SHE 1315
REACTION KINETICS
TOPIC 8: REACTION KINETIC
1.
Define the rate of reaction.
(8a)
The reaction rate is the change in the concentration of reactants or products per
unit time
2.
Explain the difference between average rate, instantaneous rate and initial rate.
(8c)
Average rate is the total change in concentration over a given period of time.
Instantaneous rate is the rate at particular instant during the reaction.
Initial rate is the instantaneous rate at the moment the reactants are mixed.
3.
Write the rate expression in terms of changes in concentration for the following reaction.
(8d)
a) 2 X + 3 Y  6 Z
Rate = - ½ ∆[X]/∆t = -1/3 ∆[Y]/∆t = 1/6 ∆[ Z]/∆t
b) E + ½ F  7 G + ¾ H
Rate = -∆[E]/∆t = -2 ∆[F]/∆t = 1/7 ∆[G]/∆t = 4/3 ∆[ H]/∆t
c) 2/3 A + 5 B  1/6 C + D
Rate = -3/2 ∆[A]/∆t = -1/5 ∆[B]/∆t = 6 ∆[C]/∆t = ∆[D]/∆t
4.
Consider the following reaction,
2 N2O5 (g)
a)
(8d)
4NO2 (g) + O2(g)
Write the rate expression in terms of changes in concentration.
Rate = - ½ ∆[ N2O5]/∆t = 1/4 ∆[ NO2]/∆t = ∆[ O2]/∆t
b)
If the rate of disappearance of N2O5 gas is 1.0 x 102 mol dm-3s-1 , determine the rate
of formation of NO2 under the same condition.
- ½ ∆[ N2O5]/∆t = 1/4 ∆[ NO2]/∆t
∆[ NO2]/∆t =- 4/2∆[ N2O5]/∆t
= 2.0 x 102 mol dm-3 s-1
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5.
REACTION KINETICS
Write the chemical equation for the rate expression below.
Rate = - ∆[ CH4]/∆t
= -½ ∆[ O2]/∆t
CH4 + 2 O2
6.
(8d)
= ½ ∆ [ H2O]/∆t = ∆[CO2]/∆t
2 H2O + CO2
The rate of formation of ammonia shown below was measured as 10.0 mmol dm−3 s−1.
N2 (g) + 3 H2 (g) → 2 NH3 (g)
Calculate the rate of disappearance of hydrogen.
(8d)
Rate = - ∆[ N2]/∆t = -1/3 ∆[ H2]/∆t = 1/2∆[ NH3]/∆t
∆[ H2]/∆t= 3/2∆[ NH3]/∆t
=3/2 x 10 mmol
=15 mmol dm−3 s−1
7.
2 NO (g) + 2 H2 (g)
2 H2O (g) + N2 (g)
(8e)
The reaction above is second order with respect to NO and first order with respect to H2.
a) Write the rate law and determine the overall order.
Rate = k[NO]2[H2]
Overall order = 3
b) Predict the reaction rate changes if:
i)
the NO concentration doubles
Rate increases by a factor of 4
ii)
the NO concentration triples
Rate increases by a factor of 9
iii)
the H2 concentration doubles
Rate increases by a factor of 2
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REACTION KINETICS
iv)
8.
both reactant concentrations are doubles
Rate increases by a factor of 8
Data for the reaction 2 NO(g) + O2 (g)  2 NO2 (g) are given in the table.
(8e, 8f, 8g)
Experiment
[NO] (M)
[O2] (M)
Initial Rate (M/s)
1
3.60 x 10-4
5.20 x 10-3
3.4 x 10-8
2
3.60 x 10-4
1.04 x 10-2
6.8 x 10-8
3
1.80 x 10-4
1.04 x 10-2
1.7 x 10-8
4
1.80 x 10-4
5.2 x 10-3
x
a) Write the rate law for the reaction.
Rate 2
Rate 3
= k[NO]2m [O2]2n
k[NO]3m [O2]3n
6.8 x 10-8 = (3.60 x 10-4 )m (1.04 x 10-2)n
1.7 x 10-8
(1.8 x 10-4)m (1.04 x 10-2)n
m
=2
Rate 2
Rate 1
= [NO]2m [O2]2n
[NO]1m [O2]1n
6.8 x 10-8 = (3.60 x 10-4 )m (1.04 x 10-2)n
3.4 x 10-8
(3.60 x 10-4)m (5.20 x 10-3)n
n
=1
Rate
= k [NO]2 [O2]1
b) Calculate the rate constant of the reaction.
k=
Rate
[NO]2[O2]1
=
3.4 x 10-8
(3.6 x 10-4)2(5.2 x 10-4)
c) Determine the initial rate for experiment 4.
