Author: Cheryl Turner

IB Chemistry – HL 
Topic 6 Questions
1.	The reaction between NO_2 and F_2 gives the following rate data at a certain temperature. 
What is the order of reaction with respect to NO_2 and F_2?
[NO_2]/mol dm^–3 | [F_2]/mol dm^–3 | Rate /mol dm^–3 min^–1
0.1 | 0.2 | 0.1
0.2 | 0.2 | 0.4
0.1 | 0.4 | 0.2

| NO_2 order | F_2 order
A. | first | first
B. | first | second
C. | second | first
D. | second | second
2.	Which step in a multi-step reaction is the rate determining step?
A.	The first step
B.	The last step
C.	The step with the lowest activation energy
D.	The step with the highest activation energy
3.	The rate expression for a reaction is shown below.
rate = k[A]^2[B]^2
	Which statements are correct for this reaction?
I.	The reaction is second order with respect to both A and B.
II.	The overall order of the reaction is 4.
III.	Doubling the concentration of A would have the same effect on the rate of reaction as doubling the concentration of B.
A.	I and II only
B.	I and III only
C.	II and III only
D.	I, II and III

4.	Values of a rate constant, k, and absolute temperature, T, can be used to determine the activation energy of a reaction by a graphical method. Which graph produces a straight line?
A.	k versus T
B.	k versus 
C.	ln k versus T
D.	ln k versus 
5.	The rate expression for a particular reaction is 
Rate = k[P][Q]
	Which of the units below is a possible unit for k?
A.	mol^–2 dm^6 min^–1
B.	mol^–1 dm^3 min^–1
C.	mol dm^3 min^–1
D.	mol^–2 dm^–6 min^–1
6.	The reaction 2X(g) + Y(g) ® 3Z(g) has the rate expression
		rate = k [X]^2[Y]^0
	The concentration of X is increased by a factor of three and the concentration of Y is increased by a factor of two. By what factor will the reaction rate increase?
A.	6			B.	9
C.	12			D.	18
7.	A reaction occurs in four steps. The steps and their rates are shown in the table
Step | Rate
1 | 0.01 mol dm^–3 s^–1
2 | 0.10 mol dm^–3 s^–1
3 | 0.01 mol dm^–3 min^–1
4 | 0.10 mol dm^–3 min^–1
	Which is the rate-determining step?
A.	Step 1			B.	Step 2
C.	Step 3			D.	Step 4

8.	The rate expression for a reaction is
		rate = k[CH_3Br][OH^–]
Which is a possible unit for k?
A.	mol^2 dm^–6 min^–1			B.	mol dm^–3 min^–1
C.	mol^–1 dm^3 min^–1			D.	mol^–2 dm^6 min^–1
9.	What happens to the rate constant (k) and activation energy (E_a) of a reaction when the temperature is increased?
A.	k increases and E_a is unaffected.		B.	k decreases and E_a is unaffected.
C.	E_a increases and k is unaffected.		D.	E_a decreases and k is unaffected.
10.	The mechanism of a reaction is
		XY_2 + XY_2 ® X_2Y_4
		X_2Y_4 ® X_2 + 2Y_2
		X_2 + Y_2 ® 2XY
What is the overall equation for the reaction?
A.	X_2Y_4 ® 2XY_2			B.	2XY_2 ® X_2 + 2Y_2
C.	2XY_2 ® 2XY + Y_2			D.	X_2Y_4 ® 2XY + Y_2
11.	Consider the reaction
		2I^-(aq) + H_2O_2(aq) + 2H^+(aq) ® I_2(aq) + 2H_2O(l)
In the presence of S_2O_3^2–(aq) and starch solution, the time taken for a blue colour to form was observed at various reactant concentrations.
Experiment | [I^–] / mol dm^–3 | [H_2O_2] / mol dm^–3 | [H^+] / mol dm^–3 | Time / s
1 | 0.10 | 0.12 | 0.01 | 25
2 | 0.05 | 0.12 | 0.01 | 50
3 | 0.10 | 0.06 | 0.01 | 100
What is the correct order with respect to I^– and H_2O_2?
| I^– | H_2O_2
A. | 1 | 2
B. | |
C. | 2 | 1
D. | 2 | 4

