #270

given:

Step 1: NO(g) + Br2(g) <--> NOBr2(g) (fast)
Step 2: NOBr2(g) + NO(g) --> 2NOBr (Slow)

Which of the following expressions gives the rate of the reaction?

The answer is k[NOBr2][NO]

On Kaplan Full Length 8, #49 on PS was a similar question that went

given:

B + B --> B2 (fast)
A + B2 --> AB2 (slow)
AB2 + A --> 2AB (fast)

and the answer is k[A]^2

Both rates are determined by the slow step in the mechanism, however Kaplan says that a species that is not in the overall reaction cannot be in the rate law (B2 is an intermediate and is dependant on B + B, therefore ^2 is in the rate law and not [B2]. However in the EK question, NOBr2 is not the in the overall reaction but is in the rate law for the correct answer. Which one is right?

Step 1: NO(g) + Br2(g) <--> NOBr2(g) (fast)
Step 2: NOBr2(g) + NO(g) --> 2NOBr (Slow)

What is the overall reaction???????

Add the first two steps to get the overall reaction.

NO (g) + Br2(g) + NOBr2(g) + NO(g) <---> NOBr2(g) + 2NOBR

when you condense the equation above down you get ...

Overall Reaction: 2NO (g) + Br2 (g) + <---> 2 NOBR

What is the RATE LAW?

rate law= k[NOBr2][NO]

But wait!!!! This is no good because we have an intermediate (NOBR2) that is not in the overall reaction in our rate law. We want to put the rate law in terms of the reactants. SO what do we do???????????

we use the
Steady state approximation in chemical kinetics.

What is that? Well we assume that the rate of change of a reaction intermediate in a reaction mechanism equals to zero. Basically we are assuming that the rate that the intermediate is being made equals the rate that it is disappearing.

So we have to use more math to figure it out.

So we go back to our two steps. And we write rate laws for each reaction in which the intermediate (NOBR2) is being made and when it is being consumed.

Step 1: NO(g) + Br2(g) <--> NOBr2(g) (fast)

NOBR2 is being made in step one so the rate that NOr2 is being made is written as.

Rate of NOBR2 formation =K1[NO][Br2]

Step 2: NOBr2(g) + NO(g) --> 2NOBr (Slow)
The rate that NOBR2 is being consumed is give in step 2
so we write

RATE of NOBR2 consumption = K2[NOBr2][NO]

Now remember the
Steady state approximation in chemical kinetics states that Rate of consumption equals rate of formation.

so we sets they two reaction equal to each other

K1[NO][Br2] = K2[NOBr2][NO]

Solve for [NOBr2]

we get K1[Br2]/K2= [NOBr2]

Notice the NO cancels out.
To make it a little simpler lets say k1/ k2 = K3
so now we have
K3 [Br2] = [NoBr2]

Now !!!!!!! Remember our rate law for the overall reaction?

1. overall rate law= k[NOBr2][NO]

2. K3 [Br2] = [NoBr2]

lets add the [NoBR2] to the first equation. so we can get rid of the NOBR2 in the overall rate law.

we then get

K(K3)[NO][Br2]= overall rate law

and to make it a little bit cleaner
lets sat K (K3) = K4

then we finally get

K4[NO][Br2] = Overall Rate law

Hence this is what we wanted.

Now both of the rate laws are fine
1. rate law= k[NOBr2][NO]
2.K4[NO][Br2] = rate law

but the second one is better because we have it in terms of the reactants.

SuperSaiyan3

7+ Year Member
you lost me completely. But very good answer nonetheless.

I think it's pretty simple how you form rate laws... you just take the slow step reaction, and all its reactants, and then write it as rate = k [a]^x^y, where x and y are the coefficients in front of a and b respectively... correct me if I am wrong.

crazybob

i understand what ezsanche did since i took chemical kinetics from physical chemistry.

but to the OP, you just have to remember that the slow step is the rate determining step in these cases.