Chemistry

Project description

The rates of reactions;
Chapter 19
student questions
–
Chemical thermodynamics
Chapter 20
student questions
–
Chemical Kinetics
Question 1 (10 marks)
By
writing the units of concentration as ‘conc’ and the units of time as ‘time’, determine the
units of the rate constants in the following rate laws
rate =
k
1 [N
2
O
5
]
rate =
k
2 [PhCH
2
Br] [DABCO]
rate =
k
3 [RH] [Cl2]
1/2
rate =
k
4 [CH

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3
CHO]
3/2
rate =
k
5
]
rate =
k
6
Question 2 (
15
marks)
Methanoic acid is oxidized in acid solution by Br
2
according to the following stoichiometric
Equation:
HCOOH + Br
2
+ 2H
2
O CO
2
+ 2Br
–
+ 2H
3
O
+
It is found that the rate of reaction depends on the
concentration of HCOOH and Br2, and is also
influenced by the concentration of added H3O
+
. If H3O
+
and HCOOH are in excess, then it is found
that the reaction is first order in [Br2]. The following rate law was therefore proposed:
rate =
k
obs
[Br
2
][HCOOH]
a
[H3O
+
]
b
;
where
a
and
b
are the orders with respect to HCOOH and H
3
O+, respectively.
If H
3
O+ and HCOOH are in excess the rate law can be written:
rate =
k

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1st
[Br
2
];
where
k
1st
is a pseudo first
–
order rate constant.
(a)
Write down an expression for
k
1st
.
(b) The following values of
k
1st
were measured for different excess concentrations of
HCOOH and H
3
O
+
:
[HCOOH] / mol dm
–
3
0.10
0.10
0.12
0.22
[H3O+] / mol dm
–
3
0.05
0.12
0.15
0.15
K
1st
/s
–
1
7.00 × 10
–
3
2.92 × 10
–
3
2.80 × 10
–
3
5.13 × 10
–
3
By comparing the data in the first two columns, determined the value of
b
. Similarly,
determine the
value of a by comparing the data in the third and fourth columns.
(c) Use all of the data in the table to determine an average value of
k
obs
; state the
units of
this
rate constant.
Question 3 (5 marks)
In aqueous solution the oxidation of Fe
II
by Pb
IV
is thought to proceed via the intermediate species
Pb
III
according to the following mechanism:
Fe
II
+ Pb
IV
?
Fe
III
+ Pb
III
Fe
III
+ Pb
III
?
Fe
II
+ Pb
IV
Fe
II
+ Pb
III
?
Fe
III
+ Pb
II
Assuming that Pb
III
can be placed in the steady state, determine the overall rate law for the
rate of change of the concentration of Fe
III
.
Question 4
(10 marks)
4 × 10
–
3
moles of an ideal gas are held inside a cylinder by a piston such that the volume of the gas is
10 cm3; the whole assembly is held in a thermostat at 298 K.
Calculate the pressure of the gas in N m
–
3
.
(a) Assume that the external pressure is fixed at 1
bar. Explain why the piston moves out
when it is released, and why it eventually comes to a stop. What will the pressure of the gas
inside the cylinder be when the piston finally stops?
(b) Calculate the volume of the gas inside the cylinder when the pis
ton has come to rest, and
hence the work for this irreversible expansion.