12.	Which statement is correct about the rate expression for a chemical reaction?
A.	It can be determined from its chemical equation.
B.	It can be predicted from the value of ΔH^Ө for the reaction.
C.	It can be calculated from the effect of temperature on the reaction.
D.	It can be determined by measuring the change in concentration of a reactant or product over time.
13.	For the reaction 2NO_2(g) + F_2(g) ® 2NO_2F(g) the accepted mechanism is
		NO_2(g) + F_2(g) ® NO_2F(g) + F(g)	slow
		NO_2(g) + F(g) ® NO_2F(g)	fast
What is the rate expression for this reaction?
A.	rate = k[NO_2]^2[F_2]			B.	rate = k[NO_2][F_2]
C.	rate = k[NO_2][F]			D.	rate = k[NO_2]^2
14.	The activation energy, of a reaction can be obtained from the rate constant, k, and the absolute temperature, T. Which graph of these quantities produces a straight line?
A.	k against T
B.	k against 
C.	ln k against T
D.	ln k against 
15.	What is the order of reaction with respect to NO_2(g) and F_2(g) given the following rate data at a certain temperature?
[NO_2(g)] / mol dm^–3 | [F_2(g)] / mol dm^–3 | Rate / mol dm^–3 min^–1
0.1 | 0.2 | 0.1
0.2 | 0.2 | 0.4
0.1 | 0.4 | 0.2

| Order with respect to NO_2(g) | Order with respect to F_2(g)
A. | first | first
B. | first | second
C. | second | first
D. | second | second
16.	Nitrogen(II) oxide reacts with hydrogen as shown by the following equation. 
2NO(g) + 2H_2(g) → N_2(g) + 2H_2O(g)
	The table below shows how the rate of reaction varies as the reactant concentrations vary. 

Experiment

Initial [NO] / 
mol dm^–3

Initial [H_2] / mol dm^–3

Initial rate / 
mol N_2 dm^–3 s^–1

1 | 0.100 | 0.100 | 2.53×10^–6
2 | 0.100 | 0.200 | 5.05×10^–6
3 | 0.200 | 0.100 | 10.10×10^–6
4 | 0.300 | 0.100 | 22.80×10^–6
(a)	Determine the order of reaction with respect to NO and with respect to H_2. 
Explain how you determined the order for NO.
NO ..............................................................................................................................
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H_2 ................................................................................................................................
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(3)
(b)	Write the rate expression for the reaction.
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(1)
(c)	Calculate the value for the rate constant, including its units.
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(2)
(d)	A suggested mechanism for this reaction is as follows.
	H_2 + NO  X fast step 
X + NO → Y + H_2O slow step
Y + H_2 → N_2 + H_2O fast step
	State and explain whether this mechanism agrees with the experimental rate expression in (b).
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(4)
(e)	Explain why a single step mechanism is unlikely for a reaction of this kind.
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(2)
(f)	Deduce how the initial rate of formation of H_2O(g) compares with that of N_2(g) in experiment 1. Explain your answer.
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(2)
(Total 14 marks)
17.	The oxidation of nitrogen monoxide takes place as follows:
		2NO(g) + O_2(g) ® 2NO_2(g)
The following experimental data was obtained at 101.3 kPa and 298 K.
Experiment | Initial [NO] / mol dm^–3 | Initial [O_2] / mol dm^–3 | Initial rate / mol dm^–3 s^–1
1 | 3.50×10^–2 | 1.75×10^–2 | 3.75×10^–3
2 | 3.50×10^–2 | 3.50×10^–2 | 7.50×10^–3
3 | 7.00×10^–2 | 7.00×10^–2 | 6.00×10^–2
(a)	Deduce the order of reaction with respect to O_2.
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(1) 
(b)	Deduce the order of reaction with respect to NO.
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(1)
(c)	State the rate expression for the reaction.
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(1)
(d)	Calculate the value of the rate constant and state the units.
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(2)
(e)	Suggest a possible mechanism that is consistent with the rate expression. Indicate which of the steps is the rate-determining step.
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(3)
(Total 8 marks)
18.	An equation for the decomposition of substance A is
2A ® 2B + C
	A graph showing the change in concentration of A against time as the reaction proceeds at a particular temperature is shown below.

