UNl( QUESTIONS
248 PHYSICAL CHEMISTRY
PHYSICAL CHEMISTRY
..,,,,....,_ ANALYSIS,
a The diagram below shows the readings on the thermometer before and after
.......... PROBLEM SOLVING
the student dissolved a salt, potassium chloride, in water.
UNIT QUESTIONS
mD
0
c
Hydrogen, H2 , and bromine, Br2 , react vigorously to form hydrogen bromide,
HBr, according to the following equation:
INTERPRETATION
l,s
H,(g) + Br2 (g) --+ 2HBr(g)
a Draw a dot-and-cross diagram t o show the arrangement of the electrons
before adding salt
in a hydrogen bromide molecule, showing electrons in the outer shell
~
00~
mD
PROBLEM SOLVING
mD
INTERPRETATION
"··'
0
c
iro
i,s
after dissolving
salt in water
irl!WiMliA'Ull1
Write down the thermometer readings and calculate the temperature
change.
(3)
b Use the bond energies given below to calculate the enthalpy change for
the reaction shown in the above equation.
(3)
Temperature before
Bond energy/kJ/mol
..••.
ro
c i
H- H
436
Br-Br
196
H-Br
368
mD
Temperature after
"""" oc
"""" oc
Temperature change
........ °C
ii Which of the following statements is true about the dissolving of
CRITICAL THINKING
potassium chloride in water (choose one answer)?
(1 )
A The process is endothermic and t;.H is positive.
B The process is exothermic and t;.H is positive.
Complete the reaction profile diagram by showing the products of the
reaction.
(1)
C The process is endothermic and t;.H is negative.
mD
D The process is exothermic and t;.H is negative.
b Another student repeated the experiment with a different salt, calcium
chloride, using the same method and recorded these results.
PROBLEM SOLVING
Volume of water= 100cm3
Starting temperature of water= 15.9°C
Maximum temperature of solution= 23.2 °C
progress of the reaction
•••
••
ii Label the diagram to show the enthalpy change, t;.H, and the activation
(2)
energy of the reaction.
"··'
(Total 8 marks)
Mass of salt= 5.00g
Calculate the heat energy released in this experiment. The specific
heat capacity of the solution c = 4.2 J/g/°C and the mass of 1 cm3 of
mixture = 1 g.
ii How many moles of calcium chloride dissolved in the solution?
A group of students wanted to investigate the energy changes when salts
dissolve in water. The teacher suggested that they should measure the
temperature changes that occur when the salts were dissolved.
This is the method they followed:
• Add 100 cm3 of water to a beaker.
• Record the temperature of the water.
• Weigh 5.00g of salt and add it to the water in the beaker.
• Stir the mixture with a glass rod vigorously until all the solid has dissolved.
• Record the maximum (or minimum) temperature of the solution.
mD
(3)
(2)
iii Work out the molar enthalpy change, in kJ/mol, for dissolving calcium
(2)
chloride in water.
REASONING
•••
••
"··'
c Another student dissolved magnesium chloride in water. She compared her
result with a data book value.
Student's value = - 105 kJ/mol
Data book value= - 141 kJ/mol
~-
There are no errors in the calculation of her result.
Suggest two reasons why the student's value differs from the data book
~
(Total 13 marks)
250 PHYSICAL CHEMISTRY
mD
PROBLfM SOLVING
PHYSICAL CHEMISTRY
UNl( QUESTIONS
In an experiment to investigate the rate of decomposition of hydrogen
peroxide solution in the presence of manganese(IV) oxide, a student mixed
1Ocm3 of hydrogen peroxide solution with 30 cm3 of water and added 0.20g
of manganese(IV) oxide. She measured the volume of oxygen evolved at 60s
intervals. The results of her experiment are recorded in the table below.
--
REASONING
120
180
240
300
30
48
57
60
60
The equation for this reaction is
They used the following method:
(2)
b Explain why the manganese(IV) oxide was added in a weighing bottle rather
(1)
than directly into the hydrogen peroxide solution.
side-ann
flask
hydrogen
peroxide
~l
Jas syringe
a) before reaction
INTERPRETATION
c Plot a graph of her results and draw a line of best fit. Make sure that you
d Use your graph to find out the following:
How long it took to produce 50cm3 of oxygen.
ii The volume of gas produced after 100 seconds.
