# CHAPTER 4 FORM5

```THERMOCHEMISTRY
1.The following equation shows the reaction
between nitrogen and oxygen to form nitrogen
oxide.

N2 (g) + O2 (g) ↔2NO (g)
H = +181 kJ
Refer to the equation, what conclusion can be
A
The temperature increased during the
reaction.
B
The energy level diagram for the reaction
is:
C
D
The energy content of the product is lower
than that of the reactants.
The heat absorbed on breaking the
bonds is greater than the heat released
on forming the bonds.
2. The following represent the volumes of 1.0
mol dm-3 silver nitrate (AgNO3) solution and 1
mol dm-3 potassium chloride solution used for
determining the heat of precipitation of silver
chloride in four experiments. Which of the
experiments (I, II, III or IV) gives the highest
rise in temperature?
I.Volume of AgNO3 = 55cm3 and volume of KCl
= 5cm3
II.Volume of AgNO3 = 50cm3 and volume of
KCl = 10cm3
III.Volume of AgNO3 = 40cm3 and volume of
KCl = 20cm3
IV.Volume of AgNO3 = 35cm3 and volume of
KCl = 25cm3
A.I
B.II
C.III
D.IV
3. The temperature of the solution increases by
5 oC, when 500 cm3 of 1.0 mol dm-3 solution
containing M+ ions is added to 500 cm3 of 1.0
mol dm-3 solution containing SO4 2- ions.
Calculate the heat of precipation of M2SO4 in
the reaction.[Specific heat capacity of water =
4.2 J g-1 O C-1]-C
A
- 4.2 X 5 kJ mol -1
B
- 500 X 4.2 X 5 kJ mol -1
C
-4.2 X 5 kJ mol -1
0.25
D
- 500 X 4.2
kJ mol -1
4.The diagram below shows the structure when
combustion of methane in oxygen releases
heat energy.
When 1 mol of methane is burnt,
A
one O = O is broken
B
one C -H bond is broken
C
one C = O bond is formed
D
four O- H bonds are formed
thermochemical.
2Ag+ (aq) + Cu (s)→ Cu(s) +ZnSO4 (aq)
∆H = - y kJ
From the thermochemical equations, it can be
concluded that
I.the colour of the solution changes from
colourless to blue
II.the heat of displacement of silver by copper
is 1/ 2y kJ mol-1
III.the concentration of silver ions decreases as
the reaction proceeds
IV.the total concentrations of the ions before
and after the reaction are not the same
A.I, II B.III,IV C.I,II,III DI, II, III ,IV
displacement of copper (II)sulphate
solution by magnesium.-B
A
- ( 40 X 4. 2 X t ) kJ
1000
B
- ( 40 X 4. 2 X t ) kJ
0.4 X 1000
C
- ( 42 X 4. 2 X t ) kJ
1000
D
- ( 42 X 4. 2 X t ) kJ
0.4 X 1000
7. An experiment is carried out to observe
the heat of precipation of silver chloride
for the reaction between silver nitrate
solution and sodium chloride solution.
The steps that must be taken to obtain an
accurate results is
I.stir the solution continuously
II.use a screen to block the wind
III.use a container that is a good
conductor of heat
IV.add the silver nitrate solution quickly to
the sodium chloride solution.
A.I, III B.I, IV C.II,III,IV D I, II, III, IV
8. Copper(II) sulphate reacts with zinc
according to the following equation.
Zn (s) + CuSO4 (aq) →Cu(s) +ZnSO4(aq)
∆H = - 218 kJ mol-1
Calculate the minimum mass of zinc
required in order to produces 10.9 kJ of
heat? [Relative atomic mass; Zn, 65.4 ]
A.1.7 g
B.3.3 g
C.5.0 g
D.6.6 g
9. An experiment is carried out to
determine the heat of displacement for
the reaction between copper and silver
nitrate.The measurement that must be
carried out in this experiment are
I.mass of silver formed
II.volume of silver nitrate solution used
III.concentration of silver nitrate solution
IV.initial temperature of silver nitrate
solution.
A.III,IV B.I,II,III C.II,III,IV D.I, II,III,IV
10.Given that the equation, which
represents the reaction between two
particular ions is as follows:
Pb2+ (aq) + CrO42-(aq)→ PbCrO4(s)
Next, 50 cm3 of 0.5 mol dm-3 lead(ll) nitrate
solution is added to 50 cm3 of 0.5 mol dm-3
potassium chromate(VI) solution, the
temperature rises by x oC. If the
experiment is repeated using 50 cm 3 of
1.0 mol dm-3 lead(II) nitrate solution and
50 cm3 of 1.0 mol dm-3 potassium
chromate(VI) solution, then which of the
following is correct pertaining the rise in
temperature?
A. 1/ 2x oC
B.x oC C.2x oC D.4x oC
11.The reaction between barium chloride
and sodium sulphate is an exothermic
reaction. Which of the following energy
level diagram shows the reaction?-A
D
12.In one experiment, 50 cm3 of 2.0 mol dm-3
sodium hydroxide solution is poured into 50
cm3 of 1.0 mol dm-3 copper(II) sulphate
solution. Given that the heat of precipitation of
copper(II) hydroxide in the reaction between
copper(II) sulphate solution is -59 kJ mol-1.
[specific heat capacity of water = 4.2 J g-1 oC-1]
Amongst the following, which is the correct rise
in temperature?
