Topic 6/16 Kinetics (Rates of Reactions)

6.1 Rates of Reaction1. What are the standard units for a reaction rate?

mols dm-3 s -1

2. Consider the following reaction:

2 MnO4-(aq) + 5 C2O4

-2(aq) + 16 H+(aq) 2 Mn+2(aq) + 10 CO2(g) + 8 H2O(l)

Describe three methods in which you could measure the rate of this reaction.

1. Change in mass of gas (production of a heavy gas) *CO2 can be soluble in water (cooler water) and therefore may not provide accurate results.

2. Change in concentration as measured by titration. Because there is a H+ present, there will be a change in acidity and can be tested through a titration

3. Change in concentration as measured by conductivity. More ions present on reactant side than product side therefore conductivity should decrease.

3. The reaction between calcium carbonate and hydrochloric acid, carried out in an open flask, can be represented by the following reaction:

CaCO3(s) + 2 HCl(aq) CaCl2(aq) + H2O(l) + CO2 (g)

Which of the following measurements below could be used to measure the rate of the reaction?

I. The mass of the flask and contents Open flask, CO2 heavy gasII. The pH of the reaction mixture start with acid

III. The volume of carbon dioxide produced could be done if water warma. I and II onlyb. I and III onlyc. II and III onlyd. I, II, and III

4. The following data were collected for the reaction. Draw a graph on concentration against time and determine the reaction rate after 60s and after 120s

2 H2O2(aq) 2 H2O(l) + O2(g) at 390°C

[H2O2]/ mold m-3 Time/s0.200 00.153 200.124 400.104 630.090 800.079 1000.070 1200.063 1400.058 1600.053 1800.049 200

0 20 40 60 80 100 120 140 160 180 2000

0.05

0.1

0.15

0.2

0.25

0.165/160 = 0.001 mols dm-3 s-1

0.14/250 = 0.00056 mols dm-3 s-1

5. State Collision Theory in your own words. Molecules need kinetic energy in order to collide for breaking and forming bonds.

6. What is an effective collision?

Successful interactions breaking then forming bonds

7. What are the requirements for a collision to be effective?

Kinetic energy of particles must be greater than or equal to the activation energy Particles must collide with the proper orientation

8. What is a catalyst? Use a labeled potential energy graph (enthalpy diagram) to show how a catalyst can speed up a reaction.

Definition: substance that lowers the activation energy and increases the rate of reaction without participating in the reaction

9. Explain the relationship between rate of reaction and temperature in terms of molecules and effective collisions.

Increase the temperature, incrase the kinetic energy in molecules therefore more molecules break the activation barrier.

Increase the number of collisions (frequency), increasing the probability of proper orientation occurring.

10. Explain the relationship between rate of reaction and concentration in terms of molecules and effective collisions.

Molecules: increase concentration, increase the amount of molecules present, increasing the potential frequency of collisions

Collisions: more molecules means more collisions, better potential for proper orientation

11. Explain the relationship between rate of reaction and surface area in terms of molecules and effective collisions.

Molecules: increase the surface area, you increase the exposure of the chemical to the reaction and collisions (similar to increasing concentration)

Collisions: increasing surface area increases number of collisions as more molecules are present during the reaction at one time

12. Which change of condition will decrease the rate of the reaction between excess zinc granules and dilute hydrochloric acid?

a. Increasing the amount of zincb. Increasing the concentration of the acidc. Pulverizing the zinc granules into powderd. Decreasing the temperature – lowers kinetic energy, fewer molecules at or above

activation energy

13. The rate of a reaction between two gases increases when the temperature is increased and a catalyst if added. Which statements are both correct for the effect of these changes on the reaction?

Increasing Temperature Adding CatalystCollision frequency increases Activation energy increasesActivation energy increases Activation energy does not changeActivation energy does not change Activation energy decreasesActivation energy increases Collision frequency increases

Temperature only affects the number of molecules at or above activation energy, not the activation energy itself

14. The graph below shows how the volume of carbon dioxide formed varies with time when a hydrochloric acid solution is added to excess calcium carbonate in a flask.

(i) Explain the shape of the curve.

rate = increase in timevolume

= slope of graph;initially/to begin with steeper slope / fastest rate / volume of gas/CO2 produced faster/quickly as concentration of HCl highest;as reaction progresses/with time, less steep slope / volume of gas production slows / rate decreases due to less frequent collisions as concentration (of HCl) decreases;curve flattens/becomes horizontal when HCl used up/consumed (as there are no more H+ ions to collide with the CaCO3 particles);Each mark requires explanation.