(1)
(1)
iii The average rate (with unit) in the first 150 seconds.
(2)
e Explain why the graph becomes horizontal after 240 seconds.
mD
The experiment was then repeated using different concentrations of
hydrochloric acid.
(4)
label the axes.
11!!1'1!111!!11 ANALYSIS,
lliMiliiil" PROBLfM SOLVING
a To ensure it was a valid (fair) test, the students kept the number of marble
chips constant in each experiment. Suggest two other properties of the
marble chips that should be kept the same in each experiment.
CRITICAL THINKING
(2)
A catalyst speeds up a chemical reaction by (choose one answer):
b What is the purpose of the cotton wool?
(1)
•••itlMillilullllll
A increasing the number of collisions between the reactant particles.
B providing an alternative reaction pathway with lower activation energy.
C increasing the energy of the reactant particles.
mD
mD
D changing the enthalpy of the reaction.
OECISION MAKING
Ii Describe a method you could use to show that the manganese(IV) oxide
(4)
is acting as a catalyst in this reaction.
INTERPRETATION
g Suppose the experiment had been repeated using the same quantities of
everything, but with the reaction flask immersed in ice.
Sketch the graph you would expect to get. Use the same grid as in c.
Label the new graph G.
(2)
(2)
(1)
c The teacher gave the students some hydrochloric acid that was labelled
100% . The table below shows the results.
Manganese(IV) oxide acts as a catalyst for this reaction.
CRITICAL THINKING
b) during reaction
The students recorded the time taken for the mass of the flask and contents to
decrease by 0.50g.
solution
mD
A group of students investigated the effect of changing the concentration of
dilute hydrochloric acid on t he rate of its reaction with marble chips (calcium
carbonate).
CaC03(s) + 2HCl(aq) ..., CaCl2 (aq) + H2 0(1) + C02 (g)
60
a Write a balanced equation for the decomposition of hydrogen peroxide.
mD
11!!1'1!111!!11 EXECUTIVE
lliMiliiil" FUNCTION
5.00
25
100
105
5.00
20
80
150
5.00
15
5.00
10
5.00
mD
PROBLfM SOLVING
h On the same grid as in c and g sketch the graph you would expect to get
if you repeated the experiment at the original temperature using 5 cm3 of
hydrogen peroxide solution, 35cm3 of water and 0.20g of manganese(IV)
(2)
oxide. Label this graph H.
60
175
15
40
272
20
20
520
The results of Student 3 are incomplete.
Calculate the volume of water the student should have used for the
result to be comparable with the other four (choose one answer).
(1)
A 5
B 10
C 15
(Total 22 marks)
mD
D 20
INTERPRETATION
ii Plot a graph of the results on a separate piece of graph paper, with
concentration of acid on the x-axis and time on the y-axis.
Draw a line of best fit.
(4)
252 PHYSICAL CHEMISTRY
ll!!'ll'lll!!li ANALYSIS,
~ PROBLEM SOLVING
11D
UNl( QUESTIONS
• ·~
..;'-f
PHYSICAL CHEMISTRY
iii Use your graph in ii to find the time taken for the loss of 0.50 g of mass
(1)
from the flask when the concentration of acid is 70%.
iv One of the points on the graph is anomalous. Identify a reason for this
(choose one answer).
(1)
REASONING
11D
HIW1MM'l@D
c Complete the reaction profile
diagram for this reaction.
INTERPRETATION
A The student started the stopwatch too late.
2N02
B The student stopped the stopwatch before mass loss reached 0.50g.
C The student added too much water at the beginning of the experiment.
D The student spilt some water before adding it into the reaction mixture.
d Another group of students repeated the experiment, but this time they
measured the mass loss after 1 minute.
progress of the reaction
11D
The table below shows the results obtained by the students.
Mass of carbon d1ox1de
given off/g
0.36
0.72
0.88
1.28
1.44
1.65
Concentration of acid
/mol/dm3
0.20
0.40
0.50
0.70
0.80
0.90
Label t he diagram clearly with the activation energy and the t!.H of the
reaction.
(3)
d Predict what would happen to the position of equilibrium and the colour of
the mixture if you:
(3)
i increase the temperature
ii increase the pressure
(3)
REASONING
e Reactions can be accelerated using a catalyst. Write down and explain the
effect of a catalyst on the position of equilibrium in this reaction.
