Rate Law Notes

Rate Law Introduction

Learning Outcomes

• Define rate law and its variables.
• Analyze how changes in reactant concentration affect the rate of a chemical reaction using the rate law.
• Deduce the value of the rate constant using the rate law.

Success Criteria

• Define a rate law and its link to reaction rate and reactant concentration.
• Use experimental data to determine the rate law of a reaction.
• Deduce the rate constant by using the rate law.

Reaction Kinetics

• Rate of Reaction
  • Average Rate
  • Initial Rate
  • Instantaneous Rate
  • Units: mol/L*s
• Factors Affecting
  • Concentration
  • Temperature
  • Surface Area
  • Catalyst
• Collision Theory
  • Particles must collide with enough energy (> E_a).
  • Particles must collide with the correct orientation.
• Rate Law
  • Rate = $k[A]^m[B]^n$
    • m, n: Order of A, B
    • Overall order
    • Deduced from experimental data (not from coefficients).
    • k: The rate constant
• Mechanism
  • Steps in reaction
  • Rate-determining step

Rate Law Definition

The rate law is an equation that describes the relationship between the rate of a chemical reaction and the concentrations of the reactants. It shows how changes in the concentration of reactants affect the speed (rate) of the reaction.

Rate Law Expression

Rate = $k[A]^m[B]^n$

• k: proportionality constant called the rate constant. Units of “k” depend on the order of reaction.
• m: order of the reaction with respect to A.
• n: order of the reaction with respect to B.
• The values of m & n are not related to the reaction stoichiometry and must be determined by analyzing experimental data

$aA + bB \rightarrow products$

Reaction Order Importance

• Predicts rate changes
• Defines rate law
• Needed to calculate rate constant k
• The power to which the concentration of a reactant is raised in the rate law, showing how it affects the reaction rate.

Experimentally Deduce Rate Law

$A + B \rightarrow C$

1. Look at two experiments where the concentration of one reactant changes, but the others stay constant.
2. See how the rate changes.
3. Use this formula to find the order: $\frac{Rate<em>2}{Rate</em>1} = \frac{[A<em>2]}{[A</em>1]}^m$
4. Repeat this step for each of the reactants
5. Write the general rate law form Rate = $k[A]^m[B]^n$
6. Use data from experiments to find the order of each reactant m and n - the orders of reaction with respect to reactants A and B
7. Write the full law with the orders
8. Find the rate constant k
   • Use data from any one experiment
   • Plug in the values for [A], [B], and Rate into the rate law
   • Solve for k
9. After finding m and n - plug them back into the rate law

Worked Example

$2NO(g) + 2H<em>2(g) \rightarrow N</em>2(g) + 2H_2O(g)$

Step 1: Find the order with respect to [NO]

Use Experiments 1 and 2 (same [H2]):

$\frac{Rate<em>2}{Rate</em>1} = \left(\frac{[NO]<em>2}{[NO]</em>1}\right)^m$

$\frac{5.00 \times 10^{-5}}{1.25 \times 10^{-5}} = \left(\frac{0.0100}{0.0050}\right)^m$

$4 = 2^m$

Order with respect to NO = 2

Step 2: Find the order with respect to [H2]

Use Experiments 2 and 3 (same [NO]):

$\frac{Rate<em>3}{Rate</em>2} = \left(\frac{[H<em>2]</em>3}{[H<em>2]</em>2}\right)^n$
$\frac{1.00 \times 10^{-4}}{5.00 \times 10^{-5}} = \left(\frac{0.0040}{0.0020}\right)^n$
$2 = 2^n$

Order with respect to $H_2$ = 1

Step 3: Write the rate law

Rate = $k[NO]^2[H_2]$

Step 4: Find the rate constant k using Experiment 1

Given:

• Rate = $1.25 \times 10^{-5}$ M/s
• [NO] = 0.0050 M
• [$H_2$] = 0.0020 M

Substitute:

$1.25 \times 10^{-5} = k(0.0050)^2(0.0020)$

$1.25 \times 10^{-5} = k(2.5 \times 10^{-5})(0.0020) = k(5.0 \times 10^{-8})$

$k = \frac{1.25 \times 10^{-5}}{5.0 \times 10^{-8}} = 250$

Rate constant k = 250 $M^{-2} s^{-1}$

Step 5: Determine the overall order of the reaction

• Order with respect to NO = 2
• Order with respect to $H_2$ = 1
• Overall order = 2 + 1 = 3

Step 6: Calculate the rate when [NO] = 0.012 M and [$H_2$] = 0.0060 M

Use:

Rate = $250(0.012)^2 (0.0060)$

$(0.012)^2 = 0.000144$

$0.000144 \times 0.0060 = 8.64 \times 10^{-7}$

$250 \times 8.64 \times 10^{-7} = 2.16 \times 10^{-4}$

Rate = $2.16 \times 10^{-4}$ M/s

Question 1

$C<em>2H</em>5Cl(g) \rightarrow C<em>2H</em>4(g) + HCl(g)$

According to the graph, the decomposition of $C<em>2H</em>5Cl(g)$ is a first-order reaction because the rate is directly proportional to the concentration of $C<em>2H</em>5Cl$.

Question 2

The rate law for a particular reaction is rate = $k[X]^2$. In an experiment, the initial rate of the reaction is determined to be 0.080 mol/(L. s) when the initial concentration of X is 0.20 mol/L.

What is the value of the rate constant, k, for the reaction?

Rate = $k[X]^2$
$0.080 = k(0.20)^2$
$0.080 = k(0.04)$
$k = \frac{0.080}{0.04} = 2.0 L/(mol \cdot s)$

Question 3

$2 ClO<em>2(aq) + 2 OH^-(aq) \rightarrow ClO</em>3^-(aq) + ClO<em>2^-(aq) + H</em>2O(l)$

A rate study of the reaction represented above yields the following data.

Based on the data, the rate law for the reaction is Rate = $k[ClO_2]^2[OH^-]$

Question 4

$X(g) + Q(g) \rightarrow R(g) + Z(g)$

The reaction represented above is found to be second order with respect to X and zero order with respect to Q.

What happens to the rate of the reaction when [X] is doubled and [Q] is halved?

It increases by a factor of 4.

Extension Task

Divide your group into two mini groups for part 1 & part 2 tasks: Research real-world applications of rate law, such as enzyme kinetics, pharmaceutical reactions, or industrial catalysis.

Exit Ticket

For the rate law: Rate = $k[A][B]$

a. What is the order with respect to A? (1)
b. What is the order with respect to B? (1)
c. What is the overall order of the reaction? (2)
d. What would the unit of "k" be in this case? ($mol^{-1} L s^{-1}$)