# 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]

Feb 27, 2018

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

#### Explanation:

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

aA+ bB → cC

$\text{Rate} = k {\left[A\right]}^{x} {\left[B\right]}^{y}$

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

The proposed mechanism is

$\boldsymbol{\left(1\right)} \textcolor{w h i t e}{m} \text{I"_2 stackrelcolor(blue)(k_1)(⇌) "2I (fast)}$
color(white)(mmmmll)stackrelcolor(blue)(k_text(-1)color(white)(m)
$\boldsymbol{\left(2\right)} \textcolor{w h i t e}{m} \text{I + I + H"_2 stackrelcolor(blue)(k_2color(white)(mm))(→) "2HI (slow)}$

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

However, $\text{I}$ 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:

$\boldsymbol{\left(4\right)} \textcolor{w h i t e}{m} {k}_{1} {\left[\text{I"_2] = k_text(-1)["I}\right]}^{2}$

$\boldsymbol{\left(5\right)} \textcolor{w h i t e}{m} \left[{\text{I"]^2 = k_1/k_text(-1)["I}}_{2}\right]$

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])