(2)
Describe the relationship between the mass of carbon dioxide given off
in 1 minute and the concentration of the acid.
(2)
ii Explain this relationship in terms of particles.
(Total 14 marks)
Nitrogen and hydrogen are used in the manufacture of ammonia (NH3).
The reaction is reversible and can reach a state of dynamic equilibrium:
(3)
(Total 15 marks)
D
During the manufacture of nitric acid from ammonia, the ammonia is oxidised
to nitrogen monoxide, NO, by oxygen in the air.
N2(9) + 3H2(9)
11D
CRITICAL THINKING
11D
REASONING
(2)
(choose two answers)?
A The rate of the forward reaction is the same as the rate of the reverse
reaction.
B The concentrations of the reactants and the products are always equal.
C The position of equilibrium can be shifted by adding a catalyst.
D The concentrations of the reactants and the products are constant.
4NH3(9) + 502(9) -> 4NO(g) + 6H20(g)
The ammonia is mixed with air and passed through a stack of large circular
gauzes made of platinum-rhodium alloy at red heat (about 900°C). The
platinum-rhodium gauzes act as a catalyst for the reaction.
a Gas particles have to collide before they can react. Use the collision theory
to help you to answer the following questions.
b The graph below shows
how the percentage of NH3
in the equilibrium mixture
varies with temperature and
pressure (in atmospheres
(atm)).
Because the gases are in contact with the catalyst for only a very short
time, it is important that the reaction happens as quickly as possible.
Explain why increasing the temperature to 900°C makes the reaction
(3)
very fast.
ii Explain what will happen to the reaction rate if the pressure is
increased.
~~~
~
b Platinum and rhodium are extremely expensive metals. Explain why the
manufacturer can justify their initial cost.
(2)
(Total 9 marks)
II
At temperatures above 150 °C, brown nitrogen dioxide gas dissociates (splits
up reversibly) into colourless nitrogen monoxide and oxygen:
2NO,(g)
"° 2NO(g) + 0 2(g)
t!.H = + 114 kJ/mol
a Write down the meaning of the symbols "' and t!.H.
(2)
b What does the positive sign of 6H indicate about the reaction?
(1)
100
80
·11
~ 60
~
'if.
40
20
o- - ~- ~ - -~ - ~ - ~ -
(2)
iii Explain why the platinum-rhodium alloy is used as gauzes rather than
"° 2NH3(9)
a What are the characteristics of a reaction at dynamic equilibrium
o
~
.....:,
€ll
.. 8..,
••••
200
400
600
800
pressure/atm
1000
Increasing pressure increases the percentage of NH3 at equilibrium.
Explain why this is the case.
ii Describe the relationship between temperature and the percentage of
NH3 at equilibrium.
iii Using your answer from 11, comment on the sign of enthalpy change
(t!.H) for the forward reaction.
iv Predict and explain what happens to the rate of the reaction when the
temperature is increased.
(3)
(1)
(2)
(3)
(Total 11 marks)
366 ANSWERS
ANSWERS
b To find out whether it speeds the reaction up: You
c Using a catalyst increases the rate of the reaction ,
as it provides an alternative pathway for t he reaction
with lower activation energy. More particles will have
energy greater than or equal to the activation energy so
there will be more successful collisions per unit time.
More products will be made quickly and relatively low
temperature and pressure can be used in the presence
of a catalyst. This saves energy and money. Adding a
catalyst has no effect on the position of equilibrium,
as the rate of the forward reaction and the rate of the
reverse reaction are increased by an equal amount.
could do this most simply by having two test-tubes
w ith equal volumes o f the same hydrogen peroxide
solution side-by-side. Add some iron(III) oxide to one
and look for the faster prod uct ion of bubbles.
To show that it is unchanged: Use a known mass of
iron(III) oxide. When the reaction stops, filter through
previously weighed filter paper, allow to dry, and
re-weigh. Show that the mass of iron(III) oxide is
unchanged. (If it had changed, and you hadn't lost any
during the separation process, it must have reacted in
someway.)