(a)	Define the term half-life of reaction.
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(1)
 
(b)	Use the graph to measure values of half-life of reaction, starting from
time = zero ................................................................................................................
time = 1000 s ............................................................................................................
(2)
(c)	Deduce the order of the reaction with respect to A, giving a reason for your choice, and write the rate expression for the reaction.
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(3)
(d)	For a different reaction, between compounds D and E, the rate expression is
		rate = k[D]^2[E]
Calculate the value of k, including units, for the reaction when the concentrations of both D and E are 1.35×10^–2 mol dm^–3 and the reaction rate is 3.75×10^–5 mol dm^–3 min^–1.

(3)
(Total 9 marks)
19.	In a particular experiment, various concentrations of HI(aq) are reacted with a constant H_2O_2(aq) concentration according to the following equation:
	H_2O_2(aq) + 2HI(aq) ® I_2(aq) + 2H_2O(I)
A graph of [HI] against time is as follows:

(a)	Use the graph to deduce the order of reaction with respect to HI. Give a reason for your answer.
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(2)
(b)	The order with respect to H_2O_2 is the same as HI. Deduce the rate expression for this reaction.
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(1)
(c)	Determine the half-life of the reaction from the graph and calculate the value for the rate constant.
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(2)
(Total 5 marks)
20.	(a)	The table below shows kinetic data for the following reaction
		C + D ® E + F

Experiment

[C] / mol dm^–3

[D] / mol dm^–3

Initial rate
/ mol dm^–3 min^–1

1 | 2.0×10^–3 | 3.0×10^–3 | 1.0×10^–6
2 | 4.0×10^–3 | 3.0×10^–3 | 2.0×10^–6
3 | 6.0×10^–3 | 6.0×10^–3 | 3.0×10^–6
(i)	Deduce the order of reaction with respect to both C and D, giving a reason in each case.
C ......................................................................................................................
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D ......................................................................................................................
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(4) 
(ii)	Deduce the rate expression for this reaction.
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(1)
(iii)	Use data from Experiment 1 to calculate a value for the rate constant for this reaction and deduce its units.
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(3)
(b)	Define the term half-life and calculate the half-life for a first-order reaction with a rate constant of 3.3´10^–2 min^–1.
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(2)
(Total 10 marks)
21.	Nitrogen(II) oxide reacts with hydrogen according to the following equation:
	2NO(g) + 2H_2(g) → N_2(g) + 2H_2O(g)
	The table shows how the rate of reaction varies as the concentrations of the reactants are changed.

Experiment

Initial [NO] /
mol dm^–3

Initial [H_2] /
mol dm^–3

Initial rate /
mol (N_2) dm^–3 s^–1

1 | 0.100 | 0.100 | 253×10^–6
2 | 0.100 | 0.200 | 5.05×10^–6
3 | 0.200 | 0.100 | 1.01×10^–5
4 | 0.300 | 0.100 | 2.28×10^–5
 
(a)	Determine the order of reaction with respect to H_2 and with respect to NO.
H_2 ................................................................................................................................
NO ..............................................................................................................................
(2)
(b)	Write the rate expression for the reaction.
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(1)
(c)	Calculate the value for the rate constant, and state its units using the data from experiment 1.
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(2)
(d)	A suggested mechanism for this reaction is as follows.
	H_2 + NO  X	fast step
X + NO → Y + H_2O	slow step
Y + H_2 → N_2 + H_2O	fast step
	State and explain whether this mechanism agrees with the experimental rate expression in (b).
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(4)
(e)	Explain why a single step mechanism is unlikely for a reaction of this kind.
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(2)
(f)	Deduce and explain how the initial rate of formation of H_2O compares with that of N_2.
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(2)
(Total 13 marks)
22.	The data below refer to a reaction between X and Y.

