Study Guide

Rate constant calculations

Chemistry· Section 11: Reaction Kinetics· 15 min read

1. Calculating \(k\) from initial rate data and finding units★★☆☆☆⏱ 4 min

📘 Definition

Units of the rate constant

Units of (k) depend on the overall order of the reaction, derived by rearranging the rate equation to isolate (k) and substituting units of concentration (mol dm⁻³) and rate (mol dm⁻³ s⁻¹).

Example:

For a zero order reaction, (k) has the same units as rate: mol dm⁻³ s⁻¹

Once you have the rate equation and experimental initial rate data, you can substitute known values of rate and reactant concentrations to solve for (k) directly. CIE examiners always award separate marks for correct units, so never omit this step.

📐 Worked Example

For the reaction (2A + B \rightarrow C), the rate equation is (\text{rate} = k[A]^2[B]). When ([A] = 0.1 \text{ mol dm}^{-3}), ([B] = 0.2 \text{ mol dm}^{-3}), initial rate is (4 \times 10^{-5} \text{ mol dm}^{-3} s^{-1}). Calculate the value and units of (k).

  1. 1

    Rearrange the rate equation to isolate (k):

  2. 2
    k=rate[A]2[B]k = \frac{\text{rate}}{[A]^2[B]}
  3. 3

    Substitute the given numerical values:

  4. 4
    k=4×10−5(0.1)2(0.2)=4×10−50.002=0.02k = \frac{4 \times 10^{-5}}{(0.1)^2(0.2)} = \frac{4 \times 10^{-5}}{0.002} = 0.02
  5. 5

    Substitute units to find the units of (k):

  6. 6
    Units=mol dm−3s−1(mol dm−3)2(mol dm−3)=mol−2dm6s−1\text{Units} = \frac{\text{mol dm}^{-3} \text{s}^{-1}}{(\text{mol dm}^{-3})^2 (\text{mol dm}^{-3})} = \text{mol}^{-2} \text{dm}^6 \text{s}^{-1}
  7. 7

    Final answer:

  8. 8
    k=0.02 mol−2dm6s−1k = 0.02 \text{ mol}^{-2} \text{dm}^6 \text{s}^{-1}

2. Calculating \(k\) for first order reactions from half-life★★☆☆☆⏱ 3 min

For all first order reactions, half-life is constant and independent of reactant concentration. This gives a simple, direct relationship between the rate constant and half-life that is very commonly tested in CIE exams.

📘 Definition

Half-life-rate constant relationship for first order reactions

k=ln⁡2t1/2k = \frac{\ln 2}{t_{1/2}}

For any first order reaction, the rate constant equals the natural logarithm of 2 divided by the measured half-life of the reactant.

Example:

If half-life = 120 s, (k = 0.693 / 120 ≈ 5.8 \times 10^{-3} \text{ s}^{-1})

📐 Worked Example

The first order decomposition of hydrogen peroxide has a half-life of 140 s at 298 K. Calculate the rate constant for this reaction, with correct units.

  1. 1

    Write the standard relationship for first order (k) and (t_{1/2}):

  2. 2
    k=ln⁡2t1/2k = \frac{\ln 2}{t_{1/2}}
  3. 3

    Substitute (t_{1/2} = 140 \text{ s}):

  4. 4
    k=0.693140≈0.00495k = \frac{0.693}{140} ≈ 0.00495
  5. 5

    First order rate constants always have units of time⁻¹, so final answer:

  6. 6
    k=4.95×10−3 s−1k = 4.95 \times 10^{-3} \text{ s}^{-1}

3. Calculating \(k\) from kinetic graphs★★★☆☆⏱ 4 min

The gradient of correctly plotted kinetic graphs directly gives the rate constant, regardless of reaction order. The relationship between gradient and (k) depends on the plot you are given:

  • Zero order: Plot of ([A]) against (t) → gradient = (-k)

  • First order: Plot of rate against ([A]) → gradient = (k)

  • Second order: Plot of rate against ([A]^2) → gradient = (k)

📐 Worked Example

A first order reaction gives a straight line rate-concentration graph with gradient 0.025 min⁻¹. What is (k) and its units?

  1. 1

    For a first order reaction, rate = (k[A]), which matches the equation of a straight line (y = mx) where (y = \text{rate}), (x = [A]) and (m = k).

  2. 2

    The gradient of the plot equals (k), and the gradient units are min⁻¹, which matches the expected units for a first order (k).

