Rates, Orders & Arrhenius (OCR A Level Chemistry)

Two compounds, X and Y, were reacted together. 

The initial rate of reaction when compound X and compound Y were reacted together was measured in a series of experiments. 

The temperature was kept constant and the results of the experiments are shown below.

The results from three different experiments, carried out at a constant temperature, to investigate the rate of reaction between compounds A and B in a different chemical reaction are shown. 

A + B → Products 

Use the data to calculate a value for the rate constant, k.

An overall rate equation is given below.

Chemists measured the rate of a chemical reaction in a series of experiments between compounds C and D at a fixed temperature as shown below.

Chemists can use kinetic studies to help suggest mechanisms for reactions. 

The following data was obtained by reacting compounds A and B together in a series of experiments at a constant temperature.

The set of data below was obtained from a series of experiments reacting compounds X and Y together at a fixed temperature. The rate equation for this reaction is:

(If you have been unable to calculate and answer for (i), you may assume a value of 6.10 x 10^-5. This is not the correct answer.)

The rate of reaction data between nitrogen monoxide (NO) and oxygen (O₂) was obtained across a series of experiments at a constant temperature, as shown below. 

The rate equation for this reaction is:

Rate = k [NO]²[O₂]

Nitrogen dioxide reacts with carbon monoxide at 100 ℃ to form nitrogen monoxide and carbon dioxide, as shown in the equation below.

2NO₂ (g) + CO (g) → NO (g) + CO₂ (g)

The initial rate of reaction was measured in a series of experiments at a constant pressure. The rate equation below was determined.

Rate = k [NO₂]²

An incomplete table of data for the reaction between NO₂ and CO is shown.

For this question look at Figure below.

Compound X and Y react together in the following equation.

X + 4Y → XY₄

The rate equation for this reaction is below.

rate = k[X][Y]²

A possible mechanism for this reaction is

   Step 1    X + 2Y → XY₂

   Step 2    XY₂ + Y → XY₃

   Step 3    XY₃ + Y → XY₄

Chemists studied the rate of reaction of ethanal dimerises in dilute alkaline solution to form 3-hydroxybutanal in the following equation at a constant temperature of 298K.

The following three step mechanism for this reaction has been suggested.

1. 1. CH₃CHO + :OH^– → :CH₂CHO + H₂O
   2. CH₃CHO + :CH₂CHO → CH₃CH(O:^–)CH₂CHO
   3. CH₃CH(O:^–)CH₂CHO + H₂O → CH₃CH(OH)CH₂CHO + :OH^–

Iodide ions, I^-, react with S₂O₈^2- ions as shown in the equation below.

2I^-(aq) + S₂O₈^2- (aq) → I₂(aq) + 2SO₄^2-(aq)

A student investigates the rate of this reaction using the initial rates method.

The student measures the time taken for a certain amount of iodine to be produced.

Outline a series of experiments that the student could have carried out using the initial rates method.

How could the results be used to show the reaction is first order with respect to both I^- and S₂O₈^2- ? In your answer you should make clear how the results are related to the initial rates.

The Arrhenius equation can be represented as k =   in its exponential form. 

State the effect on k of an increase in;

Hydrogen peroxide decomposes to form water and oxygen as shown in the equation below.

2H₂O₂ (aq) → 2H₂O (l) + O₂ (g)

The table  below shows the value of the rate constant at different temperatures for a reaction.

Rate constant k / s-1	ln k	Temperature / K
0.000493		295
0.000656		298
0.001400		305
0.002360		310
0.006120		320

Complete the table by calculating the values of ln k and at each temperature.

The Arrhenius equation when expressed as a logarithmic relationship is as follows: 

Using your results from part (b) plot a graph of ln k against

The table below shows a series of experiments carried out to investigate how temperature affects the rate of reaction

Rate constant k / s-1	ln k	Temperature / K
3.1 x 10-5	-10.4	278	0.00360
4.7 x 10-4	-7.7	298	0.00336
1.7 x 10-3	-6.4	308	0.00325
5.2 x 10-3	-5.3	318	0.00314

The logarithmic form of the Arrhenius equation is:

ln k =-(E_a / RT) + ln A

Use the completed set of results above, to plot a labelled graph of ln k against 

Methanoic acid and bromine react as in the equation below.

