Paper 2

YEAR 2001

Subtopic: Order of reaction and rate equation.

Q5. Aqueous chlorine dioxide disproportionation in an alkaline solution according to the following equation:

2ClO2 (aq) + 2OH- (aq) ClO3- (aq) + ClO2

- (aq) + H2O (l)

(B) The result of the kinetic study of the disproportionation of aqueous chloride dioxide are shown in the table below.

Experiment number [ClO2]/ mol dm-3 [OH- ] / mol dm-3Initial rate/ mol dm -³s -

¹1 0.0575 0.0216 8.21 × 10-3

2 0.0713 0.0216 1.26 × 10-2

3 0.0575 0.0333 1.26 × 10-2

i. Determine the order of reaction for the disproportionation of aqueous chlorine dioxide.ii. Calculate the rate constant, k, for the disproportionation of aqueous chlorine dioxide.

iii. Calculate the pH of the aqueous chlorine dioxide solution if its concentration is 0.100mol dm -3 and the initial rate of disproportionation is 3.56 × 10 -2 mol dm-3 s-1.

[13M]

YEAR 2003

Subtopic: Order of reaction and rate equation.

Q2. (A) From the graph, for the decomposition of N2O5 ,

i. Determine the order of the reaction.ii. Calculate the rate constant.

iii. Calculate the half-life.iv. Write the rate equation.

[6M](B) The activation energy, Ea, and enthalpy change of reaction, ∆H, for the decomposition of

N205 are given as +100kJ mol-1 and -23kJ mol-1 respectively.i. Sketch and label the energy profile for the decomposition of N2O5 on the axes below.

Energy/kJ Reaction coordinate

ii. Calculate the activation energy of the reverse reaction, Ea.[4M]

Q8. The contact process in the manufacture of the sulphuric acid involves the oxidation of sulphur dioxide to sulphur trioxide.

(B) i. Nitrogen monoxide is used as a catalyst in the Contact process. The mechanism suggested is as follow.

2NO(g) + O2(g) 2NO2(g)

NO2(g) + SO2(g) NO(g) + SO3(g)

Describe how nitrogen monoxide acts as catalyst.

ii. Platinum is also used as a catalyst in the Contact process. Describe how platinum acts as a catalyst in this process.

[8M]

YEAR 2004

Subtopic: Order of reaction and rate equation.

Q7. (A) In acid solution, bromide ions are slowly oxidised to bromines by bromate(V) ions as represented by the following ionic equation.

BrO3- + 6H+ + 5Br- 3Br2 + 3H2O

The following data gives the results of four experiments at constant temperature.

Experiment number

[BrO3-] / mol dm-

3[Br-] / mol dm-3 [H+] / mol dm-3 Initial rate of

reaction/ mol dm-3 s-1

1 0.400 0.280 0.031 1.29 × 10-3

2 0.600 0.280 0.031 1.94 × 10-3

3 0.600 0.560 0.031 3.87 × 10-3

4 0.400 0.280 0.062 5.16 × 10-3

i. Write the two half-reaction equations for the above reaction.ii. Calculate the order of reaction with respect to BrO3

-, Br-, and H+ ions.iii. Write the rate equation for the reaction.iv. Find the value of the rate constant in experiment number 4 when the pH of the solution

is 2.2 .[9M]

Paper 1

YEAR 1999

Q10. Which of the following is correct with regard to a catalyst?

A. It decreases the temperature of the reaction.B. It alters the mechanism of the reaction.C. It increases the yield of the products.D. It does not take part in the reaction.E. All catalytic actions can be explained by the Collision Theory.

YEAR 2000

Q49.The graph below shows the variation of energy against extent of reaction for the reversible reaction 2A + 2B A2B2 .

Energy

I

II

2A + 2B

E2 A2B2

E1

E3

Extent of reaction

Which of the following is/are correct regarding the reaction?

1. The reaction is exothermic by the amount of E1 – E3

2. Curve II refers to the energy profile for the catalysed reaction.3. The activation energy for the reverse, uncatalysed reaction is E2 – E3.

YEAR 2001

Q9. The distribution of molecular kinetic energy of a gas at 279K and 289K is shown by the Maxwell-Boltzman graph below.

A B C D E

1,2,3 are correct

1 & 2 only correct

2 & 3 only are correct

1 only correct

3 only correct

No.of molecules

279K

289K

Kinetic energy Ea

Which of the following statements best explain why the rate of reaction in a gas sample at 279K increases two folds when the temperature is increased to 289K?

