Rate Law Determination by the Method of Initial Rates: Worked Examples

To find the rate law from initial-rate data, you compare experiments where one reactant concentration changes while others stay constant. The ratio of rates between those two experiments equals the ratio of changed concentrations raised to the order of that reactant. Set up the ratio: (Rate2 / Rate1) = ([A]2 / [A]1)^m. Solve for m (the order in A). Repeat for each reactant. Once you have all orders, plug any single experiment's data into Rate = k[A]^m[B]^n and solve for k. The units of k depend on the overall order: for first-order overall, units are s^-1; for second-order overall, M^-1 s^-1; for third-order overall, M^-2 s^-1. Always check the orders by trying a third experiment to make sure they predict the right rate.

The initial rate is the instantaneous rate of reaction at time zero, before products accumulate or reactants meaningfully deplete. We use initial rates because the rate law Rate = k[A]^m[B]^n applies to forward reaction only — once products build up, reverse reactions complicate things. By measuring the rate during the first few percent of conversion, we get a clean picture of the forward kinetics.

Experimentally, you measure either the rate of disappearance of a reactant or appearance of a product over a very short initial time interval. For a reaction A + B → C, the initial rate equals -d[A]/dt or -d[B]/dt or +d[C]/dt at t = 0 (with stoichiometric coefficients adjusted as needed).

For the reaction 2NO(g) + O2(g) → 2NO2(g), four experiments give:

Exp 1: [NO] = 0.020 M, [O2] = 0.010 M, Rate = 2.4 × 10^-4 M/s Exp 2: [NO] = 0.040 M, [O2] = 0.010 M, Rate = 9.6 × 10^-4 M/s Exp 3: [NO] = 0.020 M, [O2] = 0.020 M, Rate = 4.8 × 10^-4 M/s Exp 4: [NO] = 0.040 M, [O2] = 0.040 M, Rate = ? (used to verify)

Step 1 — Order in NO. Compare Exp 1 and Exp 2: [O2] is constant, [NO] doubles. Rate2/Rate1 = (9.6 × 10^-4) / (2.4 × 10^-4) = 4. Concentration ratio = 2. So 2^m = 4, meaning m = 2. Reaction is second order in NO.

Step 2 — Order in O2. Compare Exp 1 and Exp 3: [NO] is constant, [O2] doubles. Rate3/Rate1 = (4.8 × 10^-4) / (2.4 × 10^-4) = 2. Concentration ratio = 2. So 2^n = 2, meaning n = 1. Reaction is first order in O2.

Step 3 — Rate law: Rate = k[NO]^2[O2]^1. Overall order = 2 + 1 = 3.

Step 4 — Find k. Plug Exp 1: 2.4 × 10^-4 M/s = k(0.020)^2(0.010) = k(4 × 10^-6). So k = 60 M^-2 s^-1.

Step 5 — Verify with Exp 4 (predicted): Rate = 60 × (0.040)^2 × (0.040) = 60 × 1.6 × 10^-3 × 0.040 = 3.84 × 10^-3 M/s. If the experiment gives this within a few percent, your orders are correct.

For the reaction A + B → products:

Exp 1: [A] = 0.10 M, [B] = 0.10 M, Rate = 0.0050 M/s Exp 2: [A] = 0.20 M, [B] = 0.10 M, Rate = 0.020 M/s Exp 3: [A] = 0.10 M, [B] = 0.20 M, Rate = 0.0050 M/s

Step 1 — Order in A. Comparing Exp 1 and Exp 2: [B] constant, [A] doubles. Rate2/Rate1 = 4. So 2^m = 4, m = 2.

Step 2 — Order in B. Comparing Exp 1 and Exp 3: [A] constant, [B] doubles. Rate3/Rate1 = 1. So 2^n = 1, n = 0. Zero order in B.

Step 3 — Rate law: Rate = k[A]^2[B]^0 = k[A]^2.

Step 4 — Find k: 0.0050 = k(0.10)^2 = k(0.010). k = 0.50 M^-1 s^-1.