Initial concentration /
mol dm^–3

Initial rate of reaction /
mol dm^–3 s^–1

Experiment | X | Y |
1 | 0.25 | 0.25 | 10×10^–2
2 | 0.50 | 0.25 | 4.0×10^–2
3 | 0.50 | 0.50 | 8.0×10^–2
 (i)	Define the term order of reaction.
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(1)
(ii)	Deduce the order of reaction with respect to both X and Y. Explain your reasoning.
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(4)
(iii)	Write the rate expression for the reaction and calculate the rate constant, including its units.
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(4)
(iv)	Calculate the initial rate of reaction when the initial concentrations of X and Y are 0.40 mol dm^–3 and 0.60 mol dm^–3 respectively.
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(2)
(Total 11 marks)
23.	Oxygen and nitrogen monoxide react together to form nitrogen dioxide. 
O_2(g) + 2NO(g) → 2NO_2(g)
	The graph below shows how the initial rate of reaction changed during an experiment in which the initial [NO(g)] was kept constant whilst the initial [O_2(g)] was varied.

(a)	Deduce, giving a reason, the order of reaction with respect to O_2
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(2)
(b)	In a series of experiments, the initial [O_2(g)] was kept constant while the initial [NO(g)] was varied. The results showed that the reaction was second order with respect to NO. Sketch a graph to show how the rate of reaction would change if the initial
[NO(g)] was increased.
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(2)
(c)	Deduce the overall order of this reaction.
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(1)
(d)	State and explain what would happen to the initial rate of reaction if the initial concentration of NO was doubled and that of O_2 was halved.
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(3)
(e)	When the initial values are [O_2(g)] = 1.0×10^–2 mol dm^–3 and [NO(g)] = 3.0×10^–2 mol dm^–3, the initial rate of reaction is 6.3×10^–4 mol dm^–3s^–1. Write the rate expression for this reaction and calculate the rate constant, stating its units.
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(4)
(Total 12 marks)
24.	The compound iodine chloride, ICl, reacts with hydrogen to form iodine and hydrogen chloride.
(i)	Deduce the equation for this reaction.
(1)
(ii)	The kinetics of this reaction were studied at a certain temperature, when all the reactants and products were in the gas phase. The table shows the initial rate of reaction for different concentrations of reactants.
Experiment | [ICl] / mol dm^–3 | [H_2] / mol dm^–3 | Initial rate / mol dm^–3 s^–1
1 | 0.100 | 0.0500 | 5.00×10^–3
2 | 0.200 | 0.0500 | 1.00×10^–2
3 | 0.200 | 0.0250 | 2.50×10^–3
Deduce and explain the order of reaction with respect to ICl and to H_2.
(4)
(iii)	Write the rate expression for the reaction.
(1)
(iv)	Use information from Experiment 1 to determine the value, with units, of the rate constant for the reaction.
(2)
(v)	Determine the rate of reaction when the concentrations of reactants in Experiment 1 are both doubled.
(1)
(Total 9 marks)
25.	(a)	The variation of the rate constant, k, for a reaction with temperature is shown by the Arrhenius equation. Two versions of this equation are shown in Table 1 of the Data Booklet.
(i)	Explain the significance of the Arrhenius constant, A, in this equation.
(1)
(ii)	Explain what is meant by the term activation energy,E_a.
(1)
(iii)	Describe how, using a graphical method, values of A and E_a can be obtained for a reaction.
(5)