Br₂(aq) + HCOOH(aq) → 2H⁺(aq) + 2Br^-(aq) + CO₂(g)

A student investigates the rate of this reaction by monitoring the concentration of bromine over time. The student uses a large excess of HCOOH to ensure that the order with respect to HCOOH will be effectively zero.

From the experimental results, the student plots the graph below.

Using the graph, determine

Your answer must show full working using the graph and the lines below as appropriate.

A student investigates the reaction between iodine, I₂, and propanone (CH₃)₂CO,  in the presence of aqueous hydrochloric acid, HCl(aq).

The results of the investigation are shown below.

Rate concentration graph

Results of initial rates experiments

Determine the orders with respect to I_2,  (CH₃)₂CO and HCl.

You must explain your reasoning.

The student then investigates the reaction of hydrogen, H₂ and iodine monochloride, ICl.

The equation for this reaction is shown below.

H₂ (g) + 2ICl (g) → 2HCl (g) + I₂ (g)

The rate equation for this reaction is shown below.

rate = k [H₂ (g)] [ICl (g)]

Predict a possible two step mechanism for this reaction. The first step should be the rate determining step.

In the presence of acid, H⁺(aq), chlorine and propanone react together:

CH₃COCH₃ (aq) + Cl₂ (aq) → CH₃COCH₂Cl (aq) + HCl (aq)

A student carried out an investigation into the kinetics of this reaction.

The student investigated how different concentrations of chlorine affect the initial rate of the reaction. A graph of [Cl₂ (aq)] against time is shown below:

The student then investigated how different concentrations of propanone and H⁺(aq) affect the initial rate of reaction.

Their results are shown below.

Use the student's results to determine the reaction orders. Explain your answer.

Deduce the rate equation and calculate the rate constant for this reaction, including the units.

The student varied the concentrations of the reactants in another experiment and they found the rate to be 0.26 x 10^-5 mol dm^-3 s^-1. They also used a pH probe and found that the reaction mixture had a pH of 2.

What would the initial rate of the reaction mixture be if the amount of acid added was altered to give a pH of 1?

Assume the temperature and the initial concentrations of the other reactants remained the same.

The experiment was repeated at a lower temperature. What would be the effect, if any, of this change on the rate and the rate constant of the reaction?

Clock reactions can be used to obtain the initial rate of a reaction by taking a single measurement instead of continuous measurements.

The time taken for a visual change to occur is timed and the initial rate is taken to be proportional to .

What assumption is made so that the initial rate can be taken to be proportional to ?

A student investigates the reaction of hydrogen peroxide, H₂O₂, and iodide ions in the presence of an acid.

H₂O₂ (aq) + 2I^- (aq) + 2H⁺ (aq) → I₂ (aq) + 2H₂O (l)

Thiosulfate ions, S₂O₃^2-, are added to the reaction which react with iodine forming tetrathionate ions, S₄O₆^2-.

2S₂O₃^2- (aq) + I₂ (aq) → S₄O₆^2- (aq) + 2I^- (aq)

Starch solution is also added.

Explain how the addition of thiosulfate ions and starch solution enables a measurement of the time taken for a colour change to occur.

A student carried out a series of experiments to investigate the reaction of hydrogen peroxide with iodide ions in the presence of an acid.

In the first series of experiments, the students varied the concentration of hydrogen peroxide with the same concentrations and volumes of iodide ions and acid. Their results are shown in the table below.

[H2O2 (aq)] / mol dm-3 s-1	Time / s	/ s-1
0.020	259
0.040	125
0.060	78
0.080	59
0.100	48

Complete the table and use the data to plot a graph of against concentration of hydrogen peroxide on the graph below using the results in the table. You should draw a line of best fit.

In the second series of experiments, different concentration of iodide ions were used with the same concentrations and volumes of hydrogen peroxide and acid.

In the final series of experiments, different concentrations of acid were used with the same concentrations and volumes of hydrogen peroxide and iodide ions.