A.7 oC
B.10 oC
C.12 oC D.14 oC
13. 10 kJ of heat is absorbed when one mole of
solid rubidium chloride, RbCl , is dissolved in 1
dm-3 of water. What is the fall in temperature
when 24.2 g of rubidium chloride is dissolved in
1 dm3 of water? [Specific heat capacity of water
= 4.2 Jg-1oC-1; Relative atomic mass; Rb, 85.5;
Cl, 35.5]
A.0.5 oC B.1.0 oC C.2 oC D.5 oC
14. When 50 cm3 of 1.0 mol dm-3 lead(II) nitrate
solution is added to 50 cm3 of 2.0 mol dm-3
sodium chloride solution, the temperature of
the reaction mixture increases by oC. The ionic
equation represent the lead(II) nitrate reacts
with sodium chloride.
Pb2+(aq) +2Cl-(aq) →PbCl2(s)
The solutions that will produce the same
temperature rise when added to 50cm 3 of 1.0
mol dm-3 lead(II) nitrate solution is
I.50 cm3 of 2.0 mol dm-3 hydrochloric acid.
II.50 cm3 of 2.0 mol dm-3 barium chloride
solution
III.50 cm3 of 1.0 mol dm-3 magnesium chloride
solution
IV.50 cm3 of 1.0 mol dm-3 aluminium chloride
solution
A.I, III B.II ,IV C.I,II,III D I, II, III ,IV.
15. The temperature is increased from 24.5 oC
to 25.2 oC when a small amount of solid
hydroxide is added to 200 cm3 of water.
Calculate the total amount of heat released in
this reaction.[Specific heat capacity of solution
= 4.2 Jg-1 oC-1]
A.588 J B.588 Kj C.1176 J D.1176 kJ
16.The thermochemical equation shows the
reaction when ammonium nitrate is dissolved in
water.
NH4NO3(s) + water →NH4+(aq) +NO3(aq) ∆H =
+26 kJ mol -1
The graphs that represents the changes in
temperature of the solution when ammonium
nitrate is dissolved in water, until the
temperature remains unchanged is -B
A
A
B
B
C
C
17. Which of the following reactions releases
heat to the surroundings?
I.Adding concentrated acid to the water.
II.Adding sodium hydrogen carbonate to acid.
III.Dissolving solid aluminium sulphate in water.
water.
A.I, III B.I,IV C.I, II,III D.II, III ,IV
6. When 2 g of a magnesium powder (in
excess) is added to 40 cm3 of copper(II)
sulphate solution containing 0.4 mol of Cu2+,
the rise in temperature is t oC. Find the heat of
1
18. The diagram below represents the energy
level diagram for the reaction between
hydrogen and iodine.
Which of the following statements about the
diagram above is true?
A.The reaction between hydrogen and
iodine is endothermic.
B.When hydrogen and iodine react, heat is
released.
C.The dissociation of hydrogen iodine to form
hydrogen and iodine is exothermic.
D.When the covalent bonds are formed in
hydrogen iodide molecules, heat is absorbed.
19. The energy level diagram for the
displacement of iron by the metal, M is
shown in the below diagram.
Find the temperature reached if excess M
is added to 50 cm3 of 0.2 mol dm-3 iron(II)
sulphate solution at 30 oC.[Specific heat
capacity of solution = 4.2 Jg-1 oC-1]
A.14.4oC B.22.2oC C.37.8 oC
D.45.6 oC
20.In one experiment, 50 cm3 of 2.0 mol
dm-3 sodium hydroxide solution is poured
into 50 cm3 of 1.0 mol dm-3 copper(II)
sulphate solution. Given that the heat of
Paper2
1. An experiment was carried out to determine the heat of displacement
for the reaction between zinc and copper(II) sulphate.
precipitation of copper(II) hydroxide in the
reaction between copper(II) sulphate solution is
-59 kJ mol-1.
[specific heat capacity of water = 4.2 J g-1 oC-]
Amongst the following, which are not the
correct rise in temperature?
I.7 oC
II.10 oC
III.12 oC
IV.14 oC
AI, II,III B.I,II,IV C.I,III,IV D.II, III,IV
21 When 25.0 cm3 of lemon juice is added to
185.0 cm3 of milk to prepare yogurt, the
temperature of yogurt increases by 2.4 oC.
Calculate the total amount of heat released?
[The specific heat capacity of yogurt is Y Jg-1
o
C; the density of the solution is 1 g per cm 3]
A.444Y J
B.444Y kJ C.504Y J D.504Y kJ
A heating element is immersed in copper(II) sulphate solution and 1.0 kJ of
heat energy is supplied.
Experiment II
3.27 g of zinc powder is added to 200 cm 3 of copper(II) sulphate solution (in
excess)
In both of these experiments,
- the copper(II) sulphate solution is placed in a polystyrene beaker.
- the initial temperature of copper(II) sulphate is 30.4 oC.
The thermometer readings for the maximum temperature reached in both
of the experiments is shown in the diagram below.
Excess zinc powder was added to 50cm 3 of 0.5 mol dm-3 copper(II)
sulphate solution. The heat of displacement for the reaction was found to
be -205.8 kJ mol-1.
(a)Write down the equation for the above reaction.
Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
[1 mark]
(b)What is meant by 'heat of displacement of copper by zinc'?
The heat of displacement is the heat evolved when 1 mol of copper is
displaced by zinc from copper(II) sulphate solution.
[1 mark]
(c)Why is it necessary for the polystyrene cup to be put into a 250 cm3
beaker as shown in the figure above. Give reason to your answer.
To improve the accuracy of the experiment, the trap air between the
polystyrene cub and the beaker will lessen heat loss from the
polystyrene cup. This is because placing the cup in the beaker
improves the insulation by lessening moving wind.
[3 marks]
(d)(i) Give 2 observations in this experiment.
The blue colour of copper(II) sulphate becomes colourless. A brown
precipitate of copper is produced.
[2 marks]
(ii) Give one precaution that must be taken while carrying out the
experiment.
