Overall Reaction Rate Laws



 

Learning Objective

  • Combine elementary reaction rate constants to obtain equilibrium coefficients and construct overall reaction rate laws for reactions with both slow and fast initial steps

Key Points

    • In a reaction with a slow initial step, the rate law will simply be determined by the stoichiometry of the reactants.
    • In a rate law with a fast initial step, no intermediates can appear in the overall rate law.

Term

  • rate-determining stepThe slowest step in a chemical reaction that determines the rate of the overall reaction.

Slow Step Followed by a Fast Step

As discussed in the previous concept, if the first step in a reaction mechanism is the slow, rate-determining step, then the overall rate law for the reaction is easy to write, and simply follows the stoichiometry of the initial step. For example, consider the following reaction:

[latex]H_2(g) + 2\;ICl(g)\rightarrow I_2(g)+2\;HCl(g)[/latex]

The proposed reaction mechanism is given as follows:

[latex](1)\;\text{slow reaction:}\quad H_2+ICl\rightarrow HI+HCl\quad\quad \text{rate}_1=k_1[H_2][ICl][/latex]

[latex](2)\;\text{fast reaction:}\quad HI+ICl\rightarrow I_2+HCl\quad\quad \text{rate}_2=k_2[HI][ICl][/latex]

Since the first step is the rate-determining step, the overall reaction rate for this reaction is given by this step: [latex]\text{rate}=k[H_2][ICl][/latex]. As it turns out, this rate law has been verified with experimental evidence.

Fast Step Followed by a Slow Step

If the rate-determining step is not the first step in the reaction mechanism, the derivation of the rate law becomes slightly more complex. Consider the following reaction:

[latex]2\;NO(g)+O_2(g)\rightarrow 2\;NO_2(g)[/latex]

The proposed mechanism is given by:

[latex](1)\;\text{Step 1, fast:}\quad 2\;NO(g)\rightleftharpoons N_2O_2(g)\quad\quad \text{rate}_1=k_1[NO]^2[/latex]

[latex](2)\;\text{Step 2, slow:}\quad N_2O_2(g)+O_2(g)\rightarrow 2\;NO_2(g)\quad\quad \text{rate}_2=k_2[N_2O_2][O_2][/latex]

Step two is the slow, rate-determining step, so it might seem reasonable to assume that the rate law for this step should be the overall rate law for the reaction. However, this rate law contains N2O2, which is a reaction intermediate, and not a final product. The overall rate law cannot contain any such intermediates, because the rate law is determined by experiment only, and such intermediates are not observable. To get around this, we need to go back and consider the first step, which involves an equilibrium between NO and N2O2. At equilibrium, the rate of the forward reaction will equal the rate of the reverse reaction. We can write this as follows:

[latex]k_1[NO]^2=k_{-1}[N_2O_2][/latex]

Rearranging for [N2O2], we have:

[latex][N_2O_2]=\frac{k_1}{k_{-1}}[NO]^2[/latex]

We can now substitute this expression into the rate law for the second, rate-determining step. This yields the following:

[latex]\text{rate}=k_2\left(\frac{k_1}{k_{-1}}[NO]^2\right)[O_2]=k[NO]^2[O_2][/latex]

This overall rate law, which is second-order in NO and first-order in O2, has been confirmed experimentally.

Rate law for a mechanism with a fast initial stepHow to determine the rate law for a mechanism with a fast initial step. Remember, the overall rate law must be determined by experiment. Therefore, the rate law must contain no reaction intermediates.