6.1
Topic 6 Kinetics Answers
Rates of Reaction and Collision Theory
1.
SL/HL
In general, the rate of a reaction can be increased by all of the following except
A.
increasing the temperature.
B.
increasing the activation energy.
C.
increasing the concentration of reactants.
D.
increasing the surface area of the reactants.
2. The reaction between excess calcium carbonate and hydrochloric acid can be followed by measuring the
volume of carbon dioxide produced with time. The results of one such reaction are shown below.
,
Volume of CO2
Time
How does the rate of this reaction change with time and what is the main reason for this change?
A.
The rate increases with time because the calcium carbonate particles get smaller.
B.
The rate increases with time because the acid becomes more dilute.
C.
The rate decreases with time because the calcium carbonate particles get smaller.
D.
The rate decreases with time because the acid becomes more dilute.
3. Reaction rates typically increase with an increase in temperature. This increase is best
explained by an increase in the
A. average energy of the particles.
B.
number of collisions between particles.
C.
number of particles with the energy required to react.
D. activation energy of the reaction.
4. Under what conditions is the rate of reaction of magnesium with HCl(aq) fastest?
5.
A.
10 cm3 of 1.0 mol dm-3 HCl(aq) at 25 °C
B.
10 cm3of 2.0mol dm-3 HCl(aq) at 25 oC
C.
10 cm3 of 2.0 mol dm-3 HCl(aq) at 35 °C
D.
10 cm3 of 1.0 mol dm-3 HCl(aq) at 35 °C
Which factor(s) will influence the rate of the reaction shown below?
NO2 (g) + CO(g) → NO(g) + CO2 (g)
I.
The number of collisions per second
II.
The energy of the collisions
III.
The geometry with which the molecules collide
A.
I only
B.
II only
C.
I and II only
D.
I, II and III
6.. As the temperature of a reaction between two gases is increased, the rate of the reaction increases. This is
mainly because
A.
the concentrations of the reactants increase.
B.
the molecules collide more frequently.
C.
the pressure exerted by the molecules increases.
D.
the fraction of molecules with the energy needed to react increases.
7. The reaction between nitrogen and oxygen in the atmosphere under normal conditions is extremely slow.
Which statement best explains this?
A.
The concentration of oxygen is much lower than that of nitrogen
B.
The molar mass of nitrogen is less than that of oxygen
C.
The frequency of collisions between nitrogen and oxygen molecules is lower than that between
nitrogen molecules themselves
D.
Very few nitrogen and oxygen molecules have sufficient energy to react
8. The addition of a catalyst to a chemical reaction alters the rate primarily by
A. changing the enthalpy of the reaction.
B.
increasing the number of collisions between the reactant molecules in a given time.
C.
increasing the fraction of reactant molecules with a given kinetic energy.
D.
providing a different reaction pathway.
9. Most reactions occur in a series of steps, one of which is the rate determining step. The rate determining
step is so called because it is the
10.
A.
first step.
B.
last step.
C.
fastest step.
D.
slowest step.
The rate-determining step of a reaction is the
A. fastest step in the reaction.
B.
first step in the reaction.
C.
final step in the reaction.
D. step with the highest activation energy.
11. Excess magnesium was added to a beaker of aqueous hydrochloric acid on a balance.
A graph of the mass of the beaker and contents was plotted against time (line 1).
Mass
Time
What change in the experiment could give line 2?
I.
The same mass of magnesium but in smaller pieces
II.
The same volume of a more concentrated solution of hydrochloric acid
III.
A lower temperature
faster rate, same amount
faster rate and more product
faster rate, same amount
12.
A.
I only
B.
II only
C.
III only
D.
None of the above
The rate of a reaction between two gases increases when the temperature is increased and a catalyst is
added. Which statements are both correct for the effect of these changes on the reaction?
Increasing the temperature
13.
Adding a catalyst
A.
Collision frequency increases
Activation energy increases
B.
Activation energy increases
Activation energy does not change
C.
Activation energy does not change
Activation energy decreases
D.
Activation energy increases
Collision frequency increases
For a given reaction, why does the rate of reaction increase when the concentrations of the reactants are
increased?
A.
The frequency of the molecular collisions increases.
B.
The activation energy increases
C.
The average kinetic energy of the molecules increases
D.
The rate constant increases.
14 (a) The graph below shows the volume of carbon dioxide gas produced against time when excess
calcium carbonate is added to x cm3 of 2.0 mol dm -3 hydrochloric acid.