C
!I
~
use of single-headed arrows to represent activation
energy (1) and enthalpy change (1), as shown
2
~
(enthalpy change)
c At a low temperature. Lower temperature favours the
forward exothermic reaction, shifting the position of
equilibrium to the right hand side to produce more
ammonia.
d At low temperatures the reaction is extremely slow
even in the presence of a catalyst. 450 •c is chosen
because it gives a compromise between the rate and
the yield of the reaction.
CHAPTER 21
1
~
a The reversible symbol •.,. ' shows that t he reaction can
go both ways. The reactants can form products and
the p roducts can react to form the reactants.
b White solid decomposes to form colourless gases.
c Ammonium chloride, ammonia and hydrogen chloride
2
~
•
~
xx
No change
a ' Dynamic' means that the reactions are still happening
and the rate of the forward reaction is equal to the
rate of the reverse reaction. ' Equilibrium' means that
the concentrations of the reactants and the products
remain constant.
sharing of 1 pair of electrons between the two atoms (1)
b
b o nd
bond energy / kJ/ mol
H- H
436
Br-Br
196
H - Br
368
increase in pressure shifts the position of equilibrium
to the right because this side has fewer moles of
moles of gas molecules.
c The position of equilibrium shifts to the right.
Bonds made:
4
~
a Less carbon monoxide and hydrogen would be
1 1
calculation of M, (1)
produced. (A high pressure favours the reverse
reaction, as the left hand side of the reaction has
fewer number of moles of gas molecules.)
b A high temperature favours the endothermic forward
reaction and shifts the position of equilibrium to t he
right hand side of the reaction.
correct answer to 2-4 significant figures (1)
06
ill 6H = \ ; = 68.1 kJ/mol to 3 significant figures.
0 4 0
The molar enthalpy change is -68.1 kJ/ mol as the
dissolving is exothermic.
~
100
c heat loss to the surrounding air through beaker (1)
20
correct balancing (1)
d
ii
70
~ 40
0
5
0 30
~ 20
~
a I
:,
10
150
200
250
300
ii
350
a smooth curve of best fit, going through all of the points (1)
2 x H-Br
= - 736 kJ
f
5
= 0.353cm3/sec
1 0
correct numerical answer (1)
correct unit (1)
Increasing the concentration of the acid increases
the rate of reaction. (1) This is because there are
more acid particles within a fixed volume (1) so the
frequency of successful collision between the acid
and the marble chip increases (1 ).
The gas particles have more kinetic energy (1)
so more particles have energy greater t han or
equal to the activation energy (1). There are more
successful collisions per unit time (1).
The rate increases (1) as gas particles are closer
together and there are more frequent collisions (or
more successful collisions per unit time) (1 ).
Ill Reactions happen on the surface of the catalyst (1)
and gauzes have greater surface area (1).
b The use of a catalyst increases the rate of the reaction
d I 125 seconds
II 43cm3
Ill
120
The mass of CO2 g iven off is directly (1)
proportional (1) to the concentration of the acid.
~6 60
i 50
100
100
a smooth curve of best fit, going through all but
one point (1)
b To prevent the loss of oxygen at the b eginning of the
50
80
Iii 150 seconds
iv C
correct formulae of reactants and products (1)
0
60
correct points plotted (2)
3 ~ a 2H2 0 2 (aq) - 2H2 0(1) + 0 2 (g)
g
40
correct labelling of axes with units (1)
some magnesium chloride is left on the weighing boaV
did not d issolve in water completely (1)
= +632 kJ
progress of the reaction
200
concentration of acid/%
total
2HBr
j
0
correct points plotted (2)
a line labelled
with product
name or formula
below the line of
reactants (1)
400
correct answer (must have a negative sign) (1)
correct labelling of axes with units (1)
]~_H
i
0
= +436kJ
H2 + Br2
-
600
500
~ 300
dividing Q by n (1)
=+ 196kJ
activation
energy
B
II
= 0.0450 mol to 3 significant figures.
1 xBr-Br
c I and II
{?
~
c I
time/seconds
Overall change = +632 - 736 = - 104 kJ (exothermic)
A decrease in temperature favours the forward
exothermic reaction.
conical flask. (1)
+; _ • )
55 2
40
0
Bonds broken : 1 x H-H
b The proportion of sulfur trioxide will increase. An
a the mass (1). and the surface area I the size (1).
b To let t he gas escape but keep the liquid inside the
n = mass of calcium chloride/M, = (
=
C
xx
H ~ Br~
correct number of outer shell electrons for both H and
Br(1)
a Right
b Left
c Left
d No change
3
~ a
~
reaction. (1)
END OF UNIT 3 QUESTIONS
1
(half the volume) of gas is produced (1).
