Reaction Kinetics Written Response
August 1999
January 2000
April 2000
June 2000
August 2000
January 2001
April 2001
June 2001
1. The mass of a burning candle is monitored to determine the rate of combustion of
paraffin. An accepted reaction for the combustion of paraffin is:
2C28H58(s) + 85O2(g) → 56CO2(g) + 58H2O(g)
The following data is observed:
Time
(min)
0.0
6.0
12.0
18.0
24.0
30.0
Mass of Candle
(g)
25.6
25.1
24.5
23.9
23.4
22.8
a) Calculate the average rate of consumption of paraffin in g/min for the time interval
12.0 to 24.0 minutes.
b) Calculate the rate of CO2 production in mol/min for the time interval
12.0 to 24.0 minutes.
August 2001
1. Consider the following proposed reaction mechanism:
Step 1
Step 2
Step 3
Step 4
Fe3+ + H2O2 → FeH2O23+
FeH2O23+ → FeOH3+ + HO
HO + H2O2 → H2O + HO2
FeOH3+ + HO2 → Fe3+ + H2O + O2
a) Write the overall reaction
b) Define the term catalyst and identify a catalyst in the above mechanism.
January 2002
1. Consider the following reaction:
C12H22O11(s) → 11H2O(g) + 12C(s)
The rate of decomposition of C12H22O11 is 0.75mol/min.
What mass of C is produced in 10.0 seconds?
2. Define the term activation energy.
April 2002
1. Consider the reaction:
C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)
At certain conditions, 0.15 mol CO2 is produced in 2.0 minutes.
What is the rate of consumption of C2H4 in g/s?
2. Define the term reaction mechanism.
June 2002
1. Consider the reaction:
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
A 10.0g sample of Al reacts completely in excess HCl in 300.0s.
What is the rate of production of H2 in mol/s ?
2. Using collision theory, give two reactions why reactions occur more rapidly at a higher
temperature.
August 2002
1. Consider the reactions:
2H2O(l) → 2H2(g) + O2(g)
The rate of production of O2 is 1.2 × 10-2 mol/s. How many seconds will it take to
compose 100.0g H2O?
2. Define the term catalyst.
January 2003
1. Using the axes below, sketch a PE diagram for the reacting system where:
2. Consider the following reaction mechanism:
Step 1
2NO → N2O2
Step 2
N2O2 + H2 → N2O + H2O
Step 3
N2O + H2 → N2 + H2O
a) Determine the overall reaction.
Overall Reaction: _________________________________________________
b) Identify a reaction intermediate.
________________________________________________________________
April 2003
1. Consider the following reaction:
3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
A piece of copper is added to a nitric acid solution in an open beaker, allowing the
NO(g) to escape. The following data was obtained:
Time
Mass of Beaker
(min)
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
and Contents (g)
200.00
197.50
195.45
193.55
191.70
189.90
188.15
186.45
184.80
a) Calculate the reaction rate for the time period 2.0 to 6.0 min.
b) Calculate the mass of copper consumed in the first 5 minutes.
2. Using collision theory, explain why reactions between two solutions occur more
rapidly than reactions between two solids.
June 2003
1. Consider the following reaction in an open flask:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
A 155.0g sample of CaCO3(s) is placed in the flask and HCl(aq) is added.
The reaction consumed HCl(aq) at an average rate of 0.200 mol/min for 10.0 min.
What mass of CaCO3(s) remains?
2. a) Write the equation for Step 3 in the following reaction mechanism.
Step 1
Step 2
Step 3
Overall Reaction
2NO → N2O2
N2O2 + H2 → N2O + H2O
?
2NO + 2H2 → N2 + 2H2O
b) Identify a reaction intermediate in the above mechanism.
August 2003
1. Consider the following reaction:
Mg(s) + 2HBr(aq) → MgBr2(aq) + H2(g) + energy
In terms of collision theory, describe how each of the factors below would influence
the reaction rate.
a) Increasing the concentration of HBr : _____________________________________
_____________________________________________________________________
b) Decreasing the temperature : ____________________________________________
_____________________________________________________________________
c) Increasing the surface area of Mg : ________________________________________
_____________________________________________________________________
2. Consider the following reaction mechanism:
Step 1
NO(g) + O2(g) → NO3(g) slow
Step 2
NO3(g) + NO(g) → 2NO2(g)
The overall reaction is exothermic. Sketch a PE diagram on the axes below to
describe the energy changes that occur as the reaction takes place.
January 2004
1. The release of O2(g) resulting from the decomposition of bleach was measured in two
different experiments. Data was collected and the following graph was drawn:
a) Calculate the average rate of reaction for each experiment.
b) Identify a variable from Experiment 1 and how it was changed to produce the
different reaction rate for Experiment 2. Explain using collision theory.
Written Response Answers
August 1999
January 2000
Question 1
Question 2
April 2000
Question 1
Question 2
June 2000
August 2000
January 2001
April 2001
June 2001
1. a)
b)
August 2001
1. a) 2H2O2 → 2H2O + O2
b) A species which speeds up a reaction by providing a lower energy pathway.
Catalyst : Fe3+
January 2002
1.
2. Activation energy is the minimum amount of energy required to form the activated
complex from the reactants.
April 2002
1.
2. A reaction mechanism is a series of steps that result in the overall reaction.
June 2002
1.
2. There is a greater fraction of collisions with sufficient energy.
There are more frequent collisions.
August 2002
1.
2. A catalyst is a substance that increases the rate of a chemical reaction and may be
recovered at the end of the reaction.
January 2003
1.
2. a) Overall reaction : 2NO + 2H2 → N2 + 2H2O
b) N2O2 OR N2O
April 2003
1. a)
b)
2. Particles must be able to collide to react.
Only the particles on the surface of a solid are available for reaction.
In a solution, all particles are available.
June 2003
1.
2. a) Step 3 : N2O + H2 → N2 + H2O
b) Either N2O2 OR N2O
August 2003
1. a) Greater concentration and more collisions. Therefore more successful collisions and
a greater rate.
b) Fewer collisions with sufficient energy to overcome PE barrier. Therefore, a lower
rate.
c) Increased surface area leads to more collisions and more successful collisions.
therefore, a higher rate.
2.
January 2004
1. a)
b)
Variable/Change
Explanation
Temperature is decreased.
Lower fraction of effective
collisions.
OR
Concentration of reactants
was decreased.
Fewer collisions.