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Chemistry LibreTexts

Rates and Mechanisms

Skills to Develop

  • Explain reaction mechanism.
  • Derive a rate law from a given mechanism.

Reaction Mechanisms - Derive Rate Laws

A reaction mechanism is a collection of elementary processes or steps (also called elementary steps) that explains how the overall reaction proceeds.

A mechanism is a proposal from which you can work out a rate law that agrees with the observed rate laws. The fact that a mechanism explains the experimental results is not a proof that the mechanism is correct.

As a mechanism, no proof is required. Proposing a mechanism is an interesting academic exercise for a mature chemist. Students in general chemistry will not be required to propose a mechanism, but they are required to derive the rate law from a proposed mechanism.

Elementary Processes or Steps

A summary of the elementary process or steps is given in a table form here. This previous link provides details about these steps. In the table below, \(\ce{A}\), \(\ce{B}\), and \(\ce{C}\) represent reactants, intermediates, or products in the elementary process.

Molecularity Elementary step(s) Rate law
1 \(\ce{A \rightarrow products}\) \(rate = k \ce{[A]}\)
2 \(\ce{A + A \rightarrow products}\)
\(\ce{A + B \rightarrow products}\)
\(rate = k \ce{[A]^2}\)
\(rate = k \ce{[A] [B]}\)
3 \(\ce{A + A + A \rightarrow products}\)
\(\ce{A + 2 B \rightarrow products}\)
\(\ce{A + B + C \rightarrow products}\)
\(rate = k \ce{[A]^3}\)
\(rate = k \ce{[A] [B]^2}\)
\(rate = k \ce{[A] [B] [C]}\)

Deriving Rate Laws from Mechanisms

You all have experienced that an accident or road construction on a free way slows all the traffic on the road, because limitations imposed by the accident and constructions apply to all cars on that road. The narrow passing stretch limits the speed of the traffic.

If several steps are involved in an overall chemical reaction, the slowest step limits the rate of the reaction. Thus, a slow step is called a rate determining step. The following examples illustrate the method of deriving rate laws from the proposed mechanism. Please learn the technique.

Example 1

If the reaction

\(\ce{2 NO2 + F2 \rightarrow 2 NO2F}\)

follows the mechanism,

  1. \(\ce{NO2 + F2 \rightarrow NO2F + F\: (slow)}\)
  2. \(\ce{NO2 + F \rightarrow NO2F\: (fast)}\)

what is the rate law?

Since step i. is the rate-determining step, the rate law is

\(\mathrm{- \dfrac{1}{2}\dfrac{d[NO_2]}{dt} = \mathit{k} [NO_2] [F_2]}\)

Since both \(\ce{NO2}\) and \(\ce{F2}\) are reactants, this is the rate law for the reaction.

Addition of i. and ii. gives the overall reaction, but step ii. does not affect the rate law. Note that the rate law is not derived from the overall equation either.

Example 2

For the reaction

\(\ce{H2 + Br2 \rightarrow 2 HBr}\),

the following mechanism has been proposed

  1. \(\ce{Br2 \underset{\Large{\mathit{k}_{-1}}}{\overset{\Large{\mathit{k}_1}}{\rightleftharpoons}} 2 Br}\textrm{ (both directions are fast)}\)
  2. \(\mathrm{Br + H_2 \xrightarrow{\Large{\mathit{k}_2}}HBr + H\: (slow)}\)
  3. \(\mathrm{H + Br_2 \xrightarrow{\Large{\mathit{k}_3}}HBr + Br\: (fast)}\)

Derive the rate law that is consistent with this mechanism.