The solution must be stirred continuously to obtain a uniform
temperature.
[1 mark]
(e)Based on the information above, calculate the increase in temperature
for the mixture.[Specific heat capacity is 4.2 J g-1 oC-1, density of the
solution is 1 g cm-3]
(e)
No. of moles of the Cu2+ ions = (50 x 0.5) / 1000
= 0.025 mol
Heat given out = 0.025 mol x 205.8 kJ mol-1
= 5.145 kJ
= 5145 kJ
50 g x 4.2 J g-1 oC-1 x θ = 5145 J
θ = 24.5 oC
[3 marks]
2. Two experiments are carried out to determine the heat of displacement
of copper by zinc.
Experiment I
(a)Give the meaning of 'heat of displacement of copper by zinc'.
The heat of displacement is the heat evolved when 1 mol of copper is
displaced by zinc from copper(II) sulphate solution.
[1 mark]
(b)Why polystrene beaker is used in Experiments I and II?
The polystrene beaker is an insulator of heat. The polystyrene beaker
is used to reduce the loss of heat to the surroundings.
[1 mark]
(c)Calculate the rise of temperatures in both of the experiments.
The highest temperature reached in Experiment I is 32.8 oC.
Increase in temperature in Experiment I = 32.8 - 30.4 oC = 2.4 oC
The highest temperature reached in Experiment II is 51.2 oC.
Increase in temperature in Experiment 1 = 51.2 - 30.4 oC = 20.8 oC
[2 marks]
(d)(i) Give two observations in Experiment II.
The blue colour of copper(II) sulphate becomes colourless. A brown
precipitate of copper is formed.
[1 mark]
(ii) Give one precaution that must be taken while carrying out Experiment II.
The solution must be stirred continuously to obtain a uniform
temperature.
[1 mark]
(e)(i) Find the heat energy change in Experiment II.
(i) Experiment I shows that the release of 1.0 kJ of heat energy
increases the temperature by 2.4 oC.
Energy change in Experiment II = 1.0 x ( 20.8 / 2.4) = 8.67 kJ.
[ 2 marks ]
(ii) Calculate the number of moles of copper metal formed in Experiment II.
(ii) Relative atomic mass of zinc = 65.4
Number of moles of zinc used =3.27/ 65.4= 0.05
Zn(s) + Cu2+ (aq)→ Zn2+ (aq) + Cu(s)
From the equation, 1 mol of Zn produces 1 mol of copper metal. 0.05
mol of zinc produces 0.05 mol of Cu.
[2 marks]
(iii) Calculate the heat of displacement of copper zinc. Refer to (e)(i) and
(ii).
[Specific heat capacity of solution ;4.2 Jg-1 oC, density of solution ; 1.0 g cm3
; relative atomic mass; Zn, 65.4 ]
Displacement of 0.05 mol of Cu releases 8.67 kJ of heat. Heat change
in the displacement of 1 mol of Cu = - 8.67 x ( 1 / 0.05) = - 173.4 kJ.
Heat of displacement of Cu by Zn = -173.4 kJ mol-1
[3 marks]
(f) Experiment II can be carried out by using polystyrene beaker A or
polystrene beaker B, as shown in the diagram below.
Which of these two polystyrene beakers should be used? State your
reason.
Polystyrene container A should be used. If polystyrene is used, the
surface area of the solution exposed to the atmosphere is reduced.
This will reduce the loss of heat to the surrounding.
[2 marks]
3. The diagram below shows an experiment set-up to determine the heat
of precipitation.
The heat released when 1 mole of hydrochloric acid reacts with 1 mole
of sodium hydroxide to form 1 mole of water.
[1 mark]
(c)Calculate
(i) the number of mole of NaOH
No. of moles of NaOH = (20 x 2.0) / 1000
= 0.04 mol
[1 mark]
(c)(ii) the number of mole of HC1
No. of moles of HC1 = (40 x 0.5) / 1000
= 0.02 mol
[1 mark]
(d)Calculate the heat of neutralisation.
Total heat liberated = mc∆θ = (40 + 20)x 4.2 x 4.1
= 1033 J
0.02 moles of HC1 released 1033J
 1 mole of HC1 will release 1033/ 0.02
= 51.65 kJ
Therefore, the heat of neutralisation is 51.65 kJ mol-1
[3 marks]
(e)Draw the energy level diagram for the reaction.
Result:
Initial temperature of lead(II) nitrate, Pb (NO3)2
27.8oC
Initial temperature of sodium sulphate, Na2 SO4 27.2oC
The highest temperature of mixture
32.6oC
[Specific heat capacity of the solution is 4.2 Jg-1oC-1, density of the solution
is 1 g cm-3]
(a)(i) State the type of reaction that happens in this experiment.
Exothermic reaction
[1 mark]
The final temperature is higher than the initial temperature.
[1 mark]
(b) What is the colour of the precipitate?
White
[1 mark]
(c) Give one step that is needed to be done when mixing the two solutions
in order to get more accurate result.
The two solution need to be mixed together as fast as possible and
then stirred to ensure thoroughly mixing.
[1 mark]
(d)(i) Calculate the number of moles of lead(II) ion.
(i)number of moles of lead(II) ion
= (25 x 0.5) / 1000
= 0.0125 mol
[1 mark]
(ii) Calculate the number of moles of sulphate ion.
number of moles of sulphate ion
= ( 25 x 0.5) / 1000
= 0.0125 mol
[1 mark]
(iii) Calculate the heat given off in this reaction.
Heat given off
= cm∆θ
= (25 + 25)cm3 x 4.2 J g-1oC-1 x (32.6 - 27.4)oC
= 1092 J
[2 marks]
(e) Calculate the heat of precipitation.