(b)	The equation for a reaction used in industry is
	CH_2CH_2 + Cl_2 ® CH_2ClCH_2CL	DH^Ө = –185 kJ
Iron(III) chloride can be used as a catalyst for the reaction.
(i)	Explain the difference between the terms homogeneous and heterogeneous when applied to a catalyst.
(1)
(ii)	Draw an enthalpy level diagram for this reaction, including labels for DH^Ө, E_a and the activation energy when a catalyst is used, E_cat.
(4)
(Total 12 marks)
26.	Nitrogen(II) oxide reacts with bromine according to the following equation.
	2NO(g) + Br_2(g) ® 2NOBr(g)	DH = negative
The data below were obtained for the reaction between NO(g) and Br_2(g) at a specified temperature and pressure.
Experiment | Initial [NO] / mol dm^–3 | Initial [Br_2] / mol dm^–3 | Initial rate / mol dm^–3 s^–1
1 | 2.00×10^–2 | 5.00×10^–3 | 3.20×10^–3
2 | 2.00×10^–2 | 2.50×10^–3 | 1.60×10^–3
3 | 4.00×10^–2 | 5.00×10^–3 | 1.30×10^–2
(a)	Determine, giving a reason, the order of reaction with respect to NO and the order of reaction with respect to Br_2.
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(2)
(b)	Derive the rate expression for the reaction between NO and Br_2.
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(1)
(c)	Calculate the rate constant for the rate expression using experiment 1 and state its units.
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(2)
(d)	If the total volume of the reaction mixture was doubled at constant temperature, state the effect, if any, on
(i)	the rate constant.
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(1)
(ii)	the rate of change of the Br_2(g) concentration.
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(1)
(e)	Draw a labelled enthalpy level diagram for the reaction between NO(g) and Br_2(g), with and without the use of a catalyst.

(3)
(Total 10 marks)

27.	(i)	The reaction between propanone, CH_3COCH_3 and bromine, Br_2 in the presence of acid, H^+, is found to be second order overall, but the rate is independent of the bromine concentration. Write three possible rate expressions for the reaction.
(3)
(ii)	The concentration of each of the three reactants was doubled in three separate experiments. Choose one of the rate expressions in (i) and predict the effect on the rate of the reaction of each of these changes.
(2)

(iii)	The graph below shows how the concentration of propanone changes with time in a reaction.

Use the graph to confirm that the reaction is first order with respect to propanone showing your working.
(2)
(iv)	The overall reaction is:
CH_3COCH_3(aq) + Br_2(aq)  CH_3COCH_2Br(aq) + HBr(aq)
Describe one observation that would allow you to follow the progress of the reaction. State and explain the role of the acid in the reaction.
(4)
(Total 11 marks)

IB Chemistry – HL 
Topic 6 Answers
1.	C
2.	D
3.	D
4.	D
5.	B
6.	B
7.	C
8.	C
9.	A
10.	C
11.	A
12.	D
13.	B
14.	D
15.	C
16.	(a)	(order with respect to) NO = 2;
(order with respect to) H_2 = 1;
rate increases×4 when [NO] doubles/OWTTE;	3
(b)	rate = k[NO]^2[H_2];	1
ECF from (a).
(c)	(2.53×10^–6 mol dm^–3 s^–1 = k (0.100 mol dm^–3)^2(0.100 mol dm^–3))
k = 2.53×10^–3;	1
	mol ^–2 dm^6 s^–1;	1
ECF from (b).
(d)	agrees/yes;
slow step depends on X and NO;
X depends on H_2 and NO;
(so) NO is involved twice and H_2 once;
Overall equation matches the stoichiometric equation;
Award [1] each for any three of the four above.
OWTTE
ECF for “no”, depending on answer for (b).

