Orders of Reaction
 The order of a reactant shows how the concentration of a chemical, typically a reactant, affects the rate of reaction
 It is the power to which the concentration of that reactant is raised in the rate equation
 The order can have a positive, negative or fractional value
 Orders that are a fraction suggest that the reaction involves multiple steps
 When the order of reaction with respect to a chemical is 0
 Changing the concentration of the chemical has no effect on the rate of the reaction
 Therefore, it is not included in the rate equation
 When the order of reaction with respect to a chemical is 1
 The concentration of the chemical is directly proportional to the rate of reaction, e.g. doubling the concentration of the chemical doubles the rate of reaction
 The chemical is included in the rate equation
 When the order of reaction with respect to a chemical is 2
 The rate is directly proportional to the square of the concentration of that chemical, e.g. doubling the concentration of the chemical increases the rate of reaction by a factor of four
 The chemical is included in the rate equation (appearing as a squared term)
 The overall order of reaction is the sum of the powers of the reactants in a 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 the reaction with respect to dinitrogen pentoxide
 Deduce the effect on the rate of reaction if the concentration of dinitrogen pentoxide is tripled
Answers
Answer 1:

 Dinitrogen pentoxide features in the rate equation, therefore, it cannot be order zero / 0
 The dinitrogen pentoxide is not raised to a power, which means that it cannot be order 2 / second order
 Therefore, the order with respect to dinitrogen pentoxide must be order 1 / first order
Answer 2:

 Since the reaction 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 tripled, then the rate of reaction will also triple
Worked Example
The following equation represents the oxidation of bromide ions in acidic solution
BrO3 (aq) + 5Br (aq) + 6H+ (aq) → 3Br2 (l) + 3H2O (l)
The rate equation for this reaction is:
Rate = k[BrO3 (aq)][Br (aq)][H+ (aq)]
 State the overall order of the reaction
 Deduce the effect on the rate of reaction if the concentration of bromate ions is doubled and the concentration of bromide ions is halved
Answers
Answer 1:

 All three reactants feature in the rate equation but they are not raised to a power, this means that the order with respect to each reactant is order 1 / first order.
 The overall order of the reaction is 1 + 1 + 1 = 3 or third order.
Answer 2:

 Since each reactant is first order, the concentration of each reactant is directly proportional to the effect that it has on rate
 If the concentration of the bromate ion is doubled, then the rate of reaction will also double
 If the concentration of the bromide ion is halved then the rate will also halve
 Therefore, there is no overall effect on the rate of reaction  one change doubles the rate and the other change halves it
Deriving Rate Equations
 Let's take the following reaction and derive the rate equation from experimental data
(CH3)3CBr + OH → (CH3)3COH + Br
Table to show the experimental data of the above reaction
 To derive the rate equation for a reaction, you must first determine all of the orders with respect to each of the reactants
 This can be done using a graph, but it doesn't have to be  you can use tabulated data provided
 Take the reactants one at a time and find the order with respect to each reactant individually
 Identify two experiments where the concentration of the reactant you are looking at first changes, but the concentrations of all other reactants remain constant
 Repeat this for all of the reactants, one at a time, until you have determined the order with respect to all reactants
Order with respect to [(CH3)3CBr]
 From the above table, that is experiments 1 and 2
 The [(CH3)3CBr] has doubled, but the [OH] has remained the same
 The rate of the reaction has also doubled
 Therefore, the order with respect to [(CH3)3CBr] is 1 (first order)
Order with respect to [OH]
 From the above table, that is experiments 1 and 3
 The [OH] has doubled, but the [(CH3)3CBr] has remained the same
 The rate of reaction has increased by a factor of 4 (i.e. increased by 22)
 Therefore, the order with respect to [OH] is 2 (second order)
Putting the rate equation together
 Once you know the order with respect to all of the reactants, you put them together to form the rate equation
 If a reactant has an order of 0, then you do not include it in the rate equation
 If a reactant has an order of 1, then you do not need to include the number 1 as a power
 If a reactant has an order of 2, then you raise that reactant concentration to the power of 2
 For this reaction, the rate equation will be:
Rate = k [(CH3)3CBr] [OH]2
Exam Tip
Be careful when reading the values in standard form! It is easy to make a mistake.
 The rate constant (k) of a reaction can be calculated using the initial rates and the rate equation
Calculating the rate constant from the initial rate
 The reaction of sodium carbonate with chloride ions (from hydrochloric acid) to form sodium chloride will be used as an example to calculate the rate constant from the initial rate and initial concentrations
 The reaction and rate equation are as follows:
Na2CO3 (s) + 2Cl (aq) + 2H+ (aq) → 2NaCl (aq) + CO2 (g) + H2O (l)
 The progress of the reaction can be followed by measuring the initial rates of the reaction using various initial concentrations of each reactant
Experimental results of concentrations & initial rates table
 To find the rate constant (k):
 Substitute the values of one of the experiments to find k (for example measurement 1)
 The values of measurement 2 or 3 could also have been used to find k
 They all give the same result of 1.40 x 102
Calculating Units
 When you are asked to calculate the rate constant, k, for a reaction you must also be able to deduce the units
 This is done by replacing the values in the rearranged rate equation with the units of that value
 The units can then be combined or cancelled as required
 For example, to calculate the units for the above reaction:
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