0.0125 moles liberate 1092 J
.: 1 mole will liberate 1092/ 0.0125
= 87.36 kJ
[2 marks]
(f)The heat of precipitation that is calculated from this experiment is less
than the theoretical value. Why?
This is because some of the heat from the reaction is
- lost to the sorrounding
- been absorbed by the thermometer and the polystyrene cup
[2 marks]
4 . 20 cm3 of 2.0 mol dm-3 aqueous NaOH is poured into plastic cup that is
filled with 40 cm3 of aqueous HC1 with a concentration of 0.5 mol dm-3. The
temperature of the mixture rose from 30 oC to 34.1 oC.
[Density of solution is 1.0 g cm-1, specific heat of the solution is 4.2 J g-1 oC1
]
(a)Write the equation for the reaction in the experiment.
HC1 (aq) + NaOH (aq) → NaC1 (aq) + H2O (l)
[1 mark]
(b)Define heat of neutralization between hydrochloric acid and sodium
hydroxide.
[3 marks]
(f)(i) If the HC1 in the experiment is replaced with sulphuric acid of the
same volume and concentration, what will be the amont of heat released?
5165 J
[1 mark]
This is because the number of hydrogen ions/hydroxide ions that
reacted is doubled and the number of moles of water formed is
doubled.
[1 mark]
4. The table below shows the heats of combustion of a few members of
the alkane family.
Alkane
Heat of combustion (kJ mol-1)
Methane
-890
Ethane
-1602
Propane
-2200
Butane
-2877
Pentane
-3500
Based on the table above, answer the following questions.
(a)Write the general formula for alkenes.
CnH2n + 2
[1 mark]
(b)What is meant by 'heat of combustion'?
(b) The heat of combustion substance is the heat produced when one
mole of the substance is completely burnt in oxygen under standard
conditions.
[1 mark]
(c)The tempereture roses by 15 oC when x g methane is used to heat 500
g of water.[Relative atomic mass: H, 1; C, 12. The specific heat capacity of
water = 4.2 J g-1 oC-1]
Calculate
(i) the heat liberated when x g of methane is burnt.
(i) Heat liberated when x g of methane is burnt
= m x C x∆θ
= 500 x 4.2 x 15 = 31 500 J = 31.5 kJ
[1 mark]
(ii) the number of moles of methane used
no. of moles of methane
= 31.5 / 890
= 0.0354 mol
[1 mark]
(iii) the value of x
Relative molecular mass of CH4
= 12 + 4 = 16
[1 mark]
(d)Draw the energy level diagram for the reaction in (c).
[2 marks]
(e)The heat of combustion of ethanol (C2H5OH) is -1370 kJ mol-1. State
one benefit of using methane as a fuel compared to ethanol. Explain.
Relative molecular mass of ethanol, (C2H5OH)
= (2 x 12) + 5 + 16 + 1= 46
Fuel value of ethanol = 1370 / 46
= 29.8 kJ g-1
Relative molecular mass of methane, (CH4) = 16
Fuel value of methane = 890 / 16 = 55.6 kJ g-1
 Methane is a better fuel than ethanol in terms of heat energy
released on combustion.
[3 marks]
5. The heat of combustion of the organic compounds W, X, Y and Z is
shown in the table below. Given that general formula for these compounds
is CnH2n+2 in which n = 1, 2, 3 and 4...
Organic
Number of carbon Heat of
compound
atoms per
combustion (kJ
molcule
mol -1 )
W
1
- 890
X
2
...
Y
3
-2220
Z
4
-2877
(a)Give the homologous series to which the above organic compounds
belong.
The alkane series
[1 mark]
(b)Using the table, plot a graph of heat of combustion against number of
carbon atoms per molecule.
[3 marks]
(c)Estimate the heat of combustion of X using your graph.
- 1550kJmol-1
[ marks]
(d)When y g of W is used to heat 500 g of water, the temperature rises by
12 oC. Find [Relative atomic mass : H, 1; C, 12. The specific heat capacity
of water = 4.2 Jg-1oC-1]
(i) the heat released when y g of W is burnt.
Heat released when y g of substance P is burnt
= m x cx∆T
= 500x 4.2x 12J
=25 200 J
= 25.2 kJ
[1 mark]
(ii) the number of moles of W used.
1 mol of W releases 890 kJ of heat when burnt
Number of moles of P used = 25.2/ 890= 0.0283
[1 mark]
(iii) the value of y.
Relative molecular mass of CH4 = 12 + 4 = 16
Mass of W used = 0.0283 x 16 = 0.453 g
[1 mark]
(e)Draw the energy level diagram for the combustion of W in air.
[2 marks]
6. Your teacher has given your a lab assignment. She has given the
following chemicals :
dilute hydrochloric acid, dilute sodium hydroxide solution, silver nitrate
solution, solid ammonium nitrate, solid sodium hydrogen carbonate, water.
(a)From the list, select two pairs of chemicals that can be used to produce
(i) endothermic reactions
-Ammonium nitrate + water
- sodium hydrogen carbonate + hydrochloric acid
[2 marks]
(ii) exothermic reactions.
- sodium hydroxide + water
- sodium hydroxide + hydrochloric acid
[2 marks]
(b)The chemical equation shows the reaction between nitrogen and
oxygen.
N2 (g) + 2O2 (g)→ 2NO2 (g)
The diagram below represents the reaction.
(i) Which of these energy changes, ∆Ha, ∆Hb or∆ Hc have positive values?
∆Ha, ∆Hb
[2 marks]
(ii) In terms of bond breaking, and bond formation, what do ∆Ha and ∆Hb
represent?
∆Ha represents the heat absorbed on breaking the covalent bonds in
nitrogen and oxygen molecules. ∆Hb represents the heat evolved on
forming the covalent bonds between nitrogen and oxygen atoms.
