1.
2.
What energy changes occur when chemical bonds are formed and broken?
A.
Energy is absorbed when bonds are formed and when they are broken.
B.
Energy is released when bonds are formed and when they are broken.
C.
Energy is absorbed when bonds are formed and released when they are broken.
D.
Energy is released when bonds are formed and absorbed when they are broken.
The temperature of a 2.0 g sample of aluminium increases from 25°C to 30°C.
How many joules of heat energy were added? (Specific heat of Al = 0.90 J g–1K–1)
A.
3.
0.36
B.
2.3
C.
9.0
D.
11
D.
–910
Using the equations below:
C(s) + O2(g) → CO2(g)
Mn(s) + O2(g) → MnO2(s)
∆H = –390 kJ
∆H = –520 kJ
what is ∆H (in kJ) for the following reaction?
MnO2(s) + C(s)  Mn(s) + CO2(g)
A.
4.
5.
6.
910
B.
130
–130
C.
Under what circumstances is a reaction spontaneous at all temperatures?
∆Hο
∆Sο
A.
+
+
B.
+
–
C.
–
–
D.
–
+
Which combination of ionic charge and ionic radius give the largest lattice enthalpy for an ionic
compound?
Ionic charge
Ionic radius
A.
high
large
B.
high
small
C.
low
small
D.
low
large
What is ∆H for the reaction below in kJ?
CS2(g) + 3O2(g)  CO2(g) + 2SO2(g)
[∆Hf /kJ mol–1: CS2(g) 110, CO2(g) – 390, SO2(g) – 290]
A.
7.
–570
B.
–790
C.
–860
D.
–1080
Which statements about exothermic reactions are correct?
I.
They have negative H values.
1
8.
9.
II.
The products have a lower enthalpy than the reactants.
III.
The products are more energetically stable than the reactants.
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III
A sample of a metal is heated. Which of the following are needed to calculate the heat absorbed
by the sample?
I.
The mass of the sample
II.
The density of the sample
III.
The specific heat capacity of the sample
A.
I and II only
B.
I and III only
C.
II and III only
D.
I, II and III
The average bond enthalpies for O—O and O==O are 146 and 496 kJ mol–1 respectively.
What is the enthalpy change, in kJ, for the reaction below?
H—O—O—H(g)  H—O—H(g) + ½O==O(g)
A.
10.
11.
B.
+ 102 C.
+ 350
D.
+ 394
Which reaction has the greatest positive entropy change?
A.
CH4(g) + 1½O2(g) → CO(g) + 2H2O(g)
B.
CH4(g) + 1½O2(g) → CO(g) + 2H2O(l)
C.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
D.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
What is the energy change (in kJ) when the temperature of 20 g of water increases by 10°C?
A.
C.
12.
– 102
20×10×4.18
20  10  4.18
1000
B.
D.
20×283×4.18
20  283 4.18
1000
The lattice enthalpy values for lithium fluoride and calcium fluoride are shown below.
LiF(s)
CaF2(s)
∆Hο = +1022 kJ mol–1
∆Hο = +2602 kJ mol–1
Which of the following statements help(s) to explain why the value for lithium fluoride is less
than that for calcium fluoride?
A.
13.
I.
The ionic radius of lithium is less than that of calcium.
II.
The ionic charge of lithium is less than that of calcium.
I only
B.
II only
C.
I and II
D.
Neither I nor II
When the solids Ba(OH)2 and NH4SCN are mixed, a solution is produced and the
temperature drops.
2
Ba(OH)2(s) + 2NH4SCN(s) → Ba(SCN)2(aq) + 2NH3(g) + 2H2O(l)
Which statement about the energetics of this reaction is correct?
14.
A.
The reaction is endothermic and H is negative.
B.
The reaction is endothermic and H is positive.
C.
The reaction is exothermic and H is negative.
D.
The reaction is exothermic and H is positive.
Using the equations below
Cu(s) +
2Cu(s) +
1
2
O2(g) → CuO(s)∆Hο = –156 kJ
1
2
O2(g) → Cu2O(s)∆Hο = –170 kJ
what is the value of ∆Hο (in kJ) for the following reaction?
2CuO(s) → Cu2O(s) +
A.
15.
16.
17.
142
B.
15
C.
1
2
O2(g)
–15
D.
–142
Which reaction occurs with the largest increase in entropy?
A.
Pb(NO3)2(s) + 2KI(s) → PbI2(s) + 2KNO3(s)
B.
CaCO3(s) → CaO(s) + CO2(g)
C.
3H2(g) + N2(g) → 2NH3(g)
H2(g) + I2(g) → 2HI(g)
D.
The ∆Hο and ∆Sο values for a certain reaction are both positive. Which statement is correct
about the spontaneity of this reaction at different temperatures?
A.
It will be spontaneous at all temperatures.
B.
It will be spontaneous at high temperatures but not at low temperatures.
C.
It will be spontaneous at low temperatures but not at high temperatures.
D.
It will not be spontaneous at any temperature.
Which of the quantities in the enthalpy level diagram below is (are) affected by the use of a
catalyst?
Enthalpy
I
II
III
Time
3
A.
18.
19.
I only
B.
III only
C. I and II only D. II and III only
Which reaction has the most negative ∆Hο value?
A.
LiF(s) → Li+(g) + F–(g)
B.
Li+(g) + F–(g) → LiF(s)
C.
NaCl(s) → Na+(g) + Cl–(g)
D.
Na+(g) + Cl–(g) → NaCl(s)
Consider the following equations.