Zero order in B means the reaction rate is independent of [B] within the concentration range tested. This often signals a multi-step mechanism where B is involved in a fast step that's not rate-limiting, or where B is in large excess (pseudo-zero-order conditions). The catalyst surface in heterogeneous catalysis is a classic case where surface saturation produces zero-order kinetics in the surface-bound species.

Sometimes a reaction has fractional order, often signaling a chain mechanism or a complex elementary step. For 2A → products:

Exp 1: [A] = 0.10 M, Rate = 0.020 M/s Exp 2: [A] = 0.40 M, Rate = 0.040 M/s

Ratio of rates = 2.0. Ratio of concentrations = 4.0. So 4^m = 2, meaning m = log(2)/log(4) = 0.5. Half-order in A.

Rate law: Rate = k[A]^0.5. Plug Exp 1: 0.020 = k(0.10)^0.5 = k(0.316). k = 0.0633 M^0.5 s^-1.

Fractional orders are physically meaningful but often confuse students. They typically arise from chain mechanisms (free-radical polymerization is the classic case), pre-equilibrium steps with subsequent rate-limiting decomposition, or surface-mediated reactions. If you find a fractional order on an exam, double-check your math — it's usually 1/2, 3/2, or another simple ratio. If you compute 0.43, you probably made an arithmetic error.

Quick reference table for order determination:

| Concentration Ratio | Rate Ratio | Order | |---|---|---| | 2× | 1× | 0 | | 2× | 2× | 1 | | 2× | 4× | 2 | | 2× | 8× | 3 | | 3× | 1× | 0 | | 3× | 3× | 1 | | 3× | 9× | 2 | | 4× | 2× | 1/2 | | 4× | 8× | 3/2 |

The general formula: order m = log(rate ratio) / log(concentration ratio).

Most exam questions are constructed with concentration ratios of 2× or 3× to give clean integer orders. If the math gets messy, recheck the rate ratio — small experimental errors can throw off fractional orders.

The rate constant k has different units depending on the total order. Rate is always in concentration per time (typically M/s). Solve for k's units:

- Zero order overall: Rate = k. Units: M/s - First order overall: Rate = k[A]. Units: s^-1 - Second order overall: Rate = k[A]^2 or k[A][B]. Units: M^-1 s^-1 - Third order overall: Rate = k[A]^2[B] or k[A][B][C]. Units: M^-2 s^-1 - nth order: Units of k = M^(1-n) s^-1

If you compute k and the units don't match the overall order, you have an arithmetic error somewhere. Always check units as a self-check.

Snap a photo of the data table and ChemistryIQ identifies which experiments to compare for each reactant, computes the rate ratios and concentration ratios, solves for each order, plugs back to find k, checks units, and verifies the rate law against any extra experiments in the table. Useful for general chemistry students working through ACS exam practice and for physical chemistry students transitioning from basic kinetics to the integrated rate law and Arrhenius equation work.

First mistake: comparing experiments where TWO concentrations change at once. The method works only when you isolate one variable. If both [A] and [B] change between Exp 1 and Exp 2, you can't determine order in A from that comparison. Always pick experiments that hold all but one concentration constant.

Second mistake: getting the ratio direction wrong. Rate2/Rate1 = ([A]2/[A]1)^m — both sides have the same direction. Inverting one but not the other flips the sign of the order.

Third mistake: forgetting that concentrations and rates have units, but in the ratio they cancel — so the order is dimensionless. The units only matter when you solve for k.

Fourth mistake: assuming the order matches the stoichiometric coefficient. The rate law for 2NO + O2 → 2NO2 turned out to be second order in NO (matching) and first order in O2 (matching), but this is coincidence. The reaction 2H2 + O2 → 2H2O has experimental rate law approximately first order in H2 and first order in O2, NOT second order in H2 as stoichiometry would suggest. The rate law comes from the rate-limiting step's molecularity, not the overall balanced equation.