Which step would be the rate-determining step?

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The entire reaction is

H2+I22HI

and the rate law is rate=k[H2][I2]. The overall order for this 2, but none of these steps have a molecularity of 2.

I'd appreciate some help!

1 Answer
Mar 17, 2018

Well, the first step generates I(g), which is used in the second step. I2(g) is quite stable compared to I(g), and needs light to become I(g).

I would say that the second step is slow, but that it is due to not enough I(g) available as a result of not enough I2(g) bonds cleaved.


But let's suppose it was just the second step, because it contains H2 as one of the reactants. Then the initially proposed rate law becomes:

r(t)=k2[I]2[H2]

However, I is an intermediate and [I] must be replaced with [I2], since I2 is the other reactant. Now, step 1 is incorrectly written. It must be an equilibrium to agree with the proposed rate law...

I2(g)k1 2I(g)
k1

2I(g)+H2(g)k2 H2I2(g)

H2I2(g)k3 2HI(g)

Then if we assume the fast equilibrium approximation, we get the equilibrium expression to be:

r1(t)=k1[I2]=r1(t)=k1[I]2

Kk1k1=[I]2[I2]

As a result,

[I]2=k1k1[I2]

And this gives:

r(t)=k2k1k1[I2][H2]

And this agrees with the proposed rate law, where k=k2k1k1.