Chemical Reactions & Equilibrium

Rates of Reaction

5th Year · 6th Year (Leaving Cert)

  • Define the rate of a chemical reaction and explain how it is measured.
  • Explain collision theory and its role in determining reaction rates, including the concept of activation energy.
  • Describe and explain the effect of concentration, temperature, surface area, and catalysts on the rate of chemical reactions.
  • Deduce the order of a reaction with respect to each reactant and the overall order from experimental data (HL).
  • Write the rate expression for a reaction and calculate the rate constant (k) with appropriate units (HL).

Key concepts

Rate of Reaction

The rate of a chemical reaction is a measure of how quickly reactants are consumed or products are formed. It is defined as the change in concentration of a reactant or product per unit time.

Rate = Δ[concentration] / Δt
Collision Theory

For a chemical reaction to occur, reactant particles must collide with each other. However, not all collisions lead to a reaction. Only 'effective collisions' result in product formation. An effective collision requires two conditions: 1. **Correct Orientation**: The colliding particles must be oriented in a way that allows the reactive parts of the molecules to come into contact. 2. **Sufficient Energy**: The colliding particles must possess a minimum amount of kinetic energy, known as the activation energy (Ea), to break existing bonds and form new ones.

Activation Energy (Ea)

The minimum amount of energy that reacting particles must possess upon collision for a chemical reaction to occur. It represents an energy barrier that must be overcome for reactants to transform into products.

Effect of Concentration

Increasing the concentration of reactants means there are more reactant particles per unit volume. This leads to a higher frequency of collisions between reacting particles, thereby increasing the frequency of effective collisions and speeding up the reaction.

Effect of Temperature

Increasing the temperature increases the average kinetic energy of the reactant particles. This has two main effects: 1. Particles move faster, leading to a higher frequency of collisions. 2. A significantly larger proportion of collisions will have energy equal to or greater than the activation energy, leading to a higher frequency of effective collisions. This second effect is the more dominant factor.

Effect of Surface Area

For reactions involving solid reactants, increasing the surface area (e.g., by crushing a lump into a powder) exposes more reactant particles to the other reactants. This increases the frequency of collisions between the solid and other reactants, thus increasing the rate of reaction.

Effect of a Catalyst

A catalyst is a substance that increases the rate of a chemical reaction without being consumed in the overall reaction. It achieves this by providing an alternative reaction pathway with a lower activation energy. This means a greater proportion of reactant particles will possess sufficient energy to react, increasing the frequency of effective collisions.

Rate Expression (Rate Law) (HL)

For a general reaction aA + bB → cC + dD, the rate expression (or rate law) mathematically relates the rate of reaction to the concentrations of the reactants. It is determined experimentally and cannot be deduced directly from the stoichiometry of the balanced chemical equation. Rate = k[A]^x[B]^y Where: * Rate is the reaction rate. * k is the **rate constant**, a proportionality constant specific to a reaction at a given temperature. * [A] and [B] are the molar concentrations of reactants A and B. * x is the **order of reaction** with respect to reactant A. * y is the **order of reaction** with respect to reactant B. * The **overall order of reaction** is the sum of the individual orders, i.e., x + y.

Rate = k[A]^x[B]^y

Key facts to remember

  • 1The rate of reaction is the change in concentration per unit time.
  • 2Collision theory requires particles to collide with sufficient energy (activation energy) and correct orientation for a reaction to occur.
  • 3Activation energy (Ea) is the minimum energy required for an effective collision.
  • 4Increasing concentration, temperature, and surface area generally increase the rate of reaction by increasing the frequency of effective collisions.
  • 5A catalyst speeds up a reaction by providing an alternative reaction pathway with a lower activation energy, without being consumed.
  • 6(HL) The rate expression (Rate = k[A]^x[B]^y) is determined experimentally, not from the stoichiometry of the balanced equation.
  • 7(HL) The rate constant (k) is temperature-dependent and has units that vary with the overall order of the reaction.

Worked examples

Example 1

In the decomposition of hydrogen peroxide, 2H₂O₂(aq) → 2H₂O(l) + O₂(g), the concentration of H₂O₂ decreases from 0.80 mol L⁻¹ to 0.40 mol L⁻¹ in 200 seconds. Calculate the average rate of disappearance of H₂O₂ during this time interval.