  3. 3

    Final answer:

  4. 4
    k=0.025 min−1k = 0.025 \text{ min}^{-1}

4. Calculating \(k\) from the integrated rate equation★★★☆☆A2 only⏱ 4 min

The integrated rate equation relates reactant concentration at time (t) to initial concentration, allowing you to calculate (k) from any single measurement of concentration at a known time. For CIE A-Level, the first order integrated rate equation is the most commonly tested.

📘 Definition

First order integrated rate equation

ln⁡[A]t=−kt+ln⁡[A]0\ln[A]_t = -kt + \ln[A]_0

Where ([A]_t) is concentration of A at time (t), ([A]_0) is initial concentration, and (k) is the rate constant. A plot of (\ln[A]_t) against (t) gives a straight line with gradient (-k).

📐 Worked Example

The initial concentration of A in a first order reaction is 0.10 mol dm⁻³. After 100 s, the concentration of A falls to 0.035 mol dm⁻³. Calculate (k).

  1. 1

    Start with the integrated first order rate equation:

  2. 2
    ln⁡[A]t=−kt+ln⁡[A]0\ln[A]_t = -kt + \ln[A]_0
  3. 3

    Rearrange to isolate (k):

  4. 4
    k=ln⁡[A]0−ln⁡[A]tt=ln⁡([A]0[A]t)tk = \frac{\ln[A]_0 - \ln[A]_t}{t} = \frac{\ln\left(\frac{[A]_0}{[A]_t}\right)}{t}
  5. 5

    Substitute the given values:

  6. 6
    k=ln⁡(0.100.035)100=ln⁡(2.857)100≈1.05100=0.0105k = \frac{\ln\left(\frac{0.10}{0.035}\right)}{100} = \frac{\ln(2.857)}{100} ≈ \frac{1.05}{100} = 0.0105
  7. 7

    Final answer with units:

  8. 8
    k=1.05×10−2 s−1k = 1.05 \times 10^{-2} \text{ s}^{-1}

5. Common Pitfalls

Wrong move:

Forgetting to include units of (k), or writing incorrect units for the overall reaction order

Why:

CIE always awards at least one separate mark for correct units, so you can lose a mark even if your numerical value is right

Correct move:

Always derive units by substituting concentration and rate units into the rearranged rate equation for (k)

Wrong move:

Using log base 10 instead of natural log for first order rate constant calculations

Why:

The relationship (k = \ln 2 / t_{1/2}) only works for natural logs; using log base 10 will give a value 2.3 times too small

Correct move:

Always confirm you are using natural log (ln) for all first order rate constant calculations

Wrong move:

Using the half-life formula (k = \ln 2 / t_{1/2}) for non-first order reactions

Why:

Half-life is only constant for first order reactions, so this relationship does not hold for other orders

Correct move:

Only use the half-life method for confirmed first order reactions, use initial rate substitution for other orders

Wrong move:

Forgetting to apply the reaction order to concentration terms when substituting into the rate equation

Why:

If a reactant is second order, you must square its concentration, not use the concentration directly

Correct move:

Always check the order of each reactant in the rate equation before substituting values to calculate (k)

6. Quick Reference Cheatsheet

Overall order

Units of (k)

Method to calculate (k)

0

mol dm⁻³ s⁻¹

k = rate, gradient of [A] vs t = -k

1

time⁻¹ (e.g. s⁻¹)

k = ln2 / t₁/₂, gradient of rate vs [A] = k

2

mol⁻¹ dm³ s⁻¹

Substitute into rate = k[A]² or k[A][B]

3

mol⁻² dm⁶ s⁻¹

Substitute into full rate equation

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

  • 2022 · 2

    Calculate k from initial rate data

  • 2023 · 4

    Find k value and units for 2nd order

  • 2021 · 1

    Deduce units of k for 3rd order

Going deeper

What's Next

Rate constant calculations are the foundation for all further topics in reaction kinetics, and mastery of this sub-topic is required to access full marks on all kinetics questions in CIE A-Level exams. Next, you will use rate constants measured at different temperatures to calculate the activation energy of a reaction using the Arrhenius equation, a common extended response question in Paper 4. Understanding (k) also supports the study of reaction mechanisms, where rate constants are used to identify the rate-determining step and compare the rate of different reaction pathways. Solid skills in calculating (k) and its units will make all subsequent kinetics topics much easier to master.