(3)

(ii) Copy the above graph on your answer sheet and sketch the curve you would obtain if double the volume of hydrochloric acid solution of half the concentration as in the example above is used instead, with all other variables kept constant from the original. Explain why the shape of the curve is different.

less steep curve;same maximum volume at later time;half/lower H+ /acid concentration less frequent collisions/slower rate;same amount of HCl, same volume CO2 produced;

(4)

(iii) Outline one other way in which the rate of this reaction can be studied in a school laboratory. Sketch a graph to illustrate how the selected variable would change with time.

mass loss/of CO2 / mass of flask + content;

OR

OR

(2)

(iv) Define the term activation energy and state one reason why the reaction between calcium carbonate and hydrochloric acid takes place at a reasonably fast rate at room temperature.

minimum energy (of colliding particles) for a reaction to occur / OWTTE;lower Ea / greater surface area/contact between CaCO3 and HCl / higher HClconcentration / (sufficient) particles/molecules have activation energy;

(2)

15. Graphing is an important method in the study of the rates of chemical reaction. Sketch a graph to show how the reactant concentration changes with time in a typical chemical reaction taking place in solution. Show how the rate of the reaction at a particular time can be determined.

labelled axes (including appropriate units);correctly drawn curve;correctly drawn tangent;

rate equal to slope/gradient of tangent (at given time) / rate = xy

at time t;

(Total 4 marks)

16.1 Rate Expression16. The reaction CH3Cl(aq) + OH-(aq) CH3OH(aq) + Cl-(aq) is found to be second order overall. Give

three possible rate expressions consistent with this finding.

Rate = k[CH3Cl][OH-]

Rate = k[CH3Cl]2

Rate = k[OH-]2

17. Give the units of k in each of the rate expressions below:a. Rate = k[NO2]2 mols-1 dm3 s-1

b. Rate = k[NH3]0 mols dm-3 s-1

c. Rate = k[CH3CH2Br] s-1

d. Rate = k[NO]2[Br2] mols-2 dm6 s-1

18. Consider the reaction and the corresponding data regarding reaction rates: A + B C + D

Initial rate mol dm-3s-1 initial [A] mol dm-3 initial [B] mol dm-3

1.35 x 10 -7 .100 .0052.7 x 10 -7 .100 .0105.4 x 10 -7 .200 .010

a) Write the generic rate law expression for the reaction.

Rate = k[A][B]

b) Determine the rate law.

2.7x10-7/1.35x10-7 = 2 (holding A constant), 1st order for B 5.4x10-7/2.7x10-7 = 2 (holding B constant), 1st order for A

Rate = k[A][B]

c) The rate law is __1__ order in the reactant A, and __1__ order in the reactant B. The overall reaction order is __2__.

d) Calculate the value of the rate constant, k.

1.35x10-7 = k[0.100][0.005]

1.35x10-7/5x10-4 – k

2.7x10 -4 mols -1 dm 3 s -1

19. The reaction between bromate ions and bromide ions in acidic aqueous solution is given by the equation

BrO3-(aq) + 5Br-(aq) +6H+(aq) 3Br2(l) + 3H2O(l)

The table below gives the results from four experiments.

Exp initial [BrO3-] mol

dm-3initial [Br-] mol dm-3

initial [H+] mol dm-

3Initial rate mol dm-

3s-1

1 .10 .10 .10 8.0 x 10 -4

2 .20 .10. .10 1.6 x 10 -3

3 .20 .20 .10 3.2 x 10 -3

4 .10 .10 .20 3.2 x 10 -3

a) Using these data, determine the orders for all three reactants.

2[BrO3-] = 1.6x10-3/8.0x10-4 = 2 [BrO3

-]1

2[Br-] = 3.2x10-3/1.6x10-3 = 2 [Br-]1

2[H+] = 3.2x10-3/8.0x10-4 = 4 [H+]2

b) Write the rate law.

Rate = k[BrO3-]1[Br-]1[H+]2

c) What is the overall reaction order?

4

d) Determine the value of the rate constant, k.

8.0x10-4 = k[.10][.10][.10]2

8.0x10.4/1.0x10-4 = k

8 mols -3 dm 9 s -1

e) Explain how a graph of concentration versus time can be used to determine reaction order by examining successive half-lives (sketch an example of a zero order reaction, 1st order reaction, a 2nd order reaction)

At half life (concentration ½ of original), ½ life increases with each order as rate of reaction increases with each order

zero order first order second order

20. Consider this graph of the change in concentration of reactant versus time for the reaction: N2O4 2NO2

Explain how the graph indicates that this is a first-order reaction.

There is a linear change in rate. Half of concentration is lost per unit of time (equal time between each ½ life)

21. The following is a proposed reaction mechanism:

Step 1: NO + NO N2O2 (fast)

Step 2: N2O2 + O2 2NO2 (slow/rds)

a) Overall: 2NO + O2 2NO2

b) Show the rate law for the reaction mechanism.

Rate = k[NO]2[O2] Creation of N2O2 is controlled by its reactants

c) What is the intermediate in the above mechanism?

N2O2

d) Use the Maxwell Boltzman diagram as part of an explanation of why increasing temperature increases rate of reaction.

Increase temperature, increase number of molecules which meet the activation energy

e) Use the Maxwell Boltzman diagram as part of an explanation of why the addition of a catalyst increases rate of reaction.

22. If the mechanism of a reaction is:

AB2 + AB2 A2B4 slowA2B4 A2 + 2B2 fast

a. What is the overall equation for the reaction?

2AB2 A2 + 2B2

b. What is the rate expression fort this reaction?