For a problem of this type, you should give the rate law according to the rate-determining (slow) elementary process. In this case, step ii. is the rate-determining step, and the rate law is

\(\mathrm{\dfrac{1}{2}\dfrac{d[HBr]}{dt} = \mathit{k}_2 [H_2] [Br]}\)

The factor 1/2 results from the \(\ce{2 HBr}\) formed every time, one in step ii. and one in step iii. Since \(\ce{[Br]}\) is not one of the reactants, its relationship with the concentration of the reactants must be sought. The rapid reaction in both directions of step i. implies the following relationship:

\(k_1 [\ce{Br2}] = k_{-1} [\ce{Br}]^2\)


\(\ce{[Br]} = \left(\dfrac{k_1}{k_{-1}} [\ce{Br2}]\right)^{1/2}\)

Substituting this in the rate expression results in

\(\ce{rate} = k_2 \left(\dfrac{k_1}{k_{-1}}\right)^{1/2} \ce{[H2] [Br2]}^{1/2}\)

The overall reaction order is 3/2, 1 with respect to \(\ce{[H2]}\) and 1/2 with respect to \(\ce{[Br2]}\).

The important point in this example is that the rapid equilibrium in step i. allows you to express the concentration of an intermediate (\(\ce{[Br]}\)) in terms of concentrations of reactants (\(\ce{[Br2]}\)) so that the rate law can be expressed by concentrations of the reactants.

The ratio k1 / k-1 is often written as K, and it is called the equilibrium constant for the reversible elementary steps.

Example 3

Derive the rate law that is consistent with the proposed mechanism in the formation of phosgene from \(\ce{Cl2}\) and \(\ce{CO}\).  (\(K_1 = \dfrac{k_1}{k_{-1}}\) and \(K_2 = \dfrac{k_2}{k_{-2}}\) may be considered as equilibrium constants of the elementary processes, and \(\ce{M}\) is any inert molecule.)

  1. \(\ce{Cl2 + M \rightleftharpoons 2 Cl + M\:\: (fast\: equilibrium,\: \mathit{K}_1)}\)
  2. \(\ce{Cl + CO + M \rightleftharpoons ClCO + M\:\: (fast\: equilibrium,\: \mathit{K}_2)}\)
  3. \(\ce{ClCO + Cl2 \rightarrow Cl2CO + Cl\:\: (slow,\: \mathit{k}_3)}\)

The overall reaction is

\(\ce{Cl2 + CO \rightarrow Cl2CO}\)

From the rate-determining (slow) step,

\(\mathrm{\dfrac{d[Cl_2CO]}{dt}} = k_3 \mathrm{[ClCO] [Cl_2]}\tag{1}\)

You should express \(\ce{[ClCO]}\) in terms of concentrations of \(\ce{Cl2}\) and \(\ce{CO}\). This is done by considering step ii.

\(\ce{[ClCO]} = K_2 \ce{[Cl] [CO]}\tag{2}\)

You should express \(\ce{[Cl]}\) in terms of \(\ce{[Cl2]}\). For this, you may use step i.

\(\ce{[Cl]} = K_1^{1/2} \ce{[Cl2]}^{1/2} \tag{3}\)

Substituting (3) in (2) and then in (1) gives the Rate,

\ce{Rate} &= k_3 K_1^{1/2} K_2 \ce{[CO] [Cl2]}^{3/2}\\
&= k \ce{[CO] [Cl2]}^{3/2}

where \(k = k_1 K_1^{1/2} K_2\), the observed rate constant. The overall order of the reaction is 5/2: strange, but that is the observed rate law.

This example shows how the concentrations of intermediates are related to those of the reactants in a two-step equilibrium.

If the third step is

  1. \(\ce{ClCO + Cl \rightarrow Cl2CO\:\: (slow,\: \mathit{k}_3)}\)

the rate law will be different from the result derived above.


Derive the rate law using the alternate step from the Discussion in Example 3.

Example 4

In an acid solution, the mechanism for the reaction

\(\ce{NH4+ + HNO2 \rightarrow N2 + 2 H2O + H+ }\)


  1. \(\ce{HNO2 + H+ \rightleftharpoons H2O + NO+ \:\: (equilibrium,\: \mathit{K}_1)}\)
  2. \(\ce{NH4+ \rightleftharpoons NH3 + H+ \:\: (equilibrium,\: \mathit{K}_2)}\)
  3. \(\ce{NO+ + NH3 \rightarrow NH3NO+ \:\: (slow,\: \mathit{k}_3)}\)
  4. \(\ce{NH3NO+ \rightarrow H2O + H+ + N2 \:\: (fast,\: \mathit{k}_4)}\)

Derive the rate law.

