Question

The reaction? H2(g)+ I2(g)---> 2HI(g) may occur by the following mechanism: k1(--->) I2<=>2I (fast, equilibrium) k-1(<---) I + I + H2 ---> 2HI (slow) k2

Select the rate law that is predicted by the mechanism.
A. rate = k'[I2] [H2]
B. rate = k'[I2] [H2]2
C. rate = k'[I2]2 [H2]
D. rate = k'[I]2 [H2]
E. rate = k'[I] [H2]

Answer 1

A. `"rate" = k'["H"_2]["I"_2]`

Explanation:

The reaction rate is a direct function of the reactants concentrations,

aA+ bB → cC

`"Rate" = k[A]^x [B]^y`

Where
`a, b, c`: are the stoichiometric coefficients
`x, y`: are the reactant orders

The proposed mechanism is

`bb((1))color(white)(m)"I"_2 stackrelcolor(blue)(k_1)(⇌) "2I (fast)"`
`color(white)(mmmmll)stackrelcolor(blue)(k_text(-1)color(white)(m)`
`bb((2))color(white)(m)"I + I + H"_2 stackrelcolor(blue)(k_2color(white)(mm))(→) "2HI (slow)"`

It illustrates that the reaction is a function of reactant concentrations (only controlled by the slower reaction)

However, `"I"` is an intermediate. We must express the rate law using the concentrations of the reactants.

From 2:

`bb((3))color(white)(m)"rate" = k["I"^2]["H"_2]`

From 1:

`bb((4))color(white)(m)k_1["I"_2] = k_text(-1)["I"]^2`

`bb((5))color(white)(m)["I"]^2 = k_1/k_text(-1)["I"_2]`

Substitute 5 into 3.

`bb((6))color(white)(m)"rate" =(kk_1)/k_text(-1)["I"_2]["H"_2]`

`color(blue)("rate" = k_text(obs)["H"_2]["I"_2])`