A. The total area under the curve increases two folds.B. The average velocity of the molecules increases two folds.C. The number of collisions increases two folds.D. The number of molecules with energy equal to or greater than Ea increases two folds.

Q10. Which of the following statements regarding the half-life of a first order reaction is true?

A. The half-life is affected by pressure.B. The half-life depends on the concentration of the reactants.C. The half-life is shorter at lower temperature.D. At the end of the fourth half-life, the percentage of elements left is 6.25% of the original

quantity.

Q42. In a chemical reaction, a catalyst

1. Increases the rate of the forward reaction and the rate of the reverse reaction by the same factor.

2. Decreases the activation energy.3. Affects the enthalpy of reaction.

A B C D1 only correct

1 & 2 only are correct

2 & 3 only are correct

1, 2 & 3 are correct

Q47. The reaction between the X(III) ion, iodide ion and the peroxodisulphate(VI) ion follows the following steps.

2X3+(aq) + 2I-(aq) 2X2+(aq) + I2(aq) Step 1

2X2+(aq) + S2O82-(aq) 2X3+(aq) + 2SO4

2-(aq) Step 2

Which of the following statements are correct regarding the reaction?

1. The X(III) ion is the intermediate.2. The X(III) ion acts as a catalyst.3. Heterogeneous catalysis occurs.

YEAR 2002

Q8. 1.0 dm3 of an aqueous solution containing 1.0mol of trichloromethane and 1.0 mol of sodium hydroxide is left to react. The ionic equation for the reaction is

2CHCI3 + 7OH- CO + HCOO- + 6CI- + 4H2O

The rate equation is

Rate = k[CHCI3][OH-] = x .

Which of the following statements is correct with regard to the reaction?

A. The reaction is second order with respect to trichloromethane.B. When trichloromethane is reacting at a rate of 0.2 mol dm-3s-1, the rate of formation of

the chloride is 1.2 mol dm-3s-1 .C. When the concentrations of trichloromethane and the hydroxide ion are both doubled,

the rate of reaction is 2x.

D. When the half of the hydroxide ion has reacted, the rate of reaction is 37 x .

Q9. The reaction: W + X Y + Z was conducted at two different temperature and the results are shown in the graph below.

Which of the following rate equations is correct based on the information provided by the graphs?

A. Rate = k[W]B. Rate = k[X]C. Rate = k[W][X]D. Rate = k[W][X]2

YEAR 2003

Q3. The reaction mechanism for the nitration of benzene is as follows.

Which statement about the above mechanism is true?

A. NO2+ion is a nucleophile.

B. K1, k-1 and k2 are not affected by temperature.C. The value of k2 is larger than that of k-1.

D. The rate of the reaction is not affected by the concentration of NO2+ ion.

Q16. Which statement about the effect of a catalyst on a reversible reaction is true?

A. It increases the rate of the forward reaction more than that of the reverse reaction.B. It decreases the value of the equilibrium constant of the forward reaction.C. It changes the mechanism of the chemical reaction.D. It increases the effective collision velocity of the particles of the reactants.

. Q47. In catalytic converters, platinum/palladium/rodium is used to catalyse

1. The oxidation of sulphur dioxide to sulphur trioxide.2. The oxidation of hydrocarbons to carbon dioxide and water.3. The reduction of the oxides of nitrogen to nitrogen and oxygen.

A B C D1 only correct 1 & 2 only are

correct2 & 3 only are

correct1, 2 & 3 are correct

YEAR 2004

Q3. The graph below shows an energy profile for the reaction X + Y Z.

Energy/kJ a

Z

b

X + Y

Reaction coordinate

Which of the following statements is true of the above energy profile?

A. The forward reaction is exothermic.B. ∆H for the forward reaction is +a kJ mol-1.C. The presence of a catalyst will reduce the value of b.D. The activation energy for the forward reaction is (a+b) kJ mol-1.

Q16 The Arrhenius equation is given as ¿ Ae−EaRT . A reaction happens 27 times faster at 52℃

than at 22℃. Calculate the activation energy, in kJ mol-1, for this reaction.