(i)
Write a balanced equation for the reaction.
[1]
(ii) State and explain the change in the rate of reaction with time. Outline how you would determine the
rate of the reaction at a particular time.
[4]
(bi) Sketch the above graph on an answer sheet. On the same graph, draw the curves you would
expect if:
I.
the same volume (x cm3) of 1.0 mol dm3 HC1 is used.
II.
double the volume (2x cm3) of 1.0 mol dm"3 HC1 is used.
Label the curves and explain your answer in each case.
[5]
14
15. When excess lumps of magnesium carbonate are added to dilute hydrochloric acid the following reaction
takes place.
MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H20(1)
(a)
Outline two ways in which the rate of this reaction could be studied. In each case sketch a graph
to show how the value of the chosen variable would change with time.
[4]
(b)
State and explain three ways in which the rate of this reaction could be increased.
(c)
State and explain whether the total volume of carbon dioxide gas produced would increase,
decrease or stay the same if
(i)
more lumps of magnesium carbonate were used
(ii) the experiments were carried out at a higher temperature.
[6]
[2]
[2]
Topic 6 Kinetics
6.2
HLONLY
The Rate Expression
1. The rate constant for a certain reaction has the units concentration time-1. What is the order of reaction?
A.
0
B.
1
C.
2
D.
3
2
The rate information below was obtained for the following reaction:
2NO2(g) + F2(g)H→ 2NO2F(g)
[NO2] mol dm-3
0.001
0.002
0.002
[F2] mol dm-3
Rate / moldm -3 s -1
0.005
0.005
0.010
2x10-4
4x10-4
8x10-4
What are the orders for NO2 and F2 ?
A.
NO2 is first order, F2 is first order
B.
NO2 is first order, F2 is second order
C.
NO2 is second order, F2 is first order
D.
NO2 is second order, F2 is second order
3. The rate of a gaseous reaction is given by the expression rate = k [P][Q]. If the volume of the
reaction vessel is reduced to 1/4 of the initial volume, what will be the ratio of the new rate to the original
rate?
A.
1:4
B.
1 : 16
C.
4:1
D.
16 :1
4. Use the information below to deduce the rate equation for the hypothetical reaction;
X + Y→ Z
A.
Rate = k[X][Y]
B.
Rate = k[X]2
C.
Rate = k[Y]2
D.
Rate = k[X]2[Y]2
[X]/moldm-3
[Y]/ moldm-3
Relative rate
0.01
0.01
1
0.02
0.01
4
0.02
0.02
4
5. The reaction between nitrogen dioxide and carbon monoxide is given by the equation below;
NO2 (g) + CO(g) → NO(g) + CO2 (g)
According to the following experimental data, what is the rate equation?
[NO2] / mol dm-3
[CO] / mol dm-3
Rate / mol dm-3s-1
0.10
0.10
1.0x10-6
0.30
0.10
9.0x10-6
0.30
0.30
9.0x10-6
A.
Rate = k[N02][CO]
B.
Rate = k[CO]2
C.
Rate = k[NO2]2
D.
Rate = k[N02]3
6. The kinetic data in the table below are for the reaction:
A + B→ C
From these data what are the orders of the reaction with respect to A and B?
[A]
0.1
[B]
0.1
Initial Rate (mol dm -3 sec -1)
1x10-5
0.1
0.2
4 x 10-5
0.2
0.1
lx10-5
A..
order of A = 1 order of B = 0
B.
order of A = 0 order of B = 2
C.
order of A = 0 order of B = 4
D.
order of A = 1 order of B = 2
7. The following experimental data was obtained for the reaction X + Y → products.
[X] / mol dm -3 [Y] / mol dm -3
Initial rate /
mol dm -3 sec -1
0.10
0.10
4.0x10-4
0.20
0.20
1.6xl0 -3
0.50
0.10
1.0xl0 -2
0.50
0.50
l.0xl0- 2
What is the order of reaction with respect to X and the order of reaction with respect to Y?
A.
2 and 0
B.
0 and 2
C.
2 and l
D.
1 and O
8. 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
9. For the chemical reaction
2NO(g) + O2(g) → 2NO2(g)
the following reaction mechanism has been proposed.
NO(g)+NO(g) ↔ N2O2(g)
N2O2(g) + O2(g) → 2NO2(g)
fast
slow
What could be the rate equation for this reaction?
