Which step would be the rate-determining step?

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Which step would be the rate-determining step?

The entire reaction is

$H_{2} + I_{2} \to 2 H I$ $H_2 + I_2 -> 2HI$

and the rate law is $rate = k [H_{2}] [I_{2}]$ $"rate" = k[H_2][I_2]$ . The overall order for this 2, but none of these steps have a molecularity of 2.

I'd appreciate some help!

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Answer 1 · 2018-03-17T22:56:57

Well, the first step generates $I (g)$ $"I"(g)$ , which is used in the second step. $I_{2} (g)$ $"I"_2(g)$ is quite stable compared to $I (g)$ $"I"(g)$ , and needs light to become $I (g)$ $"I"(g)$ .

I would say that the second step is slow, but that it is due to not enough $I (g)$ $"I"(g)$ available as a result of not enough $I_{2} (g)$ $"I"_2(g)$ bonds cleaved.

But let's suppose it was just the second step, because it contains $H_{2}$ $"H"_2$ as one of the reactants. Then the initially proposed rate law becomes:

$r (t) = k_{2} {[I]}^{2} [H_{2}]$ $r(t) = k_2[I]^2[H_2]$

However, $I$ $"I"$ is an intermediate and $[I]$ $[I]$ must be replaced with $[I_{2}]$ $[I_2]$ , since $I_{2}$ $"I"_2$ is the other reactant. Now, step 1 is incorrectly written. It must be an equilibrium to agree with the proposed rate law...

$I_{2} (g) k_{1} ⇌ 2 I (g)$ $"I"_2(g) stackrel(k_1" ")(rightleftharpoons) 2"I"(g)$
$_{- 1}^{k_{- 1}}$ $""^(" "" "" "k_(-1))$

$2 I (g) + H_{2} (g) k_{2} - - \to H_{2} I_{2} (g)$ $2"I"(g) + "H"_2(g) stackrel(k_2" ")(->) "H"_2"I"_2(g)$

$H_{2} I_{2} (g) k_{3} - - \to 2 HI (g)$ $"H"_2"I"_2(g) stackrel(k_3" ")(->) 2"HI"(g)$

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

$r_{1} (t) = k_{1} [I_{2}] = r_{- 1} (t) = k_{- 1} {[I]}^{2}$ $r_1(t) = k_1[I_2] = r_(-1)(t) = k_(-1)[I]^2$

$\Rightarrow K \equiv \frac{k_{1}}{k_{- 1}} = \frac{{[I]}^{2}}{[I_{2}]}$ $=> K -= k_1/(k_(-1)) = ([I]^2)/([I_2])$

As a result,

${[I]}^{2} = \frac{k_{1}}{k_{- 1}} [I_{2}]$ $[I]^2 = k_1/(k_(-1))[I_2]$

And this gives:

$r (t) = \frac{k_{2} k_{1}}{k_{- 1}} [I_{2}] [H_{2}]$ $color(blue)(r(t) = (k_2k_1)/(k_(-1))[I_2][H_2])$

And this agrees with the proposed rate law, where $k = \frac{k_{2} k_{1}}{k_{- 1}}$ $k = (k_2k_1)/(k_(-1))$ .

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Which step would be the rate-determining step?

The entire reaction is

$H_{2} + I_{2} \to 2 H I$ $H_2 + I_2 -> 2HI$

and the rate law is $rate = k [H_{2}] [I_{2}]$ $"rate" = k[H_2][I_2]$ . The overall order for this 2, but none of these steps have a molecularity of 2.

I'd appreciate some help!

1 Answer

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Impact of this question

Science

Math

Humanities

... and beyond

Which step would be the rate-determining step?

The entire reaction is H2+I2→2HIH_2 + I_2 -> 2HI and the rate law is rate=k[H2][I2]"rate" = k[H_2][I_2]. The overall order for this 2, but none of these steps have a molecularity of 2. I'd appreciate some help!

1 Answer

Related questions

Impact of this question

The entire reaction is

$H_{2} + I_{2} \to 2 H I$ $H_2 + I_2 -> 2HI$

and the rate law is $rate = k [H_{2}] [I_{2}]$ $"rate" = k[H_2][I_2]$ . The overall order for this 2, but none of these steps have a molecularity of 2.

I'd appreciate some help!