IIdentify the initial concentration, final concentration, and time interval:
IIInitial [H₂O₂] = 0.80 mol L⁻¹
IIIFinal [H₂O₂] = 0.40 mol L⁻¹
IVΔt = 200 s
VCalculate the change in concentration (Δ[H₂O₂]):
VIΔ[H₂O₂] = Final [H₂O₂] - Initial [H₂O₂] = 0.40 mol L⁻¹ - 0.80 mol L⁻¹ = -0.40 mol L⁻¹
VIICalculate the average rate of disappearance. Since concentration is decreasing, we use a negative sign to ensure the rate is positive:
VIIIRate = -Δ[H₂O₂] / Δt
9Rate = -(-0.40 mol L⁻¹) / 200 s
10Rate = 0.40 / 200 mol L⁻¹ s⁻¹
11Rate = 0.002 mol L⁻¹ s⁻¹

Answer

The average rate of disappearance of H₂O₂ is 0.002 mol L⁻¹ s⁻¹.

The rate of disappearance of a reactant is always expressed as a positive value.

Example 2

The following experimental data were obtained for the reaction: P + Q → R | Experiment | [P] (mol L⁻¹) | [Q] (mol L⁻¹) | Initial Rate (mol L⁻¹ s⁻¹) | |------------|---------------|---------------|----------------------------| | 1 | 0.10 | 0.10 | 1.5 x 10⁻³ | | 2 | 0.20 | 0.10 | 3.0 x 10⁻³ | | 3 | 0.10 | 0.30 | 1.35 x 10⁻² | Determine the order of reaction with respect to P and Q, the overall order of reaction, and the value and units of the rate constant (k). (HL)

I**1. Determine the order with respect to P:**
IICompare Experiment 1 and Experiment 2, where [Q] is kept constant.
IIIWhen [P] doubles (0.10 to 0.20), the rate doubles (1.5 x 10⁻³ to 3.0 x 10⁻³).
IVMathematically: (Rate₂ / Rate₁) = ([P]₂ / [P]₁)^x
V(3.0 x 10⁻³ / 1.5 x 10⁻³) = (0.20 / 0.10)^x
VI2 = 2^x, therefore x = 1.
VIIThe reaction is first order with respect to P.
VIII**2. Determine the order with respect to Q:**
9Compare Experiment 1 and Experiment 3, where [P] is kept constant.
10When [Q] triples (0.10 to 0.30), the rate increases by a factor of 9 (1.5 x 10⁻³ to 1.35 x 10⁻²).
11Mathematically: (Rate₃ / Rate₁) = ([Q]₃ / [Q]₁)^y
12(1.35 x 10⁻² / 1.5 x 10⁻³) = (0.30 / 0.10)^y
139 = 3^y, therefore y = 2.
14The reaction is second order with respect to Q.
15**3. Write the rate expression:**
16Rate = k[P]¹[Q]² or Rate = k[P][Q]²
17**4. Determine the overall order of reaction:**
18Overall order = x + y = 1 + 2 = 3.
19**5. Calculate the rate constant (k) and its units:**
20Using the rate expression and data from Experiment 1:
21Rate = k[P][Q]²
221.5 x 10⁻³ mol L⁻¹ s⁻¹ = k (0.10 mol L⁻¹) (0.10 mol L⁻¹)²
231.5 x 10⁻³ = k (0.10) (0.01)
241.5 x 10⁻³ = k (0.001)
25k = (1.5 x 10⁻³) / (0.001)
26k = 1.5
27To find the units of k: k = Rate / ([P][Q]²)
28k = (mol L⁻¹ s⁻¹) / ((mol L⁻¹) (mol L⁻¹)²)
29k = (mol L⁻¹ s⁻¹) / (mol³ L⁻³)
30k = mol⁻² L² s⁻¹

Answer

Order with respect to P = 1 Order with respect to Q = 2 Overall order = 3 Rate constant (k) = 1.5 mol⁻² L² s⁻¹

The units of the rate constant depend on the overall order of the reaction. Always derive them if you are unsure.

Common mistakes

  • Confusing activation energy with the overall enthalpy change (ΔH) of a reaction.
  • Assuming that the stoichiometric coefficients in a balanced chemical equation directly correspond to the orders of reaction in the rate expression (HL).
  • Forgetting to include units for the rate constant (k) or using incorrect units (HL).
  • Incorrectly explaining the effect of temperature, focusing only on increased collision frequency rather than the increased proportion of effective collisions.
  • Believing that a catalyst is consumed or used up during the reaction.

Exam tips

  • When explaining factors affecting reaction rate, always link your explanation back to collision theory (frequency of collisions and/or proportion of effective collisions).
  • For Higher Level questions on rate expressions, clearly show your step-by-step working for determining the order with respect to each reactant by comparing experimental data.
  • Remember that the units of the rate constant (k) are dependent on the overall order of the reaction; derive them if you are unsure.
  • Practise drawing and interpreting energy profile diagrams, especially showing how a catalyst lowers the activation energy.
  • Be precise with terminology; use 'activation energy' rather than just 'energy barrier' or 'threshold energy'.

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