[2 marks]
(iii) What is ∆Hc?
∆Hc is the heat of reaction when 1 mol of nitrogen reacts with 2 mol of
oxygen of form 2 mol of nitrogen dioxide.
[1 mark]
6. (a) (i) Give 2 examples to chemical reactions which are exothermic
and 2 examples which are endothermic.
Exothermic reaction
Combustion : C + O2 → CO2
Neutralisation : HN1 + NaOH → NaC1 + H2O
Endothermic reaction
Thermal decomposition: ZnCO3 → ZnO + CO2
Acid on carbonate: 2 CH3COOH + Na2CO3 →2 CH3COONa + CO2 + H2O
[4 marks]
(ii) Chemical reactions involve the breaking and formation of bonds. Explain
how the breaking and formation of bonds is related to the exothermic and
endothermic reaction.
- During a reaction is occurred, the chemical bonds between the
reactants need to be broken.
- Breaking of chemical bonds absorbs energy.
- New chemical bonds also need to be formed during the formation of
the products.
- Formation of chemical bonds releases energy.
- A reaction is exothermic is the energy released during bond
breaking is less than the energy released during bond formation.
- A reaction is endothermic if the energy absorbed during bond
breaking is more than energy released during bond formation.
[6 marks]
(b)The table below shows some bond energies from a series of
experiments:
Bond
Bond energy (kJ mol-1)
H-H
436
H - C1
431
C1 - C1
242
Hydrogen react with chlorine to form hydrogen chloride:
H2 (g) + C12 (g)→ 2HC1 (g)
Calculate:
(i) the energy required to break the bonds in one mole of C12
242 kJ
[2 marks]
(ii) the energy required to break the bonds in one mole of H2
436 kJ
[2 marks]
(iii) the total energy required to break a bond
242 + 436 = 678 kJ
[2 marks]
(iv) the total energy required to break 2 moles of HC1 into free ions
-(2 x 431) = -862 kJ
[2 marks]
(v) the heat of reaction for this reaction
678 + (-862) = -184 kJ
[2 marks]
7. (i) Define heat of combustion of methanol, CH2OH.
The heat energy released when 1 mole of methanol is completely
burnt in excess oxygen.
[2 marks]
(ii) The heat of combustion of methanol and ethanol are -715 kJ mol-1 and 1371 kJ mol-1 respectively. Give 2 similarities between the combustion of
methanol and the combustion of ethanol. The combustion of ethanol is
different from the heat of combustion of methanol. Explain why the
difference is there netween the two combustion.
∆H = -715 kJ mol-1
CH3OH (l) + 3/2 O2 (g) → CO2 (g) + 2H2O (l)
CH3CH2OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l)
∆H = -1371 kJ mol-1
Similarities:
- Both methanol and ethanol burn in excess air to form carbon dioxide
and water.
- Both of them release a large amount of heat energy when burnt in
excess air.
Differences:
- The heat of combustion of ethanol is higher than the heat of
combustion of methanol
- A molecule of ethanol contains one carbon atom and 2 hydrogen
atoms (i.e. -CH2 group) more than a molecule of methanol.
- Hence, the combustion of 1 mol of ethanol produces 1 mol of CO2
and 1 mole of H2O more than the combustion of 1 mol of methanol.
[5 marks]
(b) State three benefits of using liquid hydrogen as fuel for rockets.
- Hydrogen has very high fuel value, that is combustion of a little of H2
produces a large amount of heat energy.
- Does not cause pollution because the product of combustion of H2 is
water.
- H2 is very light and reduce the rocket weight.
(c)
water.
[3 marks]
Information below shows four chemical salts which are soluble in
CH3OH (I)+ 3/2O2 (g) →CO2 (g) + 2H2O(l) ∆H = - 715 kJ mol
∆H = - 1371 kJ mol.
CH3CH2OH(I) + 3O2 (g) →2CO2 (g) + 3H2O(I)
Similarities in combustion
Both the alcohol burn in excess air to form carbon dioxide and water.
Both the alcohols release a large amount of heat energy when burnt in
excess air.
(i) Classify the chemical compounds in figure above into 2 groups based
on the heat changes when they dissolve in water.
- Sodium hydroxide and anhydrous calcium chloride release heat
energy when dissolve in water. This is an exothermic reaction.
- Potassium nitrate and ammonium chloride absorb heat energy when
dissolve in water, Hence, this is an endothermic reaction.
[6 marks]
(ii) Explain your classification in (c)(i) by using suitable labelled diagrams
for each of compounds.
Difference in heat of combustion
The heat of combustion of ethanol is higher that the heat of
combustion of methanol. This is because a molecule of ethanol
contains one carbon atom and two hydrogen ( that is, the -CH2 group )
more than a molecule of methanol. Hence, the combustion of 1 mol of
ethanol produces 1 mol of CO2 and 1 mol of H2O more than the
combustion of 1 mol of methanol.
C(s) + O2 → CO2 (g) ∆H = - x kJ mol-1
H2(g) + 1/2O2 → H2O(I) ∆H = - y kJ mol-1
Hence, the heat released on the combustion of hydrogen.
[5 marks]
(c)State three advantages of using liquid hydrogen as fuel for rockets.
Hydrogen has very high fuel value (kJ/g). This means that the
combustion of 1 g of hydrogen releases a large amount of heat
energy. The combustion of hydrogen does not cause air pollution
because the product of combustion of hydrogen is water. Hydrogen is
very light and this reduces the rocket weight when it carries large
quantities of fuel.
9. (a) Describe one experiment to determine the heat of displacement in
the laboratory. Your description should include the procedure required and
the necessary steps of calculation.
