MC test: Rate expression & reaction mechanisms

Multiple choice test on 16.1 Rate expression & reaction mechanisms

Use the following 'quiz' to test your knowledge and understanding of this sub-topic. You will need access to a periodic table (Section 6 of the IB data booklet).

If you get an answer wrong, read through the explanation carefully to learn from your mistakes.

The graph shows a sketch of the rate of a reaction plotted against concentration.

What is the order of the reaction?

Changing the concentration of a reactant has no effect on the rate of a zero order reaction

 

It is found that the half-life, t½, of a reaction is independent of the concentration of the reactants.
What are the units for the rate constant of this reaction?

Since the half-life is independent of the concentration it must be a first order reaction making the units of rate constant s−1.

 

Which statements are true about the rate equation for the reaction between nitrogen monoxide and chlorine to form nitrosyl chloride, NOCl? 

rate = k[NO]2[Cl2]

I. The order of the reaction with respect to Cl2 is 1.

II. The units for the rate constant are dm3 mol−1 s−1

III. The rate equation can only have been determined by experiment.

All rate equation scan only be determined experimentally. The overall order is third order (it is first order w.r.t. Cl2 and second order w.r.t. NO) so the units of the rate constant are dm6 mol−2 s−1.

 

The initial rate of the reaction between P and Q was measured using different concentrations. The table shows the data gathered.

Experiment Initial [P] / mol dm−3 Initial [Q] / mol dm−3 Initial rate  / mol dm−3 s−1
1 0.3 0.2 2.5 x 10−3
2 0.3 0.4 5.0 x 10−3
3 0.6 0.4 2.0 x 10−2

Which is the correct rate expression for the reaction?

From 1 and 2 doubling [Q] doubles the rate so the reaction is first order w.r.t to [Q]. From 2 and 3 doubling [P] causes the rate to increase fourfold so the reaction is second order w.r.t. to [P].

 

The rate expression for the reaction between hydrogen and nitrogen monoxide to form nitrogen and steam is

rate = k[H2][NO]2

When the initial concentration of hydrogen was 1 x 10−3 mol dm−3 and the initial concentration of nitrogen monoxide was 6 x 10−3 mol dm−3 the initial rate was found to be 3.6 x 10−3 mol dm−3 s−1. What will be the rate when the initial concentrations of both hydrogen  and  nitrogen monoxide are 4 x 10−3 mol dm−3?

From the rate equation, k = rate ÷ ([H2][NO]2) = 3.6 x 10−3 ÷ (1 x 10−3 x 3.6 x 10−5) = 1 x 105 dm6 mol−2 s−1.
By substitution rate =  1 x 105 x 4 x 10−3 x (4 x 10−3)2 = 6.4 x 10−3 mol dm−3 s−1.

 

Bromine reacts with nitrogen monoxide according to the following equation

2NO(g) + Br2(g) → 2NOBr(g)

The rate expression for this reaction is rate = k[NO]2[Br2]

Which is a possible mechanism for this reaction?

Because it is a third order reaction overall the first step involving just two reacting species cannot be the slow (rate determining step).

 

The iodination of propanone proceeds in the presence of acid.

H3CCOCH3 + I2 → H3CCOCH2I + HI

The rate equation for this reaction is rate = k[H3CCOCH3][H+]

Which conclusions can be deduced from this information?

 

I. The acid is behaving as a catalyst.

II. Iodine is not involved in the rate determining step.

III. The molecularity of the rate determining step is 2.

Because the acid does not appear in the stoichiometric equation it is acting as a catalyst. Iodine cannot be involved in the rate determining step otherwise it would appear in the rate equation. Since the rate determining step involves two reacting species it is bimolecular.

 

Carbon monoxide reacts with nitrogen dioxide to form carbon dioxide and nitrogen monoxide.

CO(g) + NO2(g)→ CO2(g) + NO(g)

A mechanism has been proposed for the reaction which is consistent with the experimentally determined rate equation.

First step: 2NO2(g) → NO3(g) + NO(g) (slow)

Second step:NO3(g) + CO(g) → NO2(g) + CO2(g) (fast)

Which is the experimentally determined rate equation?

The slow step is bimolecular with two molecule of NO2 reacting together so the rate will only be proportional to [NO2]2.

 

 

The enthalpy diagram represents a two step reaction mechanism.

What is the activation energy for the reaction?

The first step will be the slowest as the transition state has the highest enthalpy so x will be the activation energy for the whole reaction. The reaction is endothermic and z is the value of ΔH.

 

A reaction is first order with respect to one of the reactants. The initial concentration of the reactant (present as a stoichiometric amount) was 1.00 mol dm−3 and after 3.25 minutes the concentration of the reactant had fallen to 0.50 mol dm−3.
What would be the concentration of the reactant after a further 13.0 minutes?

The half-life is 3.25 minutes and is constant for a first order reaction. 13 minutes is four more half-lives.  After one more half-life the concentration would be 0.25 mol dm−3, after two it will be 0.125 mol dm−3, after three 0.0625 mol dm−3 and finally after four half-lives it will be 0.03125 mol dm−3.

 

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