Kinetics
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Shortcut to Questions
Q: 1 2 3
4 5
6 7 8

1

(a)
What is the meaning of E_{A} on the diagram?
(b)
Indicate on the diagram the effect of the presence of a catalyst.


2 
Given below is a potential energy diagram for the reaction
CO(g) + NO_{2}(g)
→ CO_{2}(g) + NO(g)
(a)
What is the activation energy for the reaction?
(b) What is the activation energy for
the reverse reaction?
(c) What is the value of ΔH
for the reaction?
(d) Explain briefly the effect of an
increase in temperature on the rate of the forward reaction.


3 
For the reaction
2A + B + C → 3D + E
the rate of formation of E is found to double if [A] is doubled,
keeping [B] and [C] constant, and is found to double if [B] is doubled
keeping [A] and [C] constant. The rate is unaffected by changes
in [C]. Write the rate law for the reaction.


4 
For the reaction 2NO(g) + O_{2}(g) → 2NO_{2}(g) state the relationship between
the rate of consumption of NO, the rate of consumption of O_{2},
and the rate of production of NO_{2}.


5 
The reaction between nitrite ion and iodide ion in acidic solution
can be represented by the equation:
2NO_{2}^{}
+ 2I^{} + 4H^{+} → I_{2}
+ 2NO + 2H_{2}O
The experimental rate
law for this reaction is
R = k[NO_{2}^{}][I^{}][H^{+}]^{2}
By what factor, if any, would the rate of reaction be altered if
(a)
[H^{+}] and [I^{}] were kept constant but [NO_{2}^{}]
were doubled?
(b) [I^{}] and [NO_{2}^{}]
were kept constant but [H^{+}] were doubled?
(c) [I^{}] and [NO_{2}^{}]
were kept constant but [H^{+}] were halved?
(d) All concentrations were doubled?


6 
The rate of the reaction H_{3}O^{+} + NH_{3} → NH_{4}^{+}
+ H_{2}O was studied in 1960 and the rate expression
shown to be 
R
= k[H_{3}O^{+}][NH_{3}]
where
k = 4.3 x 10^{10} litre mole^{1} second^{1} at 298 K. Calculate the rate
of this acidbase reaction the instant 0.10 M ammonia (400 cm^{3})
is mixed with 0.050 M nitric acid (200 cm^{3}).


7 
Benzene diazonium chloride is found
to undergo hydrolysis as follows:
C_{6}H_{5}N_{2}Cl + 2H_{2}O → C_{6}H_{5}OH + H_{3}O^{+} + N_{2} + Cl^{}
At
298 K, the initial rate of hydrolysis was found to increase with
the initial concentration of benzene diazonium chloride as follows:
Initial concentration (mole litre^{1}) 
0.0200 
0.0400 
Initial rate (mole litre^{1} second^{1}) 
8.70
x 10^{7} 
1.74 x 10^{6} 
(a)
Use these results to determine the value of the reaction order,
n, in the rate equation:
Rate of formation of phenol = rate constant x [C_{6}H_{5}N_{2}Cl]^{n}
(b)
Hence calculate the rate constant, k, in the rate equation above.
(c)
The rate of the reaction can be expressed as the derivative
d[C_{6}H_{5}OH]
/ dt
Choose
an alternative derivative expression for the rate, and give the
relationship between the two differentials.


8 
Dinitrogen
pentoxide decomposes in the solvent carbon tetrachloride according
to the following stoichiometric equation
2N_{2}O_{5} → 4NO_{2} + O_{2}
The
following data concerns this decomposition at 303 K.
Reactant concentration (M) 
Decomposition rate: d
[N_{2}O_{5}] / dt (mole
litre^{1} hour^{1})^{} 
0.170 
0.050 
0.340 
0.10 
0.680 
0.20 
(a)
Write the rate equation for the decomposition.
(b)
Give the reaction order for the decomposition.
(c)
Calculate the rate constant for the decomposition reaction at 303 K.



Answers
to Kinetics Questions 

1

(a)
E_{A} is
the activation energy for the forward reaction. This is the minimum
energy that the colliding reactant molecules must possess before their
collisions can lead to the formation of product molecules.
(b)
The effect of the catalyst is to lower the height of the "hump".
By offering a new route the presence of the catalyst lowers E_{A}
(for both forward and backward reactions).


2 
(a)
From the graph, E_{A} for the forward reaction = (360 
226)
= 134 kilojoule mole^{1}
(b) From the graph, E_{A} for
the reverse reaction = (360  0)
= 360 kilojoule mole^{1}
(c) The enthalpy change, ΔH
= (0  226) = 226 kilojoule mole^{1}
(d)
At higher temperatures the proportion of molecules with the necessary
activation energy is increased. Rate of forward (and backward) reaction
is thus increased.


