Lindemann Mechanism and the Steady-State Approximation
Many reactions look simple on paper but have complicated rate laws. A gas-phase decomposition such as A → products may be first order at high pressure and second order at low pressure. The Lindemann mechanism explains this with a short, clear idea, and the steady-state approximation is the tool used to solve it. Both appear regularly in physical chemistry papers for IIT-JAM, GATE, CSIR-NET and CUET-PG. This page derives the result, shows the two limits, and works through numbers.
The steady-state approximation
In a multi-step mechanism, a reactive intermediate is made slowly and used up quickly, so its concentration stays very small and almost constant. The approximation sets its rate of change to zero:
This turns a set of differential equations into simple algebra. It is valid when the intermediate is consumed much faster than it is formed.
Step 1: Check the condition: k2/k1 = 1.0/0.010 = 100, so I is used up 100 times faster than it forms. The approximation is reasonable.
Step 2: Steady state: k1[A] = k2[I], so [I] = k1[A]/k2 = 0.010 × 0.50 / 1.0 = 5.0 × 10−3 mol L−1.
Step 3: Rate of formation of P = k2[I] = 1.0 × 5.0 × 10−3 = 5.0 × 10−3 mol L−1 s−1. This equals k1[A] = 0.010 × 0.50 = 5.0 × 10−3 ✓.
Answer: [I] ≈ 5.0 × 10−3 mol L−1, about 1 % of [A], and the rate of P formation is 5.0 × 10−3 mol L−1 s−1.
The Lindemann mechanism
Lindemann proposed that a molecule A does not decompose just because it exists. It first has to gain enough energy by collision with another molecule M (M can be A itself, or an inert gas). The energised molecule A* can then lose energy again by another collision, or can react to give the product P.
A* → P (k2)
Apply the steady-state approximation to A*:
[A*] = k1[A][M] / (k−1[M] + k2)
Rate = k2[A*] = k1k2[A][M] / (k−1[M] + k2) = kuni[A]
So the observed first-order rate constant depends on [M], which is proportional to pressure for a gas:
The two limits
| Condition | Simplification | Rate law | Order |
|---|---|---|---|
| High pressure: k−1[M] ≫ k2 | Denominator ≈ k−1[M] | Rate = (k1k2/k−1)[A] | First |
| Low pressure: k2 ≫ k−1[M] | Denominator ≈ k2 | Rate = k1[A][M] | Second overall |
At high pressure, collisions are so frequent that A and A* stay in equilibrium, and the slow step A* → P controls the rate. At low pressure, the collision that energises A is the bottleneck, so every A* that forms reacts before it can be de-energised.
Step 1: High-pressure limit: k∞ = k1k2/k−1 = (5.0 × 10−3 × 2.0 × 103) / 1.0 × 107 = 10 / 1.0 × 107 = 1.0 × 10−6 s−1.
Step 2: The [M] at which k−1[M] = k2 is [M] = 2.0 × 103 / 1.0 × 107 = 2.0 × 10−4 mol L−1. At this point kuni = k∞/2.
Step 3: Use kuni = k1k2[M] / (k−1[M] + k2) at several [M]:
| [M] (mol L−1) | kuni (s−1) | kuni / k∞ |
|---|---|---|
| 1.0 × 10−5 | 4.76 × 10−8 | 0.048 |
| 2.0 × 10−4 | 5.0 × 10−7 | 0.50 |
| 1.0 × 10−2 | 9.80 × 10−7 | 0.980 |
| 1.0 | 9.998 × 10−7 | 0.9998 |
Answer: kuni rises with [M] and levels off at k∞ = 1.0 × 10−6 s−1. At low [M] (1.0 × 10−5) it is close to the low-pressure value k1[M] = 5.0 × 10−8.
1/kuni = (k−1[M] + k2) / (k1k2[M]) = k−1/(k1k2) + 1/(k1[M]).
A plot of 1/kuni against 1/[M] is a straight line. The slope is 1/k1 and the intercept is k−1/(k1k2) = 1/k∞.
Numerical check with Example 2 values: intercept = 1.0 × 107 / 10 = 1.0 × 106 s. At [M] = 1.0 × 10−2: 1/kuni = 1.0 × 106 + 1/(5.0 × 10−3 × 1.0 × 10−2) = 1.0 × 106 + 2.0 × 104 = 1.02 × 106 s, and 1/9.80 × 10−7 = 1.02 × 106 s ✓.
Answer: slope = 1/k1 = 200 mol s L−1, intercept = 1.0 × 106 s.
Where the simple theory falls short
The Lindemann picture explains why order changes with pressure, but its predicted fall-off from the high-pressure value is steeper than what experiments usually show. The reason is that it treats k2 as a single constant, although a more energised A* should react faster. Later theories such as Hinshelwood's and RRKM theory treat this better. For examinations, you are expected to know the Lindemann result and its two limits.
- Applying steady state to a stable intermediate. The approximation is valid only for a very reactive intermediate present at low concentration.
- Dropping [M] too early. The term k−1[M] in the denominator is what makes the order change. Simplify only after you have named the limit.
- Mixing the limits. High pressure gives first order. Low pressure gives second order. Students often reverse them.
- Unit errors. k1 and k−1 are second-order (L mol−1 s−1) and k2 is first-order (s−1). Check that k−1[M] and k2 have the same units before you compare them.
- Setting d[A]/dt = 0. The steady-state condition applies to the intermediate A*, never to the reactant.
Exam relevance
| Question type | Key step |
|---|---|
| Derive the rate law from a mechanism | Set d[intermediate]/dt = 0, solve for it, substitute in the rate |
| Order at high or low pressure | Compare k−1[M] with k2 in the denominator |
| Test by a graph | Plot 1/kuni against 1/[M] |
| Validity of the approximation | Intermediate must be consumed much faster than it forms |
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