Steady State Approximation- Khan Academy’s Kinetic Modeling Guide

What the Steady State Approximation Actually Is

The steady state approximation is a shortcut for solving reaction mechanisms that have intermediate species — compounds that form during the reaction but don't appear in the overall balanced equation.

Instead of tracking every intermediate's concentration moment by moment, you assume its concentration stays roughly constant after an initial buildup. The intermediate forms at about the same rate it disappears.

This sounds like a hand-wavy assumption. It is. But it works surprisingly well for many real reaction networks.

Why You Need This Approximation

Most interesting reaction mechanisms involve multiple steps. Solving them with differential equations gets messy fast. The steady state approximation cuts through that math without losing too much accuracy.

You encounter this most often with:

If you're working through Khan Academy's kinetics unit, you'll hit this concept around the intermediate steps section. It's tested constantly on exams.

The Core Assumption in Plain Terms

For an intermediate I:

Rate of formation ≈ Rate of consumption

Mathematically:

d[I]/dt ≈ 0

This doesn't mean [I] is literally zero. It means the concentration changes so slowly relative to reactants and products that you can treat it as approximately constant during the measurement window.

When the Approximation Breaks Down

Don't apply this blindly. The steady state assumption fails when:

Enzyme-substrate complexes often violate this early in the reaction. That's why Michaelis-Menten derivations specify certain conditions.

Step-by-Step: Applying the Approximation

1. Identify the intermediate

Look for species that appear in elementary steps but not in the overall reaction. Common examples include free radicals (Cl•, Br•), enzyme complexes (ES), and transient species.

2. Write the rate law for the intermediate

Sum up all formation pathways. Sum up all consumption pathways. Set them equal.

3. Solve for the intermediate's concentration

Isolate [I] in terms of reactant concentrations. Plug back into the overall rate law.

4. Check your assumptions

Does the result match experimental data? Does the pre-equilibrium approximation give a similar answer? If yes, you're probably on track.

Comparing Approximation Methods

Method Best When Accuracy Complexity
Steady State Intermediate forms and decomposes rapidly Good for most mechanisms Medium
Pre-Equilibrium Reverse reaction is fast relative to overall rate Excellent when valid Medium
Rate-Determining Step One step is much slower than all others Depends on mechanism Low
Exact Integration All concentrations are measurable Perfect High

The steady state and pre-equilibrium approximations often give identical results when both are valid. If they diverge, something's wrong with your mechanism.

Khan Academy's Approach to This Topic

Khan Academy breaks this down into digestible chunks. The videos walk you through:

The practice problems are worth doing twice. The first pass, you learn the method. The second pass, you catch the mistakes you made the first time.

One thing Khan Academy does well: they show the algebra step-by-step without skipping steps. This is where most students get lost in textbooks.

Common Mistakes That Will Cost You Points

Forgetting to express everything in terms of observable species. Your final rate law should only contain reactants and products you can measure. If you still have intermediate concentrations in your final equation, you haven't finished.

Applying steady state to reactants or products. This only works for intermediates. Always.

Rounding too early. Keep symbols in your equations until the final step. Substituting numbers too soon introduces errors that compound through the calculation.

Confusing steady state with equilibrium. They're different assumptions. Steady state says the net change is zero. Equilibrium says forward and reverse rates are equal. Don't mix them up.

A Quick Worked Example

Consider the mechanism:

Step 1: A + B → I

Step 2: I → C

Apply steady state to I:

d[I]/dt = k₁[A][B] - k₂[I] ≈ 0

Solve for [I]:

k₁[A][B] = k₂[I]

[I] = (k₁/k₂)[A][B]

Rate of product formation:

rate = k₂[I] = k₁[A][B]

The overall rate law matches Step 1. This tells you Step 1 is the rate-determining step, which you'd expect since I is in steady state.

What to Do Next

Go to Khan Academy's kinetics section. Find the steady state approximation video. Watch it once at 1.5x speed to see the whole derivation. Then watch it again at normal speed with paper ready.

Work through at least five practice problems before moving on. The pattern clicks after you've made your own mistakes a few times.

If you're preparing for an exam, focus on being able to derive rate laws from mechanisms without looking anything up. That's the skill they're testing.