5.2 Rate Equations (A Level only)

The results of three experiments carried out at a temperature, T₁,  to investigate the rate of the reaction between compounds A and B are shown in Table 1.

Table 1

Use the data in Table 1 to deduce the rate equation for the reaction between compounds A and B.

A separate experiment is performed at a different temperature, T₂.

When the initial concentration of A is 0.27 mol dm^–3 and B is 0.80 mol dm^–3, the value of the rate constant is 1.894 x 10^-4 mol^–2 dm⁶ s^–1.

Use the information from part (a), to deduce whether T₁ or T₂ is the higher temperature.

(If you did not get an answer for (a), you may assume that rate = k [A]² [B]². This is not the correct answer.)

Use your answer to part (b) to sketch a labelled Maxwell-Boltzmann energy distribution on Figure 1 below, to show the reaction between compounds A and B at temperatures T₁ and T₂.

Figure 1

Use your knowledge of kinetic theory to explain whether increasing temperature or increasing concentration will have a greater effect on the rate of reaction.

Compounds C and D react together in three different, third order experiments at the same fixed temperature, T₁. The results of these experiments are shown in Table 1 .

Table 1

State the assumption made about [C] and [D].

Using the information in part (a), deduce the rate equation for the reaction between compounds C and D.

Using your answer to part (b), calculate the missing value in Table 1.

Table 1

In a further experiment at temperature, T₁, the initial rate of reaction was found to be 3.74 x 10^-2 mol dm^-3 s^-1.

The initial concentrations of C and D were both y mol dm^-3.

Using your answers from parts (b) and (c), calculate the concentration of C and D.

(If you did not get answers for (b) and (c), you may assume that rate = k [C] [D] and that the rate constant k = 52.7 mol^-1 dm³ s^-1. These are not the correct answers.)

Bromate (V) ions and bromide ions react in acidic conditions to form bromine and water.

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

A series of experiments were carried out to investigate how the rate of the reaction of bromate and bromide in acidic conditions 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 Table 1.

Table 1

Complete Table 1.

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,

ln  = ln A + 

Use your answers from part (b) to plot a graph of ln  against  x 10^-3 on Figure 1 below.

Figure 1

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 = 8.31 J K^–1 mol^–1.

Three experiments were carried out at a temperature, T₁,  to investigate the rate of the reaction between compounds F and G. The results are shown in Table 1.

Table 1

Use the data in Table 1 to deduce the rate equation for the reaction between compounds 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 Table 1 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.

(If you did not get an answer for (a), you may assume that rate = k [F]² [G]². This is not the correct answer.)

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

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 = 8.31 J K^–1 mol^–1.

Give your answer to the appropriate number of significant figures.

(If you did not get an answer for (a), you may assume that k has a value of 4.97. This is not the correct answer.)

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

The following data in Table 1 was obtained by reacting compounds A and B together in a series of experiments at a constant temperature. 

Table 1

The set of data below in Table 2 was obtained from a series of experiments reacting compounds X and Y together at a fixed temperature. The rate equation for this reaction is:

Rate = k[X]²[Y]

Table 2

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 in Table 3 below.

The rate equation for this reaction is:

Rate = k [NO]² [O₂]

Table 3

Nitrogen dioxide reacts with carbon monoxide at 100 C 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 in Table 4.

Table 4

For this question look at Figure 1 below.

Figure 1

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.

2CH₃CHO → CH₃CH(OH)CH₂CHO

Rate = k[CH₃CHO][:OH^–]

Table 1

The overall rate equation for a reaction that occurred is below.

Rate = k[A]

Explain why doubling the temperature of this reaction would have a much greater effect on the rate of the reaction, compared to doubling the concentration of A.

You must refer to kinetic theory in your answer.

The decomposition of hydrogen peroxide into water and oxygen occurs at a slow rate with a rate constant of k = 6.62 x 10^-3 mol dm^-3 s^-1 and at a temperature of 290 K. Use the equation ln k = ln A – E_a / RT to calculate a value, in kJ mol⁻¹, for the activation energy of this reaction.

The gas constant R = 8.31 J K⁻¹ mol⁻¹.

The constant A = 3.18 × 10¹¹ mol⁻¹ dm³.

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

State the effect on k of an increase in;

Calculate the activation energy, E_a, of a reaction at 57 C and a rate constant of 1.30 x 10^-4  mol dm^-3 s^-1.

The gas constant R = 8.31 J K⁻¹ mol⁻¹.

The constant A = 4.55 × 10¹³.

A series of experiments was carried out with nitrogen dioxide and ozone as shown in the equation below;

2NO₂ (g) + O ₃(g)  →  N₂O₅ (g) + O₂ (g)

The rate equation for the reaction between the two compounds is below.

Rate = k [NO₂] [O₃]

In one experiment at 30 C, the initial rate of reaction is 3.46 × 10⁻³ mol dm⁻³ s⁻¹ when the initial concentration of NO₂ is 0.50 mol dm⁻³ and the initial concentration of O₃ is 0.21 mol dm⁻³.

Manganate (VII) ions, MnO₄^-, oxidise hydrogen peroxide, H₂O₂, to oxygen gas in acidic conditions.

2MnO₄^- + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂

The results of the reaction mixture at a fixed temperature are shown in Table 1.

Table 1

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

2H₂O₂ (aq) → 2H₂O (l) + O₂ (g)

Table 2 below shows the value of the rate constant at different temperatures for a reaction.

Table 2

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:

ln k = + ln A

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

Table 3 below shows a series of experiments carried out to investigate how temperature affects the rate of reaction.

Table 3

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 = + ln A

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