Rate
= k [NO]2 [O2]1
= 50 M-2s-1
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REACTION KINETICS
= 8.4 x 10-9 Ms-1
9. The rate of reaction between CO and NO2 was studied at 540 K starting with various
concentrations of CO and NO2. The data was collected as shown below.
(8e, 8f, 8g)
Experiment
[CO] (M)
[NO2] (M)
Initial Rate (M/s)
1
5.10 x 10-4
0.350 x 10-4
3.4 x 10-8
2
5.10 x 10-4
0.700 x 10-4
6.8 x 10-8
3
5.10 x 10-4
0.175 x 10-4
1.7 x 10-8
4
1.02 x 10-3
0.350 x 10-4
6.8 x 10-8
5
1.53 x 10-4
0.350 x 10-4
10.2 x 10-8
a)
Predict (without calculation) the reaction order with respect to CO and NO2.
[CO] constant [NO2] doubled, rate doubled so rate order of NO 2 is 1st order
[CO] doubled [NO2] constant, rate doubled so rate order of CO is 1st order
b)
Calculate the value of k.
k=
Rate
[CO]1[NO2]1
=
3.4 x 10-8
(5.10 x 10-4)(0.350 x 10-4)
= 1.9 M-1s-1
c)
Calculate the rate of reaction if the concentration of CO is 2.1 x 10 -2 M and the
concentration of NO2 is 6.25 x 10-3M.
Rate = k[CO]1 [NO2]1
=(1.9 M-1s-1)(2.1 x 10-2)(6.25 x 10-3 )
=2.49 x 10-4 Ms-1
d)
How does the reaction rate change if the temperature is changed to 480 K, while the
concentration remains constant?
The rate of reaction decreases.
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REACTION KINETICS
10. Fill in the table below.
(8h)
Conc vs time graph
Reaction Order
Integrated
Rate Law
Unit for Rate
constant (k)
ln [A]0 = kt
[A]t
First order
s-1
ln[A]
time
Zero order
[A]t - [A]0= -kt
mol L-1s-1
[A]
time
1 – 1 = kt
[A]t [A]0
Lmol-1s-1
1/[A]
Second order
time
11. The rate law for the hydrolysis of sucrose, C12H22O11 to produce C6H12O6 is Rate = k
[sucrose]. After 2.57 hour the sucrose concentration decreases from 0.0146 molL-1 to
0.0132 molL-1.
(8h, 8i)
a) Find the rate constant.
First order reaction.
ln [A]0 – ln [A]t = kt
k = ln 0.0146 – ln 0.0132
2.57
= 0.0392 hr-1
b) Determine the half-life of the reaction.
t1/2 = ln 2
k
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12.
REACTION KINETICS
= 0.693
0.0392 hr-1
= 17.68 hr
The rate constant for the decay of
a)
What is the half-life for
Rn is 0.1810 day-1.
222
(8h, 8i)
Rn expressed in days?
222
(since the unit is day-1, so, this is the first order reaction)
t1/2 = ln 2/k
= 3.83 days
b)
Calculate the fraction of a sample of 222Rn decays in a period of exactly one week.
[Assume initial concentration is 100]
ln[A]t = ln[A]0 – kt
ln[A] t = ln[100] – (0.1810 day-1)(7days)
[A] t = 28.1675
Fraction : 28.17/100
13.
The rate constant for the reaction below is 30.0 Lmol -1min-1.
2 HI (g)
(8h)
H2 (g) +2 I2 (g)
a)
State the overall order of the reaction.
Second order
b)
Calculate the time taken for the concentration of HI to drop from 0.010 M to 0.005 M.
1– 1
= kt
[A]t [A]0
1
- 1
= (30.0) t
0.005
0.010
t
= 3.3 min
14. The rate constant for the decomposition of NO2 is 3.40 M-1min-1. Determine the time needed for
the concentration of NO2 to decrease from 2 M to 1.5 M.
(8h)
Second order reaction
1– 1
= kt
[A]t [A]0
1
1.5
-
1 = (3.4) t
2.0
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REACTION KINETICS
t
15.
= 0.049min
The table below represent a reaction with the rate law of Rate = k.
Time (min)
[A]
a)
(M)
0
1
2
3
1.0
0.79
0.58
0.38
(8h)
State the overall order of the reaction.
Zero order reaction.
b)
Calculate the rate constant.