	Or	agrees/yes;
and = constant;
rate of slow step = k [X][NO]
	  = k [H_2][NO]^2;	4
ECF for “no”, depending on answer for (b).
(e)	reaction involves four molecules;
statistically/geometrically unlikely;	2
(f)	the rate of formation of H_2O(g) = 2×rate for N_2(g);
because 2 moles H_2O formed with 1 mole N_2/OWTTE;	2
[14]
17.	(a)	first order (with respect to O_2);	1
(b)	second order (with respect to NO);	1
(c)	rate = k[NO]^2[O_2];	1
Allow ECF from parts (a) and (b).
(d)	
dm^6 mol^–2 s^–1;	2
Award [1] mark for the answer and [1] mark for units.
Allow ECF from part (c).
(e)	NO + NO  N_2O_2;
N_2O_2 + O_2 ® 2NO_2;
second step is rate determining step;
Allow ECF from part (c).
OR
NO + O_2  NO_3;
NO_3 + NO ® 2NO_2;
second step is rate determining step;	3
Allow ECF from part (c).
[8]
18.	(a)	time for reactant concentration to halve/OWTTE;	1
Accept “time for mass to halve”.
(b)	1000 s;
1000 s;	2
Accept 900-1100 s.
(c)	first order;
constant half-life;
rate = k[A];	3
Allow ECF for rate expression from stated order.
(d)	
= 15.2;
Accept answer in range 15.2 to 15.3.
mol^-2 dm^6 min^-1;	3
[9]
19.	(a)	first order;
constant half-life;	2
(b)	rate = k[HI][H_2O_2];	1
ECF from(a).
(c)	47 sec;
Accept answer in range 45 to 49.
	2
Accept answer in range 0.014-0.015.
ECF from half-life.
[5]
20.	(a)	(i)	(C)	first order;
	doubling [C] doubles rate/OWTTE;
(D)	zero order;
	changing [D] has no effect on rate/OWTTE;	4
(ii)	rate = k[C]/rate = k[C]^1[D]^0;	1
Apply ECF from (a)(i).
(iii)	k = 
= 5.0×10^-4;
min^-1;		3
Apply ECF from (a)(ii).
(b)	time for half of (amount/concentration of) reactant to react/disappear;
t ( = 0.693÷0.033) = 21 min;	2
Units needed for second mark.
[10]

21.	(a)	(order with respect to) H_2 = 1;
(order with respect to) NO = 2;	2
(b)	rate = k[H_2][NO]^2;	1
ECF from (a).
(c)	(2.53×10^−6 mol dm^–3 s^–1 = k(0.100 mol dm^−3)(0.100 mol dm^–3)^2)
k = 2.53×10^–3;
mol^−2 dm^6 s^–1;	2
ECF from (b).
(d)	agrees/yes;
	slow step depends on X and NO;
(so) NO is involved twice and H_2 once;
overall equation matches the stoichiometric equation/OWTTE;
ECF for “no”, depending on answer for (b).
	OR
	agrees/yes;
and  = constant;
rate of slow step = k[X][NO];
but X depends on H_2 and NO;
rate of slow step = k[H_2][NO]^2;	4 max
Award [1] each for any three of the four above.
ECF for “no”, depending on answer for (b).
(e)	reaction involves four molecules;
statistically/geometrically unlikely;	2
 
(f)	the rate of formation of H_2O = 2×rate for N_2;
because 2 moles H_2O formed with 1 mole N_2/OWTTE;	2
[13]
(i)	the power of a reactant’s concentration in the rate equation/sum of
powers of concentration/rate = k[X]^n, where n = order of reaction;	1
Must be in terms of powers of concentration.
(ii)	experiment 1—2 : [X] doubles and rate×4;
2nd order for X;
experiment 2—3 : [Y] doubles and rate×2;
1st order for Y;	4
(iii)	rate = k[X]^2[Y](ECF from (ii))
for experiment 1, 1.0×10^–2 = k (0.25)^2(0.25);
k = 0.64;
mol^–2 dm^6 s^–1;	4
Allow ECF from rate expression.
(iv)	rate = 0.64[0.40]^2[0.60];
= 0.061;	2
Final answer to 2 sig figs only.
Allow ECF from (iii).
[11]
23.	(a)	1/first order; 
rate is (directly) proportional to concentration of oxygen/OWTTE;	2
(b)	