Procedure:
1.50 cm3 of 1.0 mol dm-3 copper(II) sulphate solution is measured.
2.The copper(II) sulphate solution is poured into a plastic cup.
3.The initial temperature of the solution is recorded.
4.Excess zinc powder is added instantly to the solution.
5.The mixture is stirred throughout the experiment.
6.The highest temperature of the mixture is recorded.
[4 marks]
8. (a)Explain one experiment to determine the heat of precipitation of
and the energy level diagram for the precipitation of lead(II) chloride.
Using a measuring cylinder 25 cm3 of 0.5 mol dm3 sodium chloride
solution is measured into plastic cup. The initial temperature of
sodium chloride solution is measured and recorded. Using another
measuring cylinder, 25 cm3 of 0.5 mol dm3 lead(II) nitrate is measured
and recorded. Lead(II) nitrate solution is poured quickly into the
sodium chloride solution. The mixture is stirred and the highest
temperature is recorded.
Results
Initial temperature of sodium chloride solution = t1 oC
Initial temperature of lead(II) nitrate solution = t2 oC
Maximum temperature of the reaction = t3 oC
Rise in temperature = t3 – [(t1 + t2) / 2 ]= t oC
Assumptions
Spesific heat capacity for water = 4.2 J g-1 oC-1
Density of solution = 1.0 g cm-3
Calculation
Heat change = m x c x∆T
= -(25 + 25) x 4.2 x tJ
= - y J = -y / 1000kJ
Number of moles of Pb2+ ions used
= (0.50 x 25)/ 1000= 0.0125
Number of moles of Cl - ions used
= (0.50 x 25)/ 1000 = 0.0125
Pb2+ (aq) + 2Cl- ( aq ) PbCl2 (s)
The equation shows that 2 mol of Cl- ions react with 1 mol of Pb2+ ions
to produce 1 mol of PbCl2.
0.0125 mol of Cl- ions reacts with 0.00625 mol of Pb2+ ions to form
0.00625 mol of PbCl2.
Heat evolved on the precipitation of 0.00625 mol of PbCl2
= y/1000kJ
Heat evolved on the precipitation of 1.0 mol PbCl2
= (y/1000) x ( 1/ 0.00625) kJ
= - (y/1000) x ( 1/ 0.00625) kJ
=- y/ 6.25 kJ mol -1
The energy level is shown in the diagram below.
[12 marks]
(b) The heats of combustion of methanol and ethanol are -715 kJ mol-1
-1
and -1371 kJ mol respectively. Give two similarities between the
combustion of methanol and combustion of ethanol. Describe why the heat
of combustion of ethanol is different from the heat of combustion of
methanol.
Result:
Initial temperature of solution = t1
The highest temperature of mixture = t2
Changes of temperature = t2 - t1 = θ
Calculations:
no. of moles of copper displaced
= 1.0 mol dm-3 x (50 / 1000) dm-3
Heat released = 50 g x 4.2 g-1 oC-1 xθ oC
Heat of displacement = - x/ 0.05 kJ mol-1
CuSO4 (aq) + Zn (s) → ZnSO4 (aq) + Cu (s)
= 0.05
= x kJ
[11 marks]
(b)Explain why the heat of neutralisation between sulphuric acid and
sodium hydroxide is the same as the heat of neutralisation between nitric
acid and potassium hydroxide solution.
Hydrochloric acid and nitric acid are strong acids that dissociate
completely in aqueous solution.
Sodium hydroxide and potassium hydroxide are strong akalis and
dissociate completely in aqueous solution.
In the reaction between 1 mol hydrochloric acid and 1 mol of sodium
hydroxide or 1 mol of nitric acid and 1 mol of potassium hydroxide,
the changes that occurs is the combination of 1 mol of H+ ions with 1
mol of OH- ions to form 1 mol of water molecule.
Heat of neutralisation is the heat released when 1 mol of H+ ions react
with 1 mol of OH- ions to form 1 mol of water.
Hence, the hear of neutralisation for both reactions between a strong
acid and a strong base are the same, that is -57 kJ mol-1
[5 marks]
(c)Give four examples of the application of energy chage in daily life and in
industry.
Chemical cells are used to generate electricity.
The combustion of fuels produces heat energy which can be used to
operate turbines to operate electrical energy.
Hot packs are used to reduce swelling and pain caused by muscle or
joint injury.
Cold packs are used to lower the body temperature.
[4 marks]
10. (a) Explain one experiment to determine the heat of neutralisation
between dilute sulphuric acid and sodium hydroxide solution. Show the
Experiment
- Using a measuring cylinder, 100 cm3 of 1.0 mol dm-3 sodium
hydroxide solution is measured into a plastic cup. The initial
temperature of the alkali is determined and recorded.
- Using another measuring cylinder, 100 cm3 of 0.5 mol dm-3 sulphuric
acid is measured out and its initial temperature determined.
- The sulphuric acid is poured quickly into the sodium hydroxide
solution. The mixture is stirred and the highest temperature recorded.
Results.
Initial temperature of sodium hydroxide = t1 oC
o
Initial temperature of sulphuric acid = t2 C
Maximum temperature of reaction mixture = t3 oC
Average initial temperature
= ½ x ( t1 + t2 ) oC
Rise in temperature
= t3 –( ½)x (t1 + t2) oC = t oC
Assumptions
Spesific heat capacity of water = 4.2 Jg-1oC-1
Density of solution = 1.0 gcm-3
Calculation
Total amount of heat energy released
= m xc x∆t
= (100 + 100) x 4.2 xt J
= yJ = y/1000kJ
Number of mol of H2SO4 used
= (0.5x100)/1000= 0.05
Number of moles of NaOH used
= (1.0 x 100)/1000= 0.1
H2SO4(aq) + 2NaOH (aq)→ Na2SO4 (aq) + 2H2O(I)
From the equation, 1 mol of H2SO4 reacts with 2 mol of NaOH to
produce 2 mol of water. 0.05 mol of H2SO4 reacts with 0.1 mol of NaOH
to form 0.1 mol of water.