[Gas constant R= 8.31 J K -1 mol-1]

A. 1.0B. 10.5C. 38.1D. 87.5

AnswersPaper 1

YEAR 1999

Q9. D

-d [H2O2]/dt is the rate of reaction. For a first order reaction, the rate is directly proportional to the concentration of H2O2. A graph of rate against [H2O2] is a straight line (with positive gradient) that passes through the origin. The rate at T2 (higher temperature) is higher than that at T1. Hence the gradient of the T2 line is greater than that the gradient of the T1 line.

Q10.B

A catalyst increases the rate of reaction by providing an alternative pathway with lower activation energy for the reaction to occur. Hence, the mechanism of the reaction is altered. Certain homogeneous catalysis cannot be explained using the collision theory.

YEAR 2000

Q12. D

Increasing temperature displaces the peak towards the left. The height of the peak is also reduced.

Q49. A

The activation energy of a catalysed reaction is lower than the activation energy of a non-catalysed reaction.

YEAR 2001

Q9. D

Based on Arrhenius collision theory, only those collisions with energy greater than the activation energy will result in the formation of product. When the temperature increases by 10K, the fraction of molecules having energy equal to or greater than the activation energy increases two times. Hence, the rate of effective collision increases two times.

Q10. D

The half-life of a first order reaction does not depend on the pressure or the concentration of reactants.

T1/2 = ¿2k (where k =rate constant)

When temperature increases, the rate constant, k, will increase. This causes the half-life to decreases.

Using the formula {1/2} n = N/No

(where n = number of half-life. No = original amount; N = amount at time t)

{1/2} 4 = N/No

N/No = 0.0625

Or = 0.0625 × 100%

= 6.25%

Q42. B

A catalyst increases the rate of a chemical reaction by providing an alternative route with lower activation energy. However, it does not affect the enthalpy of reaction (∆H) as shown by the diagram below.

Energy

∆H

Catalysed reaction

Uncatalysed reaction

Q47. B

X3+ is regenerated at the end of the reaction. Hence, X3+ acts as catalyst.

X3+ is temporarily converted into X2+ during the reaction.

This is a homogeneous catalytic reaction because the physical state of the catalytic is the same as the physical state of the reactants.

YEAR 2002

Q8. D(A) From the rate equation, it is first order with respect to trichloromethane.

(B) The rate of production of the chloride ion = 3 × 0.2 = 0.6 mole dm-3s-1 .

(C) It is a second order reaction overall. Hence, when the concentrations of both the reactants are doubled, the rate should increase four times, i.e. 4x.

(D)

2CHCI3 + 7OH- products

Initially mole dm-3 1 1

Final/mole dm-3 1-(27 ×

12 ) =

67 ( 1 -

12 ) =

12

Hence the new rate = k [ 67 ×

12 ] =

37 x

Q9. A The graph of [X] against time is a straight line graph showing that the reaction is zero order with respect to [X]. hence, the rate equation cannot contain the [X] term.

YEAR 2003

Q3. C NO2

+ is an electrophile and not a nucleophile. The rate constant is temperature dependent. NO2

+ is involved in the slow step (rate-determining step). Hence, its concentration will affect the rate of reaction. The rate constant k is a measure of the rate of a reaction. The larger the rate constant, the faster is the rate. Rate = k [concentration]

Q16. CA catalyst increases the forward rate and the reverse rate by the same factor.The equilibrium constant is affected by temperature only.Only temperature can alter the effective collision velocity of the particles.The catalyst takes part in the reaction by allowing the reaction to occur by another pathway (i.e. by a different mechanism) with lower activation energy.

Q47. CPalladium and platinum catalyses the oxidation of carbon monoxide and the unburnt hydrocarbon.

2CO(g) + O2(g) 2CO2(g)

CxHy + ( x + y2 ) O2(g) xCO2(g) +

y2 H2O(g)

Rhodium catalyses the reduction of the oxides of nitrogen to nitrogen and carbon dioxide or oxygen.