4
correct answer for O (1)
ii
two reasons: all the reaction vessels and pipework
have to be built much more strongly, and it takes a
lot more energy to drive the compressors producing
the very high pressures. The extra ammonia produced
isn't worth the extra cost, so a compromise of 200
atmospheres pressure is used.
progress of the reaction
h Sha llower curve tha n the original (1) and o n ly 30cm3
using the correct mass (100g, award mark if 105g
is used) (1)
b Very high p ressures are expensive to produce for
products
Q = mc6 T = 100.0 x 4.2 x (23.2 - 15.9) = 3.066 kJ
calculation of temperature change (1)
fewer moles of gas, shifting the position of equilibrium
to the right hand side.
.·1;H
g Shallower curve than the original (1). but the end
volume remains the same (1 ).
ii
a A high pressure favours t he reaction producing
activation energy
. without catalyst
Re-weigh the solid and the mass should be the
same as before if it acts as a catalyst (1).
19.3 (1), 16.6 (1), 2.7 (1)
A
a
b
367
without itself being used up (1). It saves money which
would be spent on increasing temperature or pressure (1).
6 ~ a reversible reaction (1), enthalpy change (1)
b The forward reaction is endothermic. (1)
C
a Reaction has stopped (1) because all the hydrogen
peroxide is used up (1 ).
I
6(1)
ii
Weigh a sample of manganese(IV) oxid e and add
to hydrogen peroxide (1). Oxygen is produced at a
faster rate with manganese(IV) oxide than without
(1 ). Filter the reaction mixture and dry the solid (1 ).
progress of the reaction
368 ANSWERS
d I
ANSWERS
Increasing the temperature favours the forward
endothermic reaction. (1)
The reaction m ixture becomes more colourless. (1)
Increasing the pressure favours the left hand side of the
reaction as it has fewer moles of moles of gases. (1)
V
The position of equilibrium shifts to the left. (1)
The reaction m ixtures becomes more brown. (1)
e No change in the position of equilibrium (1) The use of
a catalyst increases the rate of both the forward and
reverse reaction equally (1 ).
7~a AandD
b I Increasing the pressure favours the right hand side
of the reaction which has fewer moles of moles of
gas molecules. (1)
The posit ion of equilibrium shifts to the right. (1)
The yield of NH3 increases. (1)
ii As temperature increases, the percentage of NH3
at equilibrium decreases. (1)
iii t;.H for the forward reaction is negative. (1)
Increasing temperature favours the reverse
endothermic reaction. (1)
iv Increasing temperature increases the rate of
reaction. (1)
More particles will have energy greater than or
equal to the activation energy. (1)
The frequency of successful collisions increases. (1)
H
H
H
H
3
H
I I I I I I
C - C- C- C- C- CI HI HI HI HI HI
H
H-
The position of equilibrium shifts to the right. (1)
II
H
Iv
~
H
II
I
CH3CHCH20H
iv
I I I
H- C -C- CI
I
H
H
I
I
OH
H
b CH30CH2CH2CH3 , CH3CH2 0CH2CH3 , CH3 0CHCH3
H- C -H
I
I
H
I
4 ~ a addition
I
C = C-
C-
I
!
H
b combustion
c addition
d substitution
e combustion
f substitution
H- C- H
I
H
viii
H
I
H
H
H
5
I I I
IH HI HI HI
~
2
~
formula but different structural formulae.
UNIT 4 ANSWERS
CH2-
a
propane
iii
hexane
Iv
propene
V
ethene
H
ii
H
H-
H
H
I I
/ H
I
"-H
C11 H24 -;, C2H4 + C9H20
or C11 H24 -+ C2 H4 + C3 H6 + C6 H,4
or lots of other variants. In each case, at least one
hydrocarbon should be an alkane (C0H20 • ,J, and at
least one an alkene (C, H2.J.
4 ~ This is entirely open to your imagination and ability
to think both logically and laterally. It is impossible to
suggest 'right' answers.
