Chemistry · General chemistry II · Worked example

Find the orders and k for 2NO + O₂ from initial rates

For 2NO(g) + O₂(g) → 2NO₂(g) at a fixed temperature, three trials give these initial rates. Find the order in NO, the order in O₂, the rate law and k with units.

2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g), rate = k {[NO]}^{m} {[O_{2}]}^{n}

Lay out the data

Look for pairs of trials in which only one concentration changes. Trials 1 and 2 change only [NO]; trials 1 and 3 change only [O₂].

Initial-rate data (illustrative)
Trial	[NO] (M)	[O₂] (M)	Initial rate (M/s)
1	0.020	0.010	0.028
2	0.040	0.010	0.112
3	0.020	0.020	0.056

Order in NO from trials 1 and 2

Doubling [NO] multiplies the rate by 4. Since 2^m = 4, the reaction is second order in NO.

\frac{0.112}{0.028} = {(\frac{0.040}{0.020})}^{m} \Rightarrow 4 = 2^{m} \Rightarrow m = 2

Order in O₂ from trials 1 and 3

Doubling [O₂] doubles the rate, so the reaction is first order in O₂.

\frac{0.056}{0.028} = {(\frac{0.020}{0.010})}^{n} \Rightarrow 2 = 2^{n} \Rightarrow n = 1

Write the rate law and solve for k

The overall order is 3, so k has units of M⁻² s⁻¹. Trial 2 gives the same value, 0.112 ÷ (0.040² × 0.010) = 7.0 × 10³, which confirms the orders.

k = \frac{0.028 M s^{- 1}}{{(0.020 M)}^{2} (0.010 M)} = 7.0 \times 10^{3} M^{- 2} s^{- 1}

Result

Rate = k[NO]²[O₂], third order overall, with k = 7.0 × 10³ M⁻² s⁻¹.

rate = 7.0 \times 10^{3} M^{- 2} s^{- 1} {[NO]}^{2} [O_{2}]

Your turn

For A + B → C: trial 1 has [A] = 0.10 M, [B] = 0.10 M and rate 3.0 × 10⁻⁴ M/s. Trial 2 has [A] = 0.30 M, [B] = 0.10 M and rate 9.0 × 10⁻⁴ M/s. Trial 3 has [A] = 0.10 M, [B] = 0.30 M and rate 3.0 × 10⁻⁴ M/s. Find the rate law and k.

Show the answer and explanation

Rate = k[A], with k = 3.0 × 10⁻³ s⁻¹.

Tripling [A] triples the rate, so the order in A is 1. Tripling [B] leaves the rate unchanged, so the order in B is 0. Then k = 3.0 × 10⁻⁴ ÷ 0.10 = 3.0 × 10⁻³ s⁻¹, in units of s⁻¹ because the reaction is first order overall.

k = \frac{3.0 \times 10^{- 4} M s^{- 1}}{0.10 M} = 3.0 \times 10^{- 3} s^{- 1}

Keep exploring

Open the Kinetics studio with trials 1 and 2, then change the second rate and watch the order it reports.

Return to the concept →

Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

Tro, Chemistry: A Molecular Approach, 4th ed., §14.3 The Rate Law: The Effect of Concentration on Reaction Rate, pp. 629–633 (method of initial rates, p. 630)
Tro, Chemistry: A Molecular Approach, 4th ed., §14.4 The Integrated Rate Law: The Dependence of Concentration on Time, pp. 634–641
OpenStax Chemistry 2e — Rate laws
OpenStax Chemistry 2e — Integrated rate laws