3 
Note that unlike equilibrium
constants, a rate law expression cannot be deduced from the stoichiometric
equation. Instead it requires a series of separate experiments to
deduce in turn the effect that the concentration of each reactant
has on the overall rate of the reaction. In a series of experiments,
only one reactant's concentration is altered in each experiment
while the concentrations of all the other reactants are held unchanged.
By these means, the effect of changing the concentration of each
reactant in turn on the rate of the overall reaction can be determined.
In the first experiment, doubling the concentration of A (written
as [A]) alone doubles the rate of formation of E, so the reaction
rate is directly proportional to [A].
i.e. rate = constant x [A]^{1}.
In the second experiment, keeping [A] and [C] fixed while doubling
[B] alone causes the rate of formation of E to double, so the reaction
rate is directly proportional to [B] as well.
i.e. rate= constant x [B]^{1}.
In the third experiment to establish the effect of changing the
concentration of C alone, the rate of the reaction is not altered
so [C] does not appear in the rate law for the reaction,
or rate = constant x [C]^{0}.
Combining these results,
rate = k[A]^{1}[B]^{1}[C]^{0} , or
rate = k[A][B]


4

In this question, the
rate law is not requested and in fact the data needed to deduce
it is not given. Here the simple stoichiometric relationship between
the various reactants and products is all that is required to establish
their relative rates of consumption or formation. In general terms,
suppose that 1 molecule of a reactant A uses 2 molecules of another
reactant B to form 1 molecule of product C, then the rate of consumption
of B is twice that of A and that the rate of formation of C is identical
to the rate of consumption of A.
In this question, 2 molecules of NO reacts with 1 molecule of O_{2}
so the rate of consumption of NO must be twice that of O_{2}
while each molecule of NO forms 1 molecule of NO_{2}, so
the rate of consumption of NO is the same as the rate of formation
of NO_{2}.
By convention, the consumption of a reactant is shown by attaching
a negative sign to its rate while the formation of a product is
indicated by a + sign associated with its rate.
In words, the rate of consumption of NO equals the rate of formation
of NO_{2}, and the rate of disappearance of O_{2}
is onehalf this rate.
Or, using the calculus notation of d[A]/dt for the rate of change
of concentration of any species A,
2 x d[NO]/dt = d[O_{2}]/dt = + 2 x d[NO_{2}]/dt


5 
(a)
If only the [NO_{2}^{}] is altered and is doubled,
then from the rate law, this would cause the rate to double as rate
∝
[NO_{2}^{}]^{1}.
(b)
If only the [H^{+}] is altered and is doubled, then from
the rate law, this would cause the rate to increase by a factor
of 4 as rate ∝ [H^{+}]^{2}.
(c)
If only the [H^{+}] is altered and is halved, then from
the rate law, this would cause the rate to decrease to ¼ of the
original rate as rate ∝ [H^{+}]^{2}, i.e. ( ½. )^{2}
= ¼.
(d) If all concentrations were doubled, then the doubling
of [NO_{2}^{}] would double the rate and doubling
the [I^{}] would double the rate again (now increased by
a factor of 4) and doubling the [H^{+}] would increase the
rate by a further factor of 4 to give a total increase in rate of
16 times the original rate.


6 
The rate law is
rate = 4.3 x 10^{10} [H_{3}O^{+}][NH_{3}].
Calculate the instantaneous
[H_{3}O^{+}] and [NH_{3}] which have been
diluted upon mixing as follows:
Total volume.=
600 mL.
For dilutions,
the expression M_{1}V_{1} = M_{2}V_{2}
is valid.
∴
for the diluted [H_{3}O^{+}],
0.050 x 200 = M_{2} x 600
so [H_{3}O^{+}]
= 1.67 x 10^{2} M
and for the diluted
[NH_{3}], 0.10 x 400 = M_{2} x 600
so [NH_{3}]
= 6.66 x 10^{2} M
Substituting these
values for [H_{3}O^{+}] and [NH_{3}] in
the rate law
Rate = 4.3 x 10^{10}
[H_{3}O^{+}][NH_{3}]
= 4.3 x 10^{10} (1.67 x 10^{2})(6.66 x 10^{2})
= 4.8 moles litre^{1} second^{1} (4.8 M s^{1})


7 
(a) By inspection of the data, doubling
the [C_{6}H_{5}N_{2}Cl] causes the rate to
increase from 8.70 x 10^{7} M s^{1} to 1.74 x 10^{6}
M s^{1}, i.e. to double.
∴
rate is directly proportional to the [C_{6}H_{5}N_{2}Cl],
so n = 1.
(b) Substitute either set of data in the rate equation.
8.70 x 10^{7} = k x 0.0200
∴ k = 4.35 x 10^{5}
s^{1}.
(c) See answer to Question
4 above for a discussion of the background to this part.
From
the stoichiometry of the equation, each mole of C_{6}H_{5}OH
produced corresponds to the consumption of one mole of C_{6}H_{5}N_{2}Cl.
Hence an alternative differential expression for the rate would
be
d
[C_{6}H_{5}N_{2}Cl] /dt (The  sign indicates that species is being consumed)
The
relation between them is
d [C_{6}H_{5}OH] /dt = d [C_{6}H_{5}N_{2}Cl]
/dt


8 
(a)
By inspection of the data, doubling the concentration
of N_{2}O_{5} causes the rate to double, so the rate
law is rate = k[N_{2}O_{5}]^{1}.
(b) As n = 1, reaction order = 1.
(c) To evaluate k, substitute any of the sets of
data.
0.10
= k x 0.340
∴ k = 2.9 x 10^{1} hr^{1}.