A.
rate =k[NO][O2]
B.
rate =k[NO]2
C.
rate =k[N2O2][O2]
D.
rate =k[NO]2[O2]
10. The rate of a decomposition is studied by measuring the reactant
concentration at certain times.
Time/min
Conc /10-2 mol dm -3
0
1
20
0.69
40
0.48
60
0.34
80
0.24
100
0.16
120
0.11
(i) Plot a graph of concentration against time-
[3 marks]
Conc
time
(ii) How could the rate be determined from the graph at any selected time?
Explain the shape of the graph in terms of rate.
[5 marks]
Tangent
Gradient
At start rate is high
high frequency of collision
As reaction proceeds rate decreases
Due to reactive particles being used up so chance of collision decreases
(iii) What is meant by half life? Measure three half life values from the
graph. Deduce the order of this reaction.
t ½ = time taken for conc to half its initial value
measures X3, t ½ from graph
t ½ = 37-40 mins
first order since t ½ is constant
[5 marks]
(iv) Calculate a value of the rate constant from the half life and one from
the initial rate of reaction.
K = 0.693
t½
[5 marks]
= 0.017 to 0.019 min-1
using gradient = initial rate = 0.000175 moldm-3/min
Using gradient = initial rate = 0.000175 moldm-3/min
(accept 0.00015 to 0.0002)
Rate = K [A]1 substitute value for initial rate and initial conc = 0.0175 min-1
11.Oxygen and nitrogen monoxide react together to form nitrogen dioxide.
O2(g) + 2NO(g) → 2NO2(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 [O2 (g)] was varied.
Rate
[O2(g)]
(a)
Deduce, giving a reason, the order of reaction with respect to O2.
[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.
[2]
(c)
Deduce the overall order of this reaction.
(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.
[3]
(e)
When the initial values are [O2(g)] = 1.0x10-2moldm -3 and [NO(g)] = 3.0xl0 -2moldm -3, the initial rate
of reaction is 6.3x10-4 moldm-3s.
Write the rate expression for this reaction and calculate the rate constant, stating its units. [4]
[1]
(f)
Nitrogen monoxide may also be converted into nitrogen dioxide at high temperature according to the
equation below.
NO(g) + CO(g) + O2(g) → NO2(g) + CO2(g)
(i)
Sketch a graph of concentration of N02(g) produced against time for this reaction and annotate the graph to
show how the initial rate of reaction could be deduced.
[3]
(ii)
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.
Experiment [NO(g)]
/ mol dm
(g)
[CO(g)] /
mol dm-3
-3
-3
1.00x10
[02(g)]
/ mol dm-3
-3
[6]
Initial rate /
-3
mol dm-3 s -1
4.40 x10-4
1
1.00x10
1.00x10
2
2.00x10-3
1.00x10-3
1.00x10-3
1.76 x10-3
3
2.00x10-3
2.00x10-3
1.00x10-3
1.76 x10-3
4
4.00 x10-3
1.00x10 -3
2.00x10-3
7.04 x10-3
Explain why the order of a reaction cannot be obtained directly from the stoichiometric
equation.
[1]
Topic 6 Kinetics
6.3
Reaction Mechanism and Activation Energy
HL ONLY
1.The reaction between NO2 and CO to give NO and CO2 is thought to occur by the following mechanism:
NO2 + NO2 → NO + NO3
slow
NO3 + CO → NO2 + CO2
fast
What is the rate equation?
A.
Rate = k[NO2][CO]
B.
Rate = k[NO3][CO]
C
Rate = k[NO2]2[CO]
D.
Rate = k[NO2]2
2.
Consider the following statements.
I.
The rate constant of a reaction increases with increase in temperature.
II.
Increase in temperature decreases the activation energy of the reaction.
III
The term A in the Arrhenius equation (k = AeRT) relates to the energy requirements of the
collisions.
-Ba
Which statement(s) is/are correct?
A.
I only
B.
II only
C.
I and III only
D.
II and III only
3. Which statement is correct for the reaction below?
4P + Q→ 2R+2S
A.
The rate of formation of R is one half the rate of the disappearance of Q.
B.
The rate of disappearance of Q is one quarter of the rate of disappearance of P.
C.
The rates of formation of R and S are not equal.
D.
The rate of formation of S is double the rate of disappearance of P.
4. Nitrogen (II) oxide reacts with hydrogen as shown by the following equation.
2NO(g)+2H2(g) → N2(g)+2H2O(g) The table below
shows how the rate of reaction varies as the reactant concentrations vary.