Rate = k[AB2]2

c. What units will the rate constant have in this expression?

mols-1 dm3 s-1

23. Nitrogen (II) oxide reactions with hydrogen as shown by the following equation:

2 NO(g) + 2 H2 (g) N2(g) + 2 H2O(g)

The table below shows how the rate of reaction varies as the reactant concentrations vary.

Experiment Initial [NO]/mold dm-3 Initial [H2]/ mold dm-3 Initial rate/ mol N2 dm-3 s-1

1 0.100 0.100 2.53 x 10-6

2 0.100 0.200 5.05 x 10-6

3 0.200 0.100 10.10 x 10-6

4 0.300 0.100 22.80 10-6

a. Determine the order of the reaction with respect to NO and with respect to H2. Explain how you determined the order for NO.

2[H2] = 5.05/2.53 = 2 1st order

2[NO] = 10.10/2.53 = 4 2nd order (since when the concentration was doubled, and the rate quadrupled, that is representative of a 2nd order reaction – exponential relationship)

b. Write the rate expression for the reaction.

Rate = k[NO]2[H2]

c. Calculate the value for the rate constant, including its units

2.53x10-6 = k[0.100]2[0.100]

2.53x10-6/0.001 = k

2.53x10-3 mols-2 dm6 s-1

d. A suggested mechanism for this reaction is as follows.H2 + NO ↔ X fastX + NO Y + H2O slowY + H2 N2 + H2O fast

State and explain whether this mechanism agrees with the experimental rate expression in (b)

Yes, since slow step is second, the X reactant (product of first reaction) is dependent on reaction 1 reactants which include H2 and another NO

e. Explain why a single step mechanism is unlikely for a reaction of this kind.

Reaction involving four molecules is statistically unlikey

f. Deduce how the initial rate of formation of H2O(g) compares with that of N2(g) in experiment 1. Explain your answer.

H2O should be produced double that of N2 due to 2 moles of H2O is created for every mole of N2

24. The results from a series of experiments for this reaction are shown below. Deduce, giving a reason, the order of reaction with respect to each of the reactants.

6 Zn + As2O3 + 12 H+ 2 AsH3 + 6 Zn+2 + H2O

Experiment [As2O3] [ Zn ] [H+ ] Initial Rate of Reaction (M /sec)

1 0.0040 0.0010 0.0020 2.3 x 10-5

2 0.0080 0.0010 0.0020 9.2 x 10-5

3 0.0040 0.0010 0.0020 2.3 x 10-5

4 0.0040 0.0030 0.0020 2.3 x 10-5

5 0.0040 0.0010 0.0060 6.9 x 10-5

2[AssO3] = 9.2x10-5/2.3x10-5 = 4 [As2O3]2 since the concentration doubled and the rate quadrupled, the change is exponential, the order is 2

3[Zn] = 2.3x10-5/2.3x10-5 = 1 [Zn]0 since the concentration tripled and the rate did not change, the order is 0

3[H+] = 6.9x10-5/2.3x10-5 = 3 [H+]1 since the concentration is tripled and the rate tripled, that is a proportional change, the order is 1

Rate = k[As2O3]2[H+]1

16.2 Activation Energy23. Consider the following statements

I. The rate constant of a reaction increases with increase in temperatureII. Increase in temperature decreases the activation energy of the reaction no effect

III. The term A in the Arrhenius equation (k = Ae-Ea/RT) relates to the energy requirements

of the collisions.

Which statement(s) is/are correct?a. I onlyb. II onlyc. I and III onlyd. II and III only

24. The conversion of CH3NC into CH3CN is an exothermic reaction which can be represented as follows.

CH3–N≡C transition state CH3–C≡N

This reaction was carried out at different temperatures and a value of the rate constant, k, was obtained for each temperature. A graph of ln k against 1/T is shown below.

(i) Define the term activation energy, Ea.

Minimum energy needed for reaction to occur

(ii) Construct the enthalpy level diagram and label the activation energy, Ea, the enthalpy change, ∆H, and the position of the transition state.

(iii) Describe qualitatively the relationship between the rate constant, k, and the temperature, T.

As increase temperature, the rate constant k increases exponentially (Arrenhius plot)

(iv) Calculate the activation energy, Ea, for the reaction, using Table 1 of the Data Booklet.

m=−EaR

∆ lnk∆ 1T

= −5.8−−10.60.00191−0.00216

=−19200

−19200= −Ea8.314

−159628.8 J=Ea

−159 kJ=Ea

25. Rate constants for the reaction NO2(g) + CO(g) NO(g) + CO2(g) are given below.At 700 K, k = 1.3 mol dm-3 s-1

At 800 K, k = 23.0 mol dm-3 s-1

Calculate the value of the activation energy in kJ mol-1

ln ( k1k2 )= Ea

R ( 1T 2

− 1T 1 )

ln ( 1.323.0 )= Ea

8.314 ( 1800

− 1700 )

−2.87= Ea8.314

(−1.785 e−4 )

Ea=133815 J mol−1∨133kJ mol−1