Reactivity How fast? The rate of chemical change : R2.2.11 Rate constant IB DP Chemistry Study Notes - New Syllabus 2025
Reactivity How fast? The rate of chemical change – IB DP Chemistry- Study Notes
IITian Academy excellent Introduction to the Particulate Nature of Matter – Study Notes and effective strategies will help you prepare for your IB DP Chemistry 2025 exam.
- IB DP Chemistry 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Chemistry 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Chemistry 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Chemistry 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Reactivity 2.2.11 – The Rate Constant \( k \)
Reactivity 2.2.11 – The Rate Constant \( k \)
Definition:
The rate constant, \( k \), is a proportionality constant in the rate equation of a chemical reaction. It links the rate of reaction to the concentrations of the reactants raised to their respective orders.
Temperature Dependence of the Rate Constant
- The value of the rate constant \( k \) is not fixed—it varies with temperature.
- As temperature increases, more particles have kinetic energy equal to or greater than the activation energy (\( E_a \)), increasing the frequency of effective collisions and thus increasing the value of \( k \).
- Rate roughly doubles for every 10 °C increase, though this can vary.
- Higher rate ≠ higher yield (depends on enthalpy and equilibrium).
Arrhenius Equation:
\( k = A e^{-E_a/RT} \)
- \( k \) = rate constant
- \( A \) = frequency factor (related to collision frequency and orientation)
- \( E_a \) = activation energy (J/mol)
- \( R \) = gas constant = 8.314 J·mol⁻¹·K⁻¹
- \( T \) = temperature in Kelvin
As temperature increases, the exponent \( -E_a/RT \) becomes less negative, which increases the value of \( k \).
Solving Problems Involving the Rate Equation and Units of \( k \)
1. General Rate Equation
The rate of a reaction is described by the rate equation:
\( \text{Rate} = k[A]^m[B]^n \)
- \( \text{Rate} \) – typically in mol·L⁻¹·s⁻¹
- \( k \) – rate constant (units depend on overall order)
- \( [A], [B] \) – concentrations in mol·L⁻¹
- \( m, n \) – orders of reaction determined experimentally
2. Determining the Rate Constant \( k \)
Once the rate, concentrations, and reaction orders are known, you can calculate the rate constant:
\( k = \frac{\text{Rate}}{[A]^m[B]^n} \)
3. Units of \( k \) for Different Orders
The units of \( k \) depend on the overall order of the reaction. They are derived by rearranging the rate equation:
\( k = \frac{\text{Rate}}{[A]^m[B]^n} \)
| Overall Order | Units of \( k \) |
|---|---|
| Zero | mol·L⁻¹·s⁻¹ |
| First | s⁻¹ |
| Second | L·mol⁻¹·s⁻¹ |
| Third | L²·mol⁻²·s⁻¹ |
4. Rearranging the Rate Equation to Solve for Other Quantities
- To find rate: Use known \( k \), concentrations, and orders.
- To find concentration: Rearrange to isolate the unknown concentration.
- To find units of \( k \): Use dimensional analysis from \( \text{Rate} = k[A]^x[B]^y \).
Key Points
- The rate constant \( k \) changes with temperature but not with concentration.
- The larger the value of \( k \), the faster the reaction (assuming the same reactant concentrations).
- Units of \( k \) depend entirely on the overall reaction order and are critical for interpreting experimental rate laws.
Example:
The reaction between A and B follows the rate law: \( \text{Rate} = k[A]^2[B] \) In an experiment, the rate of reaction is \( 3.60 \times 10^{-4} \text{ mol·L}^{-1}\text{s}^{-1} \) when \( [A] = 0.20 \text{ mol·L}^{-1} \) and \( [B] = 0.10 \text{ mol·L}^{-1} \). Calculate the value and units of the rate constant \( k \).
▶️Answer/Explanation
Use the rate law: \( \text{Rate} = k[A]^2[B] \)
Substitute the values: \( 3.60 \times 10^{-4} = k(0.20)^2(0.10) \) \( 3.60 \times 10^{-4} = k(0.04)(0.10) = k(0.004) \)
Now solve for \( k \): \( k = \frac{3.60 \times 10^{-4}}{0.004} = 0.090 \)
Reaction order = 2 + 1 = 3 → units of \( k = \text{L}^2\text{mol}^{-2}\text{s}^{-1} \)
Final Answer: \( k = 0.090 \text{ L}^2\text{mol}^{-2}\text{s}^{-1} \)
Example:
The decomposition of gas C follows first-order kinetics. At 400 K, the rate constant \( k \) is \( 1.5 \times 10^{-3} \text{ s}^{-1} \). If the initial concentration of C is \( 0.80 \text{ mol·L}^{-1} \), calculate the concentration of C after 8 minutes.
▶️Answer/Explanation
For a first-order reaction, use the integrated rate law:
\( \ln\left( \frac{[C]_0}{[C]} \right) = kt \)
Given: \( [C]_0 = 0.80 \text{ mol·L}^{-1} \) \( k = 1.5 \times 10^{-3} \text{ s}^{-1} \) \( t = 8 \text{ min} = 480 \text{ s} \)
\( \ln\left( \frac{0.80}{[C]} \right) = (1.5 \times 10^{-3})(480) = 0.72 \) \( \frac{0.80}{[C]} = e^{0.72} \approx 2.05 \) \( [C] = \frac{0.80}{2.05} \approx 0.39 \text{ mol·L}^{-1} \)
Final Answer: \( [C] \approx 0.39 \text{ mol·L}^{-1} \)
Example:
The following reaction is proposed to follow the rate law: \( \text{Rate} = k[X]^2[Y] \) An experiment was carried out at constant temperature using the following initial concentrations:
| Trial | [X] (mol·L⁻¹) | [Y] (mol·L⁻¹) | Initial Rate (mol·L⁻¹·s⁻¹) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻⁴ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻⁴ |
| 3 | 0.20 | 0.20 | 1.6 × 10⁻³ |
Using the above data, calculate the rate constant \( k \) and its units.
▶️Answer/Explanation
Step 1: Use the rate law:
\( \text{Rate} = k[X]^2[Y] \)
Use Trial 1:
\( 2.0 \times 10^{-4} = k(0.10)^2(0.10) = k \times 0.001 \)
\( k = \frac{2.0 \times 10^{-4}}{0.001} = 0.20 \)
Check with Trial 2:
\( \text{Rate} = k(0.20)^2(0.10) = 0.20 \times 0.004 = 8.0 \times 10^{-4} \)
Check with Trial 3:
\( \text{Rate} = k(0.20)^2(0.20) = 0.20 \times 0.008 = 1.6 \times 10^{-3} \)
Step 2: Units of \( k \)
Overall order = 2 (from X) + 1 (from Y) = 3
\( [k] = \frac{\text{mol·L}^{-1}\text{s}^{-1}}{(\text{mol·L}^{-1})^3} = \text{L}^2\text{mol}^{-2}\text{s}^{-1} \)
Final Answer: \( k = 0.20 \text{ L}^2\text{mol}^{-2}\text{s}^{-1} \)