[A]t - [A]0= -kt
k = 1.0 – 0.79
1
k = 0.21 M min-1
16.
Concentration versus time data for the thermal decomposition of ethyl bromide at 700K is
given below.
(8h)
C2H5Br(g)
Time (min)
0
1
2
3
4
C2H4 (g) + HBr (g)
[C2H5Br]
1.00 M
0.82 M
0.67 M
0.55 M
0.45 M
ln[C2H5Br]
0
-0.20
-0.40
-0.60
-0.80
1/[C2H5Br]
1
1.22
1.49
1.82
2.22
a)
Complete the table by filling in the ln [C2H5Br] and 1/[C2H5Br] values.
b)
Determine the order of the reaction by plotting graph.
time
[C2H5Br]
ln
[C2H5Br]
time
1
[C2H5Br]
time
time
time
time
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REACTION KINETICS
The linear graph is observed from the second graph prove that the reaction is 1st order
reaction.
c)
Calculate the rate constant from the graph.
Slope = -k
-k = -0.4 – (-0.6)
2-3
= -0.2
k = 0.2 min-1
d)
17.
Calculate the half life of ethyl bromide in this reaction.
t1/2 = ln 2
k
k = ln 2
0.2
= 3.47 min
Define reaction mechanism and molecularity.
(8o)
Reaction mechanism is a sequence of single reaction steps that sum to overall
chemical reaction.
Molecularity is the number of reactant particles in the step.
18.
Consider the energy profile diagram below.
a)
Write the mechanism and overall equation for the reaction.
Step 1
Step 2
Step 3
H2+I2
H2 + I + I
H2I + I
Overall
H2+I2
H2 + I + I
H2I + I
HI +HI
2HI
(8o, 8p, 8q, 8r)
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REACTION KINETICS
b)
State the molecularity for each elementary step.
c)
Step 1
Unimolecular
Step 2
Bimolecular
Step 3
Bimolecular
Define intermediate.
Substances that is formed and used up during the overall reaction.
d)
Identify intermediate(s).
I and H2I
e)
Write the rate law for the reaction.
Rate = k[H2][I2]
19.
The rate law for the endothermic reaction below is Rate = k[H2][N],
2 H2(g) + 2 NO(g)
(8o, 8p, 8q, 8r)
2 H2O(g) + N2 (g)
The following equations represent a proposed mechanism for the reaction.
Step 1:
Step 2:
Step 3:
a)
H2(g) + NO(g)
N(g) + NO(g)
O(g) + H2(g)
H2O(g) + N(g)
N2(g) + O(g)
H2O(g)
Ea = 1690 kJ
Ea = 625 kJ
Ea = 436 kJ
Define elementary step.
The individual steps that make up a reaction mechanism.
b)
State the rate determining step
Step 1
c)
Determine the validity of the above mechanism valid and state the reason.
Not valid because the given rate law is not tally with the rate law from RDS
in the proposed mechanism above.
20.
State the factors that influence reaction rate.
Concentration, temperature, catalyst and physical state.
21.
Based on Arrhenius equation, state how the rate of reaction can be increased by:
(8b)
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REACTION KINETICS
(8k)
a)
increasing the temperature
The higher the temperature,the larger the value of k and leads to the higher
reaction rate.
b)
adding a catalyst
Catalyst causes lower activation energy, the larger the value of k and leads
to the higher reaction rate.
22.
Define catalyst and list all types of catalyst.
(8m)
Catalyst is substances that speed up the reaction without being consumed. Two
type of catalyst are homogenous and heterogeneous catalyst
23.
Explain how the rate of reaction can be affected by:
a)
adding a catalyst. (Add a suitable diagram)
(8n)
i)
Catalyst effects by providing an alternative pathway with lower Ea.
The fraction of reactant molecules with energy same or greater than
Ea increases leads to increasing of reaction rates.
ii)
Catalyst effects by providing an alternative pathway with lower Ea.
Rate constant increases leads to increasing of reaction rates.
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b)
REACTION KINETICS
increasing the reactant concentration.
(8j)
As the concentration increase the number of reactant particles increases.
Frequency of effective collision increase leads to increasing of reaction
rates.
c)
increasing the temperature. (Add a suitable diagram)
iii)
iv)
(8l)
As the temperature increases, the kinetic energy of the reactant
molecules increases, the frequency of collision increases. The
frequency of effective collision increases leads to increasing of
reaction rates.
As the temperature increases, The fraction of reactant molecules
with energy same or greater than Ea increases leads to increasing of
reaction rates.