Introduction to Rate Law (College Board AP Chemistry)

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Rate Law

Experimental data and the rate of reaction

One way to study the effect of concentration on the rate is monitoring the rate of reaction against the concentration of the reactant
The following general reaction will be used as example

D → E + F

The rate of reaction at different concentrations of D was tabulated and plotted

[D] (M)	Rate (M s^-1)
3.00	2.000 x 10^-3
2.00	1.334 x 10^-3
1.00	6.670 x 10^-4

A line is plotted and the gradient of the line is calculated using the formula below

$gradient = \frac{∆ y}{∆ x}$

Graph to show rate against [D]

rate-vs-concentration-graph

Rate of reaction over various concentrations of D

As shown in the diagram, there is a directly proportional relation between the rate of reaction and the concentration of D
This relation can be written as a mathematical equation as shown below

Rate ∝ [D]

Rate = k [D]

This equation means that if the concentration of D doubles, the rate will double
The constant of proportionality (k) is called the rate constant
- The units of the rate constant can be used to determine the order of the reaction

The rate equation

The following general reaction will be used to discuss the rate equation:

A +B → C+D

The general rate equation is:

Rate = k [A]^m[B]ⁿ

Where:

[A] and [B] are the concentration of reactants
m and n are the orders with respect to each reactant

The rate of the reaction will depend on the mechanism of reaction and it can only be found experimentally
Intermediate products do not feature in rate equations

The order of reactants

The order of reactants shows how the concentration of a reactant affect the rate
- It is represented as the power to which the concentration is raised in the rate equation
The most common orders are listed below:
Zero order occurs when the order respect to a chemical is 0
- This means that changing the concentration has no effect on the rate
- Therefore, it is not included in the rate equation
First order occurs when the order respect to a chemical is 1
- This means that the concentration is directly proportional to the rate
- E.g. Doubling the concentration will double the rate
Second order occurs when the order respect to a chemical is 2
- This means the that square of the concentration is directly proportional to the rate
- E.g. Doubling the concentration will increase the rate by a factor of 4 since 22 is 4

Overall order is just the sum of the powers of the reactants that appear in the rate equation

Worked example

The chemical equation for the thermal decomposition of dinitrogen pentoxide is:

2N₂O₅ (g) → 4NO₂ (g) + O₂ (g)

The rate equation for this reaction is:

Rate = k [N₂O₅ (g)]

State the order of reaction with respect to nitrogen pentoxide
State the overall order of reaction
Deduce the effect on the rate of reaction if the concentration of dinitrogen pentoxide is doubled
Determine the units of the rate constant

Answers:

Answer 1:

Dinitrogen pentoxide features in the rate equation, therefore, it cannot be order zero
The dinitrogen pentoxide is not raised to a power, which means that it cannot be order 2
Therefore, the order with respect to dinitrogen pentoxide must be order 1

Answer 2:

The overall order is just the sum of all the powers in the rate equation. Since, dinitrogen pentoxide is the only reactant and it is raised to the power of 1. The overall order is first order

Answer 3:

Since the overall order is first order, the concentration of dinitrogen pentoxide is directly proportional to the rate
This means that if the concentration of the dinitrogen pentoxide is doubled, then the rate of reaction will be the double

Answer 4:

$Units of rate = (units of rate constant) \times (units of concentration of N_{2} O_{5})$
Rearranging the equation,
$Units of rate constant = \frac{Units of rate}{Units of concentration of N_{2} O_{5}}$
$Units of rate constant = \frac{M s^{- 1}}{M} = s^{- 1}$

Exam Tip

The overall order of the reaction can be inferred from the units of the rate constant. If the rate is measure per second:

Rate constant units of a zero order reaction are M s^-1
Rate constant units of a first order reaction are s^-1
Rate constant units of a second order reaction are M^-1 s^-1

Initial Rates & Orders of Reaction

The order of the reactants can be determined with experimental data from the the initial rates of reaction
- This information is usually displayed in a table
The following steps must be followed to determine the order for each reactant:

1. Identify two experiments where the concentration of this reactant changes, but the concentration of the other reactants is the same
2. Identify what has happened with the concentration
3. Determine the effect of the concentration in the rate of reaction
4. Deduce the order of reaction
5. Repeat this procedure with all the reactants one at the time

Once all the orders are determined the rate equation can be built
- Zero order reactants are not included
- First order reactants are included and do not require a power (it is the s
- Second order reactants are raised to the power of 2

Exam Tip

Determining the orders for each reactant is one of the most assessed tasks regarding kinetics, so it's worth spending time to get to know this topic well

Worked example

(CH₃)₃CBr + OH^- → (CH₃)₃COH + Br^-

The table show experimental data for the reaction above

Experiment	Initial [(CH₃)₃CBr] / M	Initial [OH^-] / M	Initial rate of reaction / M s^-1
1	1.0 x 10^-3	2.0 x 10^-3	3.0 x 10^-3
2	2.0 x 10^-3	2.0 x 10^-3	6.0 x 10^-3
3	1.0 x 10^-3	4.0 x 10^-3	1.2 x 10^-2
4	1.5 x 10^-3	4.0 x 10^-3	4.5 x 10^-3

Determine the rate equation for this reaction

Answer:

Order with respect to (CH₃)₃CBr

In experiments 1 and 2, the concentration of (CH₃)₃CBr changes while the concentration of OH^- remains constant
The [(CH₃)₃CBr] has doubled
The rate of the reaction has also doubled
Therefore, the order with respect to [(CH₃)₃CBr] is 1 (first order)

$[Change in concentration]^{order} = change in rate$

${[2]}^{order} = 2$

${[2]}^{1} = 2$

order = 1

Order with respect to OH-

In experiments 1 and 3, the concentration of OH^- changes while the concentration of (CH₃)₃CBr remains constant
The [OH^-] has doubled
The rate of the reaction has increased by a factor of 4
Therefore, the order with respect to [OH^-] is 2 (second order)

$[Change in concentration]^{order} = change in rate$

${[2]}^{order} = 4$

${[2]}^{2} = 4$

order = 2

Building the rate equation

$Rate = k {[{({CH}_{3})}_{3} CBr]}^{m} {[{OH}^{-}]}^{n}$

Replacing the orders,

$Rate = k [{({CH}_{3})}_{3} CBr] {[{OH}^{-}]}^{2}$

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