Formation of 0.1 mol of water releases y/1000J of heat energy.Heat
energy released when 1 mol of water is formed = (y/1000) x ( 1/0.1) kJ.
(iii) The heat of precipitation remained unchanged when sodium chloride
solution is replaced by hydrochloric acid.
Sodium chloride, hydrochloric acid and silver nitrate are strong
electrolytes that dissociate completely in water.
Both sodium chloride and hydrochloric acid dissociate to give
chloride ions.
Precipitation of AgC1 involves the reaction of free aqueous Ag+ and
C1- ions.
The ionic equation for both reactions is the same, that is Ag+ (aq) +
C1- (s)
[4 marks]
(b)The reaction between hydrogen peroxide, H2O2 and hydrazine N2H4 will
produce steam and nitrogen. This reaction is an exothermic reaction that
occurs at a very fast rate.
(i) Briefly explain what is meant by exothermic reaction.
Exothermic reactions are reactions which give out heat.
[1 mark]
(ii) Why do you think the launching of rockets needs to use fast chemical
reactions?
- A rocket needs to accelerate very fast so that it escape the pull of
gravity.
- This can be done by shooting a jet of hot mixture from its exhaust
end.
- The shooting jet of exhaust can be achieved by fast chemical
reaction.
[3 marks]
(iii) A rockets is filled with 1500 kg of hidrogen peroxide. How much
hydrazine is required to completely react with it?
1 mole of H2O2 = 2 + (2x 16)
= 34 g
1 mole of N2H4 = (2 x 14) + (4 x 1)
= 32 g
2H2O2 + N2H4 → 4H2O2 + N2
Heat of neutralisation for the reaction between sulphuric acid and
sodium hydroxide.
= -(y/1000) x ( 1/0.1) kJ mol-1
[12 marks]
(b)Describe why the heat of neutralisation between sulphuric acid and
sodium hydroxide is the same as the heat of neutralisation between nitric
acid and potassium hydroxide solution.
Hydrochloric acid and nitric acid are strong acids that dissociate
completely in aqueous solution.
 1500 kg of H2O2 will react with 1500x
HCl(aq) →H + Cl
HNO3 (aq) → H+ (aq) + NO3-
Paper3
+
Thus, 2 mols of hydrogen peroxide will react with 1 moles of
hydrazine.
Therefore, 68 g of hydrogen peroxide will react with 32 of hydrazine.
1 g of H2O2 will react with 32/68 = 0.47 g hydrazine
1 kg of H2O2 will react with 0.47 kg of hydrazine
0.47 kg= 705 kg hydrazine
-
In the reaction between 1 mol of hydrochloric acid and 1 mol of
sodium hydroxide of 1 mol of nitric acid and 1 mol of potassium
hydroxide, the changes that occurs is the combination of 1 mol of H+
ions with 1 mol of OH- ions to form 1 mol of water molecules.
H+(aq) + OH- → H2O (l)
Heat of neutralisation for all reactions between a strong acid and a
strong base are the same, that is, - 57 kJ mol-1.
[5 marks]
(c)State four examples of the application of energy change in daily life and
in industry. Describe each application.
Example 1 : Cold packs are use to lower the body temperature of the
human body. In the cold packs, ammonium nitrate is mixed with water
to produce an endothermic reaction. Heat is absorbed from the
surrounding and this cools the body.
Example 2 : Hot packs are used to reduce swelling and pain caused
by muscle or joint injury. In some hot pack, iron powder is left to react
with air, in the presence of water and sodium chloride, to produce an
exothermic reaction. Heat released warm the injured part of the body.
Example 3 : The combustion of fuels (diesel), produces heat energy
which can be used to operate turbines to generate electrical energy.
Example 4 : Chemical cells are used to generate electricity. In the
chemical cells, redox reaction occurs. Chemical energy is converted
into electrical energy.
[3 marks]
11. (a) Figure below shows the thermochemical equation for a
precipitation reaction.
(i) Give the definition of the heat of precipitation.
The heat of precipitation is that heat change when one mole of a
precipitate is formed from its ions in aqueous.
[2 marks]
(ii) Draw a labelled energy level diagram for the reaction in the figure
above.
1.An experiment was carried out to find out the heat of displacement of
copper using zinc, mangnesium and iron respectively. 50 cm 3 of 0.5 mol
dm-3 copper(II) chloride solution was poured into a plastic cup. The
temperature of copper(II) chloride solution is taken after 2 minutes. Excess
metal was added into copper(II) chloride solution.
The following results was collected.
Zn
Mg
Iron
Initial temperature of copper(II) chloride solution (oC)
28.5 28.5 28.5
Highest temperature of mixture (oC)
51.5 54.0 46.0
[Specifice heat capacity = 4.2 Jg-1 oC-1, density of solution = 1.0 g cm-3]
(a)
Why was a plastic cup used in the experiment?
To reduce the heat less to surrounding.
[1 mark]
(b)
State three observations of the experiment.
(i) The temperature of the mixture increases.
(ii) The blue coloured solution becomes colourless.
(iii) Brown solid is formed.
[3 marks]
(c)
Name the three variables of the experiment.
(i) Manipulated variable: Different metals used
(ii) Responding variable: Heat of displacement.
(iii) Controlled Variables: Excess metal added, volume and
concentration of copper(II) chloride.