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(c) State the change in the internal energy,
?
U
, of the gas when it undergoes this isothermal
expansion. Hence, using the First Law, calculate the heat associated with the expansion
,
explaining its sign.
(d) Calculate the work associated with reversible isothermal expansion between the same
initial and final states as the irreversible expansion described above; hence find the heat.
Comment on these values in relation to those for th
e irreversible expansion.
(e) Determine the enthalpy change of the gas in (i) the reversible and (ii) the irreversible
expansion.
Question 5
(10 marks)
For the
A
B equilibrium, sketch a graph showing how the molar Gibbs energy of a mixture
of A and
B
varies as the composition varies from pure reactant A to pure product B for the
following cases:
(a) G°
m,A
= G°
m,B
(b)
G°
m,A
< G°
m,B
On your graph indicate the equilibrium
composition.
Explain why it is that if the proportions of A and B are not at
their equilibrium values,
there is always a spontaneous process which will result in the composition moving to its
equilibrium value, but that once this point is reached, no further change is possible.
k
1
k
–
1
k
2
Question 6 (10 marks)
A sample of methane gas of m
ass 4.50 g has volume 12.7 dm
3
at 310 K. It expands isothermally
against a constant external pressure of 200 Torr until its volume has increased by 3.3 dm
3
. Assuming
methane to be a perfect gas, calculate the work and heat associated with the process, alon
g with the
change in the internal energy and enthalpy of the gas.
Calculate these same quantities if the expansion is carried out reversibly.
Question 7 (
15
marks)
The reaction between propanone and bromine in aqueous acid conditions was studied by
measuring
the absorbance at 400 nm due to Br
2
. The initial concentrations of propanone and acid were both 0.50
mol dm
–
3
, with both reagents being in excess compared to the bromine.
The following data of absorbance at 400 nm as a function of time were obta
ined at 298 K
t
/ s
0
60
120
180
240
300
360
420
abs
0.995
0.964
0.903
0.830
0.772
0.739
0.679
0.605
The path length provided by the cuvette was 1 cm, and the extinction coefficient of Br
2
at
400 nm is 168 mol
–
1
dm
3
cm
–
1
.
(a) Convert the absorbance
data to concentrations, and then plot concentration as a function of
time. You should find that the graph is a straight line, implying that the reaction is zero
–
order
in bromine.
(b) Determine a value for the (pseudo) zero
–
order rate constant, stating its
units.
(c) Assuming that the reaction is first order in both propanone and acid, determine the second
–
order rate constant for the overall reaction, stating its units.
Question 8 (15 marks)
The oxidation of methanoic acid by bromine
HCOOH + Br
2
+ 2H
2
O
?
CO
2
+ 2Br
–
+ 2H
3
O
+
can be studied by setting up the reaction in an electrochemical cell. If it is arranged that in the reaction
mixture the bromide ion is in excess, the
potential (voltage) E generated by the cell is given by
where F is the Faraday constant (96,485 C mol
–
1
) and C is a constant. The consumption of
bromine
can therefore be followed by measuring the potential as a function of time.
The rate law of this reaction is thought to be first order in bromine. An
experiment was
designed to
investigate the order with respect to methanoic acid (
a
) and acid (
b
)
rate =
k
[Br
2
][HCOOH]
a
[H
3
O
+
]
b
The reaction mixture is set up so that HCOOH and H3O+ are in excess, making the rate law
pseudo first order
rate =
k
1st
[Br
2
]
k
1st
=
k
[HCOOH]
a
[H
3
O
+
]
b
The above expression for the cell potential can be rearranged to
(a) For a first
–
order process a plot of ln [Br
2
] against time will have slope
–
k
1st
. Show that
it
therefore follows that a plot of E
against time will have slope (

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k
1st
R
T)
/
(2F).
The following data were obtained at 25
°
C for a reaction mixture with the following
initial
c
oncentrations:
[Br
2
] = 3
.
0
×
10
–
3
mol dm
–
3
, [HCOOH] = 0.10
mol dm
–
3
, [H
3
O
+
]
= 0.05
mol dm
–
3
Time (s)
0
60
120
180
240
300
360
E (V)
–
0.772
–
0.766
–
0.761
–
0.757
–
0.752
–
0.745
–
0.741
(b) Plot these data as described above, and hence obtain a value of the first
–
order rate
constant.
Similar experiments, with different initial (excess) concentrations of methanoic
acid and acid
gave the
following data (the value indicated by the * is obtained from the graph above)
Run
1
2
3
4

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5
6
[HCOOH]
(
mol dm
–
3
)
0.10
0.10
0.10
0.12
0.16
0.22
[H
3
O
+
]
(
mol dm
–
3
)
0.05
0.12
0.25
0.15
0.15
0.15
10
3
×
k
1st
(s
–
1
)
*
2.77
1.33
2.66
3.55
4.88
Runs 1, 2 and 3 have the same (excess) concentration of methanoic acid, but different
(excess)
concentrations of acid. Runs 4, 5 and 6 have the same (excess) concentration of
acid, but different
(excess) concentrations of methanoic acid.
Taki
ng logarithms of the expression for
k
1st
gives
ln
k
1st
= ln
k
+
a
ln [HCOOH] +
b
ln [H
3
O
+
]
(c) For runs 1, 2 and 3 plot ln
k
1st
against ln [H
3
O
+
] and hence obtain
a
value for the order
b
from the slope of the graph.
(d) Similarly, use the data
from runs 4, 5 and 6 to obtain a
value for the order
a
. Comment
on
what your values for these orders imply about the mechanism for this reaction.