2NO(g) + 2CO(g) N2(g) + 2CO2(g)

2NO(g) N2(g) + O2(g)

YEAR 2004

Q3. D The forward reaction is endothermic as the product has higher energy content than the reactants.∆H for the forward reaction is +b kJ mol-1.Catalyst has no effect on the ∆H of a reaction.The activation energy for the forward reaction (E1) is (a+b) kJ mol-1, while the activation energy for the reverse reaction (E2) is a kJ mol-1.

a E2

E1

b

Q16. D

In k = In A - EaRT OR log k = log A -

E2.303R ( 1T )

log k 2k 1 = Ea

2.303R ( 1T 1− 1T 2 )

With T1 = 22℃ or 295K and T2 = 52℃ or 325K

log 27 = Ea

2.303R ( 1295− 1325 )

Ea = 87 461 J mol-1

= 87.5 kJ mol-1

Paper 2

YEAR 2001

Q5 (B) (i) Let the rate equation be:

Rate = k[CIO2]x [OH-]y

Experiment IIExperiment I : 1.26×10−28.21×10−3 = 0.0713x0.0575x × 0.0216 y0.0216 y

1.535 = 1.24x

lg (1.535) = x lg (1.24)

∴ y = 1

The disproportionation reaction is second order with respect to [CIO2] and first order with respect to [OH- ] .

The rate equation is:Rate = k[CIO2]2 [OH-]The overall order is 3.

(ii) From experiment I : 8.21 × 10-3 = k(0.0575)2(0.0216) ∴ k = 114.97 mole -2 dm6 s-1

= 115 mole -2 dm6 s-1

(iii) Using the rate equation: Rate = k [CIO2]2 [OH-] 3.56 × 10-2 = (114.97)(0.100)2(OH-) ∴ [OH-] = 3.096 × 10-2 mole dm-3 pOH = -lg (3.096 × 10-2) = 1.51 ∴ pH = 14.00 – 1.51 = 12.49

YEAR 2003

Q2 (A) (i) First order.[The equation for a first order reaction is: InC = -kt + InC0 . A plot of InC against t will give a straight line with a negative slope. The gradient of the line is equal to the rate constant, k]

(ii) The gradient = 4.585−3.920

400

= 1.663 x 10-3s-1

The rate constant = 1.663 x 10-3s-1

(iii) Using the formula:

half-life = ¿2k

= 0.693

1.663×10−3

= 414.7 s

(B) (i) Energy/kJ

100kJ

23kJ products

Reaction coordinate

(ii) Activation energy for the reverse reaction = 100 + 23 kJ mol-1

= 123kJ mol-1

Q8. (B) (i) Nitrogen monoxide acts as a homogenous catalyst in the reaction. It catalyses the reaction by changing its oxidation state temporarily to +4 (in NO2) and then back to +2 (in NO). The two steps involved have lower activation energy than the uncatalysed reaction.

Energy

SO2(g) +12 O2(g)

SO3(g)

(ii) Platinum acts as a heterogeneous catalyst. In heterogeneous catalysis, the catalyst and the reactant are in different phases. Platinum, a transition element, has many empty orbitals in its valence shell which it uses to absorb the SO2 and O2 molecules on its surface. This not only increases the concentration of SO2 and O2(on the surface of platinum) but also weakens the intermolecular covalent bonds in the reactant molecules, making them easier to break, thereby lowering the activation energy. Furthermore, adsorption also holds the reactant molecules in the correct orientation for new bonds to be formed.

YEAR 2004

Catalysed reaction

Uncatalysed reaction

Q7 (A) (i) 2Br- Br2 + 2e 2BrO3

- + 12H+ + 10e Br2 + 6H2O

(ii) Let the rate equation be: Rate = k[BrO3

-]x [Br-]y [H+]z

Experiment 2Experiment 1 =

1.94×10−31.29×10−3 =

(0.6 ) x(0.4 ) x

1.50= (1.50)x

∴ x = 1

Experiment 3Experiment 2 =

3.87×10−31.94×10−3 =

(0.56 ) y(0.28 ) y

2.00=(2.00)y

∴ y = 1

Experiment 4Experiment 1 =

5.16×10−31.29×10−3 =

(0.62 ) z(0.31 ) z

4.00= (2.00)z

∴ z = 2The reaction is first order with respect to BrO3

- and Br-, and is second order with respect to H+ .

(iii) Rate equation: Rate = k[BrO3

-][Br-][H+]2

(iv) When the pH of the solution = 2.2 [H+] = 6.31 × 10-3 mol dm-3

{NOTE: pH = -log [H+] OR log 1¿¿ }

However, the rate constant is a constant at constant temperature and does not depend on the concentration. Hence, we can use experiment 4 directly tocalculate the rate constant. 5.16 × 10-3 = k(0.400)(0.280)(0.062)2

5.16 × 10-3 = 4.31 × 10-4 k ∴ k= 8.35 × 10-2 mol-3 dm9 s-1