[3 marks]
(d) Calculate the heat of displacement copper by the three different metals
mentioned in the experiment.
Number of moles of copper displaced (by the three metals)
= 0.5 mol dm-3 x (50/100) dm-3
zinc
= 50 g x 4.2 Jg-1 oC-1x (51.5 - 28.5) oC
= 4830 J
magnesium
= 50 g x 4.2 Jg-1 oC-1 x (54.0 - 28.5) oC
= 5355 J
Heat of displacement
= -5.355/ 0.025
= -214.2 kJ mol-1
iron
= 50 g x 4.2 Jg-1 oC-1 x(46.0 - 28.5) oC
= 3675 J
[2 marks]
Heat of displacement
= -3.675kJ/0.025 mol
= -147.0 kJ mol-1
[6 mark]
(e) If the experiment is repeated using aluminium, predict the heat of
displacement of copper by aluminium based on the answer in (d).
Any value between -193.2 kJ mol-1 to -214.2 kJ mol-1
[1 mark]
2.A student bought two canisters of fuels (brand P and R) for school
camping. It is found that 200 g of fuel of brand P can boil 200 cm3 of water
in 12 minutes, whereas 200 g of fuel of brand R can boil 200 cm3 of water in
20 minutes.
Based on the above observations, you are asked to plan an experiment to
compare the fuel value (heat of combustion expressed in kJ g-1) of fuel P
and R. Your description must include the following:
(a)
Problem statement.
How is the heat of combustion of fuel P compared with that of fuel R?
(b)
Statement of hypothesis.
Fuel P has a higher heat of combustion that fuel R.
(c)
Manipulated, responding and fixed variables.
Manipulated variable : Type of fuel
Responding variable : Heat of combustion
Controlled variables : Volume of water and the metal container.
(d)
Experimental procedure.
1. 200 cm3 of water is measuring out using a measuring cylinder and
poured into a metal calorimeter.
2. The initial temperature of water is read and recorded.
3. The metal calorimeter is put on a tripod stand
4. A canister containing fuel of brand P is weighed and its mass
recorded.
5. The canister with fuel of brand P is then placed below the metal
calorimeter and lighted. The flame is protected from air movement by
putting a screen around it as shown in the diagram above.
6. The water in the metal calorimeter is stirred continuously with a
thermometer.
7. The flame from the burning of fuel P is extinguished again and its
mass recorded.
8. The canister containing fuel of brand P is weighed again and its
mass recorded.
9. The experiment is repeated using canister of brand P to replace
canister of brand R.
(a) Based on the spaces provided in the above diagram, record the
temperature.
(T1) : 28.5 oC
(T2) : 28.5 oC
(T3) : 31.5 oC
[3 marks]
(b) (i) Based on the changes in temperature in the experiment, name the
type of reaction that has occurred.
Exothermic reaction
[2 marks]
(ii) Name one precaution step that must be taken while mixing the two
solutions in order to get a more accurate results.
Silver nitrate solution must be added quickly to the sodium chloride
solution
[2 marks]
(iii) State the name and colour of the precipitate.
Silver chloride, white precipitate
[2 marks]
(c) (i) Tabulate the experiment above.
Initial temperature of silver nitrate solution = 28.5 oC
Initial temperature of silver chloride solution = 28.5 oC
Maximum temperature reached = 31.5 oC
Rise in temperature = 31.5 oC - 28.5 oC = 3.0 oC
[2 marks]
(ii) Calculate the heat change in the experiment.
Heat energy released
= Mass of solution x spesific heat capacity xrise in temperature
= (25 + 25)x 4.2 x 3.0
= 630 J = 0.63 kJ
[2 marks]
(iii) Calculate the heat of precipication for this experiment.
Step 1 : Calculate number of moles of AgCl precipitated. Number of
moles of AgNO3 used
=(0.5 x 25) / 1000 = 0.0125
Number of moles of NaCl used
=(0.5 x 25) / 1000 = 0.0125
Ag +(aq) + Cl- (aq)→ AgCl (s)
Number of moles of AgCl precipitated = 0.0125
Type of fuel
Initial mass of canister
Final mass of canister
Mass of fuel used (g)
Mass of water in metal
container (g)
Initial temperature (oC)
Final temperature (oC)
Temperature rise (oC)
Brand P
(e)
Tabulation of data.
Experiment Solution
1
2
3
4
Brand R
Step 2 : Calculate the heat of precipitation of silver chloride.
Precipitation of 0.0125 mol of AgCl releases 0.63kJ of heat.
Precipitation of 1.0 mol dm AgCl releases
0.63 x (1/0.0125) = 50.4 kJ mol-1
Heat of precipitation of AgCl is - 50.4 kJ mol-1
Observation on
oily stain
Inference
Soap + soap water
Soap + hard water
Detergent + soft
water
Detergent + hard
water
[17 marks]
3.The diagram below shows the set-up of apparatus for an experiment to
determine the heat of precipication. 25.0 cm3 of 0.5 mol dm-3 sodium
chloride solution.
The temperature before and after the reaction is shown in the diagram
below.
[2 marks]
(d) This experiment is repeated using 25.0 cm 3 of 0.5 mol dm-3 silver nitrate
solution to react with 25.9 cm 3 of 0.5 mol dm-3 hydrochloric acid.
(i) Based on the information given, predict the temperature change in this
experiment.
Rise in temperature = 3.0 oC
[2 marks]
(ii) State a reason of your prediction.
The volume of the reaction mixture remains constant, that is, 50.0 cm3.
The concentration of Ag+ ions and Cl- ions remain constant.Hence, the
temperature rise remains constant, that is 3.0 oC .
[2 marks]
```
Arab people

15 Cards

Pastoralists

20 Cards