Steady-State Approximation
Mechanisms with an Initial Fast Step
- When the first step of a reaction mechanism is the rate-limiting step, the rate law can be easily determined
- However, it is less straightforward to derive rate laws for a mechanism where an intermediate is a reactant in the rate-limiting step
- This happens when the rate-limiting step follows an initial first step
- For example, consider the gas-phase reaction between nitric oxide, NO and bromine gas, Br2
2NO (g) + Br2 (g) → 2NOBr (g)
- The experimental rate law for this reaction is given as:
Rate = k[NO]2[Br2]
- A possible reaction mechanism consistent with this rate law could be an elementary reaction, involving three reactant molecules— termolecular reaction
NO (g) + NO (g) + Br2 (g) → 2NOBr (g) Rate = k[NO]2[Br2]
- This however does not seem likely because such termolecular reactions are rare
- So let’s consider an alternative mechanism which does not involve termolecular reactions:
NO (g) + Br2 (g) ⇋ NOBr2 (g) (fast step)
NOBr2 (g) + NO (g) → 2NOBr (g) (slow step)
- From the proposed elementary reactions above, we see that
- The first step is fast and reversible
- The rate-determining step involves an intermediate, NOBr2
- Based on the equation of the rate-limiting step, the rate law for the reaction will be:
Rate = k[NOBr2][NO]
- This is not consistent with the experimental rate law and involves an intermediate
- To eliminate the intermediate and convert it into one of the reactants, we use the steady-state assumption or approximation
- This assumption states that if the second step is the rate-limiting step, then the first step must be relatively fast and reversible
- This means that the rate at which the intermediate is formed— forward reaction— is equal to the rate at which it is consumed — backward reaction
- Using the fast step elementary equation proposed for the reaction between nitric oxide and bromine:
NO (g) + Br2 (g) ⇋ NOBr2 (g) (fast step)
- The rate laws for the forward and backward reverse reactions are:
Rateforward = kf[NO][Br2]
Ratebackward = kb[NOBr]
- Since the Given that rate of forward and backward reactions are equal / at steady-state, we can write:
kf[NO][Br2] = kb[NOBr2]
- By rearranging the above expression, we can express the concentration of the NOBr2 intermediate, NOBr2 in terms of the concentration of the reactants:
[NOBr2] = kf/kb[NO][Br2]
- We then rewrite the rate law equation for the rate-limiting step by substituting the above expression for the intermediate:
Rate = k[NOBr2][NO]
Rate = k(kf/kb)[NO][Br2][[NO]
- All the k terms are constants, which means that they can be combined
This gives a rate law expression consistent with the experimental rate law:
Rate = k[NO]2[Br2]
- In general, whenever a fast step precedes a slow one, we can solve for the concentration of an intermediate by assuming that an equilibrium is established in the fast step
Worked example
The experimental rate law for the reaction between hydrogen gas and iodine gas to produce hydrogen iodide is first order with respect to both hydrogen and iodine gas. The overall balanced chemical equation is:
H2 (g) + I2 (g) → 2HI (g)
A chemistry student was asked to propose a possible reaction mechanism and provided the following elementary reactions:
I2(g) ⇋ 2I(g) (fast)
H2(g) + I(g) +I(g) → 2HI(g) (slow)
Show that the rate law is consistent with the proposed mechanism
Answer:
- Step 1: Deduce the experimental rate equation for the reaction. Since the reaction is first order with respect to both reactants then:
Rate = k[H2][I2]
- Step 2: Using the coefficient of the reactants in the rate-limiting step, write the rate law equation:
Rate = k[H2][I]2
- Step 3: Using the fast elementary step and the concept of steady state, express the concentration of the intermediate, I(g) in terms of the reactant, I2 (g)
At equilibrium: kf[I2] = kb[I]2
[I]2 = kf/kb[I2]
- Step 4: Substitute for [I]2 in the rate law equation
Rate = k(kf/kb)[I2][H2]
On simplifying,
Rate = k[I2][H2]
