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Chemistry · General chemistry II · Concept

Finding reaction orders and the rate constant from experimental data

Find each reactant’s order and the rate constant k from initial-rate data, then use integrated rate law plots and half-life to find the order and k from concentration–time data.

What the rate law asks you to find

A rate law says how the rate depends on each reactant’s concentration. The exponents are the orders, and k is the rate constant. Neither can be read from the balanced equation: coefficients describe how much reacts, not how the rate responds. Orders come from experiments, usually small whole numbers such as 0, 1 or 2, though fractions and negative orders occur.

Once the orders are known, k follows from any single measurement. Its units depend on the overall order, so they are part of the answer.

rate=k⁢[A]m[B]n,overall order=m+n

Method of initial rates: one reactant at a time

Run the reaction several times, changing one starting concentration while holding every other concentration and the temperature fixed, and measure the rate at the very start. Divide the two rate laws: k and every unchanged concentration cancel, leaving only the ratio for the reactant you changed.

If doubling [A] doubles the rate, m = 1. If it quadruples the rate, m = 2. If the rate does not change, m = 0. When the ratios are not tidy, take logarithms.

rate2rate1=([A]2[A]1)m⇒m=ln(rate2/rate1)ln([A]2/⁢[A]1)
Reading an order when one concentration doubles
Rate changes byOrder in that reactantWhy
× 102⁰ = 1
× 212¹ = 2
× 422² = 4
× 832³ = 8
× 2.831.5 (3/2)2^1.5 ≈ 2.83

Finding k and its units

Substitute one trial’s rate and concentrations into the complete rate law and solve for k. Using a second trial should give the same k within measurement error, which is a useful check on the orders you chose. The units of k are whatever makes rate come out in M/s.

k=rate[A]m[B]n
Units of k (time in seconds)
Overall orderRate law formUnits of k
0rate = kM s⁻¹
1rate = k[A]s⁻¹
2rate = k[A]² or k[A][B]M⁻¹ s⁻¹
3rate = k[A]²[B]M⁻² s⁻¹

Concentration over time: which plot is straight?

A second kind of experiment follows one reactant as it is used up. Integrating the rate law gives an equation that is linear in time for exactly one choice of y-axis. Plot [A], ln[A] and 1/[A] against t: the plot that is straight tells you the order, and its slope gives k.

Compare the three plots, not one R² value. Over a short time range a curved plot can still have R² near 0.98, so look for the plot whose points have no systematic bend.

ln[A]t=−k⁢t+ln[A]01[A]t=k⁢t+1[A]0[A]t=−k⁢t+[A]0
Integrated rate laws
OrderStraight-line plotSlopeHalf-life
0[A] vs t−k[A]₀ / 2k
1ln[A] vs t−k0.693 / k
21/[A] vs t+k1 / (k[A]₀)

Half-life as a quick check

For a first-order reaction the half-life does not depend on concentration: every successive half takes the same time. If the time to fall from 0.50 M to 0.25 M equals the time to fall from 0.25 M to 0.125 M, the reaction is first order and k = 0.693/t½. For zero and second order, successive half-lives shrink or grow.

t1⁢/2=ln2k=0.693k

Common mistakes

  • Taking orders from the coefficients of the balanced equation.
  • Comparing two trials in which more than one concentration changed.
  • Leaving k without units, or giving units that don’t match the overall order.
  • Reading the slope of a ln[A] plot as +k instead of −k.
  • Choosing the integrated rate law from one R² near 1 instead of comparing all three plots.
  • Using the first-order half-life formula for a zero- or second-order reaction.

Work through an example

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.

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

Find the order and k from concentration–time data →

Sources and scope

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