From the rate-determining step, you have

\(\mathrm{\dfrac{d[NH_3NO^+ ]}{dt}} = k_3 \mathrm{[NO^+] [NH_3]} \tag{4}\)

Neither \(\ce{NO+}\) nor \(\ce{NH3}\) is a reactant. You must express their concentrations in terms of \(\ce{[NH4+]}\) and \(\ce{[HNO2]}\) from elementary processes i. and ii.

From i, \(\ce{[NO+]} = K_1 \ce{\dfrac{[HNO2] [H+]}{[H2O]}} \tag{5}\)

From ii, \(\ce{[NH3]} = K_2 \ce{\dfrac{[NH4+]}{[H+]}} \tag{6}\)

Substituting (6) and (5) in (4) gives,

\ce{Rate} &= k_3 K_1 K_2 \ce{\dfrac{[HNO2] [NH4+]}{[H2O]}}\\ \\
&= k \ce{[HNO2] [NH4+ ]}

where \(k = \dfrac{k_3 K_1 K_2}{[\ce{H2O}]}\) is the overall rate constant.


  1. If the reaction

    \(\ce{2 A + B2 \rightarrow 2 AB}\)

    follows the mechanism,
    1. \(\ce{A + B2 \rightarrow AB + B\: (slow)}\)
    2. \(\ce{A + B \rightarrow AB\: (fast)}\)
    What is the order with respect to \(\ce{[B2]}\)?
    Hint: First order

    Skill -
    Figure out the order from a given mechanism.
    What is the overall order?

  2. Kinetic studies of the reaction

    \(\ce{A2 + B2 \rightarrow 2 AB}\)

    suggested the mechanism as follows:
    1. \(\ce{A2 \rightleftharpoons 2 A\: (fast\: equilibrium)}\)
    2. \(\ce{A + B2 \rightarrow AB + B\: (slow)}\)
    3. \(\ce{B + A2 \rightarrow AB + A\: (fast)}\)
    What is the order of the reaction with respect to \(\ce{[A2]}\)?
    Hint: The order is half (1/2).

    Discussion -
    If you got \(\ce{rate} = k \mathrm{[A_2]^{1/2} [B_2]}\), congratulations.

  3. If the mechanism for the reaction

    \(\ce{2 NO + O2 \rightarrow 2 NO2}\)

    1. \(\ce{NO + NO \rightleftharpoons  N2O2\: (fast\: equilibrium)}\)
    2. \(\ce{N2O2 + O2 \rightarrow 2 NO2\: (slow)}\)
    what is the power of \(\ce{[NO]}\) in the differential rate law?
    Hint: The order is 2 with respect to \(\ce{[NO]}\).

    Skill -
    Figure out the order from a given mechanism.
    You should get: \(rate = k \ce{[NO]^2 [O2]}\)

  4. If the reaction mechanism consists of these elementary processes,
    1. \(\ce{A \rightleftharpoons 2 B\: (fast,\: equilibrium)}\)
    2. \(\ce{B + 2 C \rightarrow E\: (slow)}\)
    3. \(\ce{E \rightarrow F\: (fast)}\)
    choose the correct differential rate law for the reaction

    \(\ce{A + 4 C \rightarrow 2 F}\)

    1. \(\mathrm{\dfrac{1}{2}\dfrac{d[F]}{dt} = \mathit{k}[A] [C]^4}\)
    2. \(\mathrm{rate = \mathit{k}[A] [C]^2}\)
    3. \(\mathrm{-\dfrac{d[A]}{dt} = \mathit{k}[A]^{1/2} [C]}\)
    4. \(\mathrm{\dfrac{d[F]}{dt} = \mathit{k}[A] [C]}\)
    5. \(\mathrm{\dfrac{d[F]}{dt} = \mathit{k}[A]^{1/2}[C]^2}\)
    Hint: e.

    The method -
    The rate determining step is ii. Express \(\ce{[B]}\) in terms of \(\ce{[A]}\) from i.