	correct axes;
correct shape curve; 	2
(c)	3/third order;	1
Allow ECF from (a) and (b).
(d)	overall effect on rate = 4×/doubled/×2;
[NO(g)] doubled, rate =×4/quadrupled;
[O_2(g)] halved, rate =×1/halved; 	3
Allow ECF from (a) and (b).
(e)	rate = k[NO(g)]^2 [O_2(g)];
;
	= 70;
mol^–2 dm^6 s^–1;	4
Allow ECF.
State symbols not needed.
[12]
24.	(i)	2ICl + H_2 ® I_2 + 2HCl;	1
(ii)	ICl order
1;
because doubling [ICl] doubles rate (when [H_2] constant);
H_2 order
2;
because halving [H_2] quarters rate (when [ICl] constant);
or doubling [H_2] quadruples rate (when [ICl] constant);	4
(iii)	rate = k [ICl][H_2]^2;	1
ECF from (ii).
(iv)	k = 5.00×10^-3÷0.100×0.0500^2  = 20;
mol^-2 dm^6 s^-1;	2
ECF from (iii).
(v)	rate = 20×0.200×0.100^2 = 4.00×10^-2 (mol dm^-3 s^-1);	1
ECF from (iii).
[9]
25.	(a)	(i)	it relates to the geometric requirements of the reaction/orientation
of reactants on collision/OWTTE;	1
(ii)	minimum energy needed for reactants to react (on collision)/OWTTE;	1
(iii)	k measured at different values of temperature;
graph plotted of ln k against 1/T;
intercept on y-axis is ln A;
A = e^intercept;
measured slope of graph = - E_a/R;
E_a = – R×gradient;	5
Award [1] each for any five.
(b)	(i)	homogeneous catalyst is in same phase as reactants and heterogeneous
catalyst is in different phase from reactants; 	1
(ii)		4

OR

	reactants line higher than product line (labels not needed);
ΔH label;
E_a label;
E_cat label;
[12]
26.	(a)	order of NO: second/2 - [NO] doubled, rate×4/OWTTE;
order of Br_2: first/1 - as [Br_2] doubled, rate of reaction doubled/OWTTE;	2
Reason needed for each mark.
(b)	rate = k [NO]^2[Br_2];	1
Allow ECF from (a).
(c)	3.20×10^-3 = k(2.00×10^-2)^2×5.00×10^-3
k = 1.60×10^3;
dm^6 mol^-2 s^-1;	2
Allow ECF from (b).
(d)	(i)	no effect/K changes only with temperature/OWTTE;	1
(ii)	decrease (by a factor of 2);	1

(e)	

curve clearly showing E_a without catalyst (E_a);
curve clearly showing E_a with catalyst (E_a(cat));
labelling for x axis;	3
Accept time/progress of reaction/course of reaction/OWTTE.
Award [2 max] if an enthalpy level diagram for an endothermic reaction has been correctly drawn.
[10]
27.	(i)	rate = k[CH_3COCH_3][H^+];
rate = k[CH_3COCH_3]^2;
rate = k[H^+]^2;	3
(ii)	[CH_3COCH_3] doubles, rate doubles and [H^+] doubles, rate doubles;
[Br_2] double, no effect on rate;
OR
[CH_3COCH_3] doubles, rate quadruples;
[Br_2] doubles/[H^+] doubles, no effect on rate;
OR
[H^+] doubles, rate quadruples;
[Br_2] doubles/[CH_3COCH_3] doubles, no effect on rate;	2
The answer given must correspond to the selected expression in (i).
(iii)	constant half-life;
at least two sets of data to justify statement;	2
e.g. [ ] from 1.6 to 0.8 mol dm^-3, 10s; 0.8 to 0.4, 10s; 0.4 to 0.2, 10s.
(iv)	decrease in the colour of the bromine/OWTTE;
catalyst;
increases rate/speeds up reaction;
by lowering E_a/activation energy (by providing an alternate pathway);	4
[11]