H- c - c= c
CH3
H
2-methylpropane
d CnH2n+2
e Same functional group / similar chemical properties;
Shows a gradation in physical properties; Each
member d iffers from the next by a -CH2-.
CHAPTER 24
t
~
a Contains only C-C single bonds and has no double or
triple bonds.
b I
C11 H24
Ii
Liquid
C 5H12 + 802 - 5C02 + 6H20
Ill C 11 H24(1) + 1702(g)- 11C02(g) + 12H20(1)
Iv 2C 11 H24 + 2302 - 22CO + 24H 2 0
Carbon monoxide is poisonous, as it reduces the
ability of the blood to carry oxygen around the
body.
3
hexane
2·methy1pentane
CH3
CH3
I
I
CH3 CH2CHCH2CH3
CH, CHCHCH,
I
H
H
H
3-methylpentane
I
I
CH3C=CH2
I
CH3
but-1-ene
H
I
I
C- C = C - CI HI HI HI
H
CH,
CH3CCH2CH3
CH,
CH,
2,3-dimethylbutane 2,2-dimethylbutane
d CH3 CH2CH = CH2 CH3CH = CHCH3
H
I I I
H- C - C - C - H
I I HI
OH H
iii
CH -
I
CH
I I
C- C I HI
H
H
I
CH3 -
CH3CHCH2CH2CH3
vi but-1-ene
vii propan-1-ol
viii butan-2-ol
H-
CH3
(In these and subsequent formulae, if you aren't asked
specifically for displayed formulae, these quicker
forms are acceptable.)
methane
ii
b
CH2 -
butane
CHAPTER 22
~
c
CH3
CH3 -
t
a C, it contains element other than carbon and hydrogen.
b
a The existence of molecules with the same molecular
b
c C 11 H24 -+ 2C2 H4 + C7 H16
d Any other valid cracking equation starting with C 11 H24 .
For example:
H
H
•>< •x
H:C:C:H
•>< •><
HH
H- C = C -C- C - C -H
a Crude oil produces too many larger hydrocarbons /
not enough of the more useful smaller ones. Smaller
alkanes can be used as fuel for cars. Cracking also
produces alkenes that can be used to make polymers.
b Heat the vaporised fraction in the presence of a silicon
dioxide or aluminium oxide catalyst at 600-?00 ' C.
These compounds are known as ethers.
H
but-2-ene
CH3
H
I
CH
/
\
CH -CH
2
2
(2-methylpropene)
CHAPTER 23
t
~
2C4 H 10 + 1302 - 8CO, + 10H,O
Carbon monoxide is poisonous as it reduces the
blood's ability to carry oxygen around the body.
2 ~ a The sulfur (or sulfur compound) burns to make sulfur
dioxide. The sulfur dioxide reacts with water and
oxygen in the atmosphere to produce sulfuric acid that
falls as acid rain.
b The spark in the engine causes nitrogen to react wit h
oxygen to give various oxides of nitrogen.
c Sulfur dioxide reacts with water and oxygen in the
atmosphere to produce sulfuric acid. The sulfuric acid
in acid rain can react with calcium carbonate and
corrode the buildings.
3 ~
CH3
H
H-
CH 3
CH3CCH3
I
vii
CH 3
Iii
H
H
ii
I
I
H-C-0-H
I
H
H
Any between n ; 5 - 10 for CnH2n,2
g I
CH3CH 2CHCH3
OH
H
vi
(first answer is shown in the textbook as an
example)
a
369
a Carbon and hydrogen.
b The crude oil (mixture of hydrocarbons) is heated until
it boils. The vapour passes into a fractionating column.
The temperature is higher at the bottom of the column
than at the top. Different fractions condense and
are drawn off at different heights in the column. The
hydrocarbons with the highest boiling points (longer
chains) condense towards the bottom of the column.
The smaller hydrocarbon molecules travel further up
the column until they condense and are drawn off.
c Gasoline - petrol for car; Diesel - fuel for lorries or
buses
d Any two from: refinery gas, kerosene, fuel oil or
bitumen.
e The average size of the molecules in gasoline is
smaller than in diesel. Diesel is darker in colour and
more viscous than gasoline.
2 ~
a C5H12
b C,H 10
C
3
~
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