Experiment
1
2
3
4
Initial [NO] /
mol dm-3
0.100
0.100
0.200
0.300
Initial [H2] /
mol dm-3
0.100
0.200
0.100
0.100
Initial rate /
molN2 dm-3s-1
2.53x10-6
5.05x10-6
10.10x10-6
22.80x10-6
[3]
(a) Determine the order of reaction with respect to NO and with respect to H2. Explain how you
determined the order for NO.
NO
Rate = [NO]x
4 = [2]x
x=2
Rate increase by X 4 when [NO] doubles
Rate = [H2]y
H2
2 = [2]y
Y=1
(b) Write the rate expression for the reaction.
[1]
Rate = [NO]2[H2]
[2]
(c) Calculate the value for the rate constant, including its units.
2.53 x 10-6 = K[0.1]2 [0.1]
K = 2.53 x 10-3 mol-2dm-6s-1
(d) A suggested mechanism for this reaction is as follows.
H 2 + NO ↔X
fast step
X + NO → Y+H2O slow step
Y+H2→ N2 + H2O
fast step
State and explain whether this mechanism agrees with the experimental rate expression in (b).
 Yes agrees
 First step involves H2 and NO
 Second step involves NO again
 [NO]2[H2]1
(last step not important as after the rate determining step)
[4]
(e) Explain why a single step mechanism is unlikely for a reaction of this kind.


Reaction would require collision of more than 2 molecules simultaneously
Statistically unlikely
[4]
5 (a) The rate constant, k, for any reaction is related to the activation energy, E, by the Arrhenius
equation.
(b)
(i)
Explain the meaning of the terms; rate constant and activation energy.
[4]
(ii)
A particular reaction is found to be second order overall. Specify the units for both the rate
of reaction and the rate constant
[2]
Rates of reactions depend on temperature.
(i)
Using the same sets of axes, draw two clearly labelled curves to show the distribution of
energies of the molecules of a fixed mass of gas at different temperatures, T1 and T2 where
T2 is the higher temperature.
[4]
(ii)
With reference to the curves in (i) explain why an increase in temperature increases the
rate of a gaseous reaction.
[2]
(iii) Explain briefly how a catalyst can increase the rate of a chemical reaction.
(c)
[2]
The rate constant, k, was determined for the decomposition of hydrogen iodide at various
temperatures. The results giving ln k for a range of temperatures are given below.
Temperature T/K
550
600
650
700
750
ln k:
-15.6
-12.2
-9.4
-7
-4.9
T-1/K-1
1.82xl0-3
1.67 xl0-3
1.54 xl0-3
1.43 xl0-3
1.33xl0-3
Plot a graph of ln k: against T -1.
(Take the lnk axis from -20 to 0 and the T-1 axis from 1.2xl0-3to 1.9xl0-3 K-1)
(i)
(ii)
(4)
Calculate the gradient (slope) of your graph and use it to determine a value for the
activation energy, Ea stating its units.
(5)
(iii) Without obtaining the actual value, state the A in the Arrhenius equation could be
determined.
(2)




Rate constant = constant number from the rate expression
= which describes how temperature, activation energy and geometry affect it
Activation energy = minimum energy
= of collisions which cause reaction
(ii) 2nd order K = mol-1dm3s-1
rate = moldm-3s-1
no
molecules
T1
(4marks)
T2
EA
ii)
Increase proportion of molecules that exceed EA
iii)
Alternative pathway
Lower Ea
ci) 
ln K
(2 marks plot)
1/T

Gradient = -7.3/3.4 x 10-4
=

-2.15 x 104 K
-Ea/R = gradient
Ea = 8.314 X 2.15 x 104
179 kJmol-1
iii)
iv)
(1 mark for answer and 1 mark for units)
extrapolate to intercept =ln A
Insert/substitute values into Arrhenius equation
6. Ammonia is manufactured by the Haber Process, for which the reaction is shown below.
N2(g)+3H2(g) → 2NH3(g) ΔH = -92kJ
A catalyst of iron is used.
(i)
Sketch an enthalpy level diagram showing the enthalpy change and the activation energy of
the reaction.
(3)
(ii)
State and explain the effect of the catalyst on the activation energy of the reaction.
Reactants
Ea
Enthalpy
H
Products
Reaction pathway
ii) catalysts reduce EA
by using an alternative pathhway
(2)