Deriving rate laws: Lindemann and Michaelis–Menten

You do not need to memorise these rate laws. Both come from the same short recipe, and in an exam it is faster and safer to derive them than to recall them.

The recipe

Write the rate you are asked for using the elementary steps (mass action).
List the intermediates: species made and used up within the mechanism.
For each one, either set $\mathrm{d}[\mathrm{X}]/\mathrm{d}t = 0$ (steady state) or set the forward and backward rates of a fast reversible step equal (pre-equilibrium).
If a catalyst is involved, add its mass balance.
Solve for the intermediates, substitute, and simplify.
Check the limits: let each term of the denominator dominate in turn.

Lindemann–Hinshelwood

A gas molecule A only reacts once a collision with any molecule M has given it enough energy (A*):

\mathrm{A} + \mathrm{M} \xrightleftharpoons[\;k_{-1}\;]{\;k_1\;} \mathrm{A^*} + \mathrm{M} \qquad \mathrm{A^*} \xrightarrow{\;k_2\;} \mathrm{P}

A* is the intermediate, so put it in steady state:

\frac{\mathrm{d}[\mathrm{A^*}]}{\mathrm{d}t} = k_1[\mathrm{A}][\mathrm{M}] - k_{-1}[\mathrm{A^*}][\mathrm{M}] - k_2[\mathrm{A^*}] = 0

[\mathrm{A^*}] = \frac{k_1[\mathrm{A}][\mathrm{M}]}{k_2 + k_{-1}[\mathrm{M}]}

\frac{\mathrm{d}[\mathrm{P}]}{\mathrm{d}t} = k_2[\mathrm{A^*}] = \frac{k_1k_2[\mathrm{M}]}{k_2 + k_{-1}[\mathrm{M}]}\,[\mathrm{A}]

The reaction is first order in A, with an effective constant that depends on pressure. When $k_{-1}[\mathrm{M}] \gg k_2$ (high pressure), it levels off at $k_1k_2/k_{-1}$ . When $k_{-1}[\mathrm{M}] \ll k_2$ (low pressure), it falls to $k_1[\mathrm{M}]$ and the reaction becomes second order overall.

Lindemann–Hinshelwood (unimolecular gas reaction)

Step 1 $\mathrm{A} + \mathrm{M} \xrightleftharpoons[k_{-1}]{k_{1}} \mathrm{A^*} + \mathrm{M}$
Step 2 $\mathrm{A^*} \xrightarrow{k_{2}} \mathrm{P}$

Steady state:

Pre-equilibrium:

Rate from the elementary steps
$\frac{\mathrm{d}[\mathrm{P}]}{\mathrm{d}t} = k_{2}[\mathrm{A^*}]$
Steady state on $\mathrm{A^*}$
$\frac{\mathrm{d}[\mathrm{A^*}]}{\mathrm{d}t} = k_{1}[\mathrm{A}][\mathrm{M}] - k_{-1}[\mathrm{M}][\mathrm{A^*}] - k_{2}[\mathrm{A^*}] = 0$
Solve for $[\mathrm{A^*}]$
$[\mathrm{A^*}] = \dfrac{k_{1}[\mathrm{A}][\mathrm{M}]}{k_{2} + k_{-1}[\mathrm{M}]}$

Rate law

\frac{\mathrm{d}[\mathrm{P}]}{\mathrm{d}t} = \dfrac{k_{1}k_{2}[\mathrm{A}][\mathrm{M}]}{k_{2} + k_{-1}[\mathrm{M}]}

Limiting cases

If $k_{2} \gg k_{-1}[\mathrm{M}]$ :
$\frac{\mathrm{d}[\mathrm{P}]}{\mathrm{d}t} \approx k_{1}[\mathrm{A}][\mathrm{M}]$
If $k_{-1}[\mathrm{M}] \gg k_{2}$ :
$\frac{\mathrm{d}[\mathrm{P}]}{\mathrm{d}t} \approx \dfrac{k_{1}k_{2}[\mathrm{A}]}{k_{-1}}$

Michaelis–Menten

An enzyme E binds its substrate S, and the complex ES gives the product:

\mathrm{E} + \mathrm{S} \xrightleftharpoons[\;k_{-1}\;]{\;k_1\;} \mathrm{ES} \qquad \mathrm{ES} \xrightarrow{\;k_2\;} \mathrm{E} + \mathrm{P}

Steady state on ES:

k_1[\mathrm{E}][\mathrm{S}] = (k_{-1} + k_2)[\mathrm{ES}]

Free [E] is unknown, but the total is fixed: $[\mathrm{E}] = [\mathrm{E}]_0 - [\mathrm{ES}]$ . Substituting and collecting [ES]:

[\mathrm{ES}] = \frac{k_1[\mathrm{E}]_0[\mathrm{S}]}{k_{-1} + k_2 + k_1[\mathrm{S}]}

Divide the top and bottom by $k_1$ and multiply by $k_2$ :

v = k_2[\mathrm{ES}] = \frac{V_\mathrm{max}[\mathrm{S}]}{K_\mathrm{M} + [\mathrm{S}]}, \qquad V_\mathrm{max} = k_2[\mathrm{E}]_0, \quad K_\mathrm{M} = \frac{k_{-1} + k_2}{k_1}

Here [S] means the substrate concentration, taken as its initial value, since $[\mathrm{S}]_0 \gg [\mathrm{E}]_0$ .

Michaelis–Menten enzyme kinetics

Step 1 $\mathrm{E} + \mathrm{S} \xrightleftharpoons[k_{-1}]{k_{1}} \mathrm{ES}$
Step 2 $\mathrm{ES} \xrightarrow{k_{2}} \mathrm{E} + \mathrm{P}$