Graphs of  against concentration are plotted for these two series of experiments below.

Use your graph from part (c) and the two graphs above to deduce the rate equation for the reaction.

Explain what evidence there is that the reaction takes place in more than one step.

Deduce a possible three-step mechanism for this reaction. Assume the first step is the rate-determining step.

Nitrogen monoxide, NO, reacts with oxygen, O₂. The rate equation for this reaction is shown below.

rate = k[NO(g)]²[O₂(g)]

State what would happen to the rate of the reaction if the initial concentration of NO is tripled and the initial concentration of O₂ is halved.

Nitrogen monoxide is produced by combustion in car engines and released into the atmosphere. 

In the lower atmosphere, nitrogen monoxide is responsible for the formation of ozone in a series of reactions shown below.

NO (g) + ½O₂ (g) → NO₂ (g)

NO₂ (g) → NO (g) + O (g)

O₂ (g) + O (g) → O₃ (g)

When nitrogen monoxide is present in the upper atmosphere, is it involved in the removal of ozone, O₃. The overall equation for the reaction is shown below.

O (g) + O₃ (g) → 2O₂ (g)

The rate equation is:   rate = k[NO(g)][O₃(g)]

Ozone in the lower atmosphere can react with ethene to produce methanal, CH₂O (g) which contributes to low-level smog. The equation is shown below.

O₃ (g) + C₂H₄ (g) → 2CH₂O (g) + ½O₂ (g)

The order of the reaction with respect to both reactants is first order. The initial rate of formation of methanal was found to be 2.0 x 10^-11 mol dm^-3 s^-1.

The initial concentration of ozone was doubled and the initial concentration of ethene was tripled.

Calculate the initial rate of oxygen formation.

Bromate(V) ions and bromide ions react in the presence of acid to form bromine and water.

Write the half equations and overall balanced symbol equation, including state symbols, for this reaction.

A  student carried out a series of experiments to investigate how the rate of the reaction of bromate and bromide in the presence of an acid varies with temperature.

The time taken, t, was measured for a fixed amount of bromine to form at different temperatures. The results are shown in the table below.

Temperature (T ) / K	/ K-1	Time (t ) / s	/ s-1	ln
408	2.451	21.14	0.0473	-3.051
428		10.57	0.0946	-2.358
448	2.232	5.54
468	2.137			-1.106
488	2.049	1.71	0.5851	-0.536

Calculate the missing values to complete the table above.

The Arrhenius equation relates the rate constant, k, to the activation energy, E_a, and temperature, T.

ln k = ln A + 

In this experiment, the rate constant, k, is directly proportional to . Therefore, 

Use your answers from part (b) to plot a graph of ln  against  x 10^-3 on the graph below.

Use your graph and information from part (c) to calculate a value for the activation energy, in kJ mol^–1, for this reaction. To gain full marks you must show all of your working. 

The value of the gas constant, R = 8.314 J K^–1 mol^–1

Three experiments were carried out at 20 ^oC to investigate the rate of the reaction between compounds F and G. The results are shown in the table below.

Use the data in the table to deduce the rate equation for the reaction between F and G.

F and G react together to form FG₂ according to the following two step mechanism.

   Step 1: F + G → FG

   Step 2: FG + G → FG₂

Using your answer to part (a), explain which step is the rate-determining step.

Use the information in the table in part (a) to calculate a value for the rate constant, k, for this reaction between 0.0534 mol dm^-3 F and 0.0624 mol dm^-3 G. 

Give your answer to the appropriate number of significant figures.

State the units for k.

The Arrhenius equation shows how the rate constant, k, for a reaction varies with temperature, T.

k = Ae^-Ea/RT

For the reaction between 0.0534 mol dm^-3 F and 0.0624 mol dm^-3 G at 25 °C, the activation energy, E_a, is 16.7 kJ mol^–1. 

Use your answer to part (c) to calculate a value for the Arrhenius constant, A, for this reaction. 

The gas constant R = 8.314 J K^–1 mol^–1. Give your answer to the appropriate number of significant figures.