Mg(s) +
1
2
O2(g) → MgO(s)
H2(g) +
1
2
O2(g) → H2O(g)
∆Hο = –602 kJ
∆Hο = –242 kJ
What is the ∆H° value (in kJ) for the following reaction?
MgO(s) + H2(g) → Mg(s) + H2O(g)
A.
20.
21.
22.
–844
B.
–360
C.
+360
D.
+844
For which of the following is the sign of the enthalpy change different from the other three?
A.
CaCO3(s) → CaO(s) + CO2(g)
B.
Na(g) → Na+(g) + e–
C.
CO2(s) → CO2(g)
D.
2Cl(g) → Cl2(g)
Which reaction has a positive entropy change, ∆Sο?
A.
H2O(g) → H2O(l)
B.
2SO2(g) + O2(g) → 2SO3(g)
C.
CaCO3(s) → CaO(s) + CO2(g)
D.
N2(g) + 3H2(g) → 2NH3(g)
Separate solutions of HCl(aq) and H2SO4(aq) of the same concentration and same volume
were completely neutralized by NaOH(aq). X kJ and Y kJ of heat were evolved respectively.
Which statement is correct?
A.
X=Y
B.
Y = 2X
C.
X = 2Y
D.
Y = 3X
23. Which statements are correct for an endothermic reaction?
A.
C.
24.
The system absorbs heat.
II.
The enthalpy change is positive.
III.
The bond enthalpy total for the reactants is greater than for the products.
I and II only
B.
I and III only
II and III only
D.
I, II and III
The mass m (in g) of a substance of specific heat capacity c (in J g–1 K–1 ) increases by t°C.
What is the heat change in J?
A.
25.
I.
mct
B.
mc(t + 273)
C.
mct D.
1000
mc(t  273)
1000
The average bond enthalpy for the C―H bond is 412 kJ mol–1. Which process has an enthalpy
change closest to this value?
4
26.
27.
28.
A.
CH4(g) → C(s) + 2H2(g)
B.
CH4(g) → C(g) + 2H2(g)
C.
CH4(g) → C(s) + 4H(g)
D.
CH4(g) → CH3(g) + H(g)
For a certain reaction at 298 K the values of both ∆Hο and ∆Sο are negative. Which statement
about the sign of ∆Gο for this reaction must be correct?
A.
It is negative at all temperatures.
B.
It is positive at all temperatures.
C.
It is negative at high temperatures and positive at low temperatures.
D.
It cannot be determined without knowing the temperature.
Which type of reaction is referred to in the definition of standard enthalpy change of formation?
A.
the formation of a compound from its elements
B.
the formation of a crystal from its ions
C.
the formation of a molecule from its atoms
D.
the formation of a compound from other compounds
The following equation shows the formation of magnesium oxide from magnesium metal.
2Mg(s) + O2(g)2MgO(s)
HӨ = –1204kJ
Which statement is correct for this reaction?
29.
A.
1204 kJ of energy are released for every mol of magnesium reacted.
B.
602 kJ of energy are absorbed for every mol of magnesium oxide formed.
C.
602 kJ of energy are released for every mol of oxygen gas reacted.
D.
1204 kJ of energy are released for every two mol of magnesium oxide formed.
The following equations show the oxidation of carbon and carbon monoxide to carbon dioxide.
C(s) +O2(g)  CO2(g)
CO(g) +
1
2
O2(g) CO2(g)
HӨ = –x kJ mol–1
HӨ = –y kJ mol–l
What is the enthalpy change, in kJ mol–1, for the oxidation of carbon to carbon monoxide?
C(s) +
A.
30.
x+y
B.
1
2
O2(g) CO(g)
–x–y
C.
y–x
D.
x–y
A simple calorimeter was used to determine the enthalpy of combustion of ethanol. The
experimental value obtained was –920 kJ mol–1. The Data Booklet value is –1371 kJ mol–1.
Which of the following best explains the difference between the two values?
A.
incomplete combustion of the fuel
B.
heat loss to the surroundings
C.
poor ventilation in the laboratory
measurements
D.
inaccurate temperature
5
31.
32.
What is the correct order of decreasing entropy for a pure substance?
A.
gas  liquid  solid
B.
solid  liquid  gas
C.
solid  gas  liquid
D.
liquid  solid  gas
For the reaction
2H2(g) + O2(g)  2H2O(g)
the bond enthalpies (in kJ mol–1) are
H–H
x
O=O
y
O–H
z
Which calculation will give the value, in kJ mol–1, of HӨ for the reaction?
33.
34.
A.
2x + y –2z
B.
4z – 2x – y
C.
2x + y – 4z
D.
2z –2x – y
Which statement about bond enthalpies is correct?
A.
Bond enthalpies have positive values for strong bonds and negative values for weak
bonds.
B.
Bond enthalpy values are greater for ionic bonds than for covalent bonds.
C.
Bond breaking is endothermic and bond making is exothermic.
D.
The carbon–carbon bond enthalpy values are the same in ethane and ethene.
An equation for a reaction in which hydrogen is formed is
CH4 + H2O  3H2 + CO
HӨ = +210 kJ
Which energy change occurs when 1 mol of hydrogen is formed in this reaction?
35.
A.
70 kJ of energy are absorbed from the surroundings.
B.
70 kJ of energy are released to the surroundings.
C.
210 kJ of energy are absorbed from the surroundings.
D.
210 kJ of energy are released to the surroundings.
The equations and enthalpy changes for two reactions used in the manufacture of sulfuric acid
are:
S(s) + O2(g)  SO2(g)
2SO2(g) + O2(g)  2SO3(g)
HӨ = –300 kJ
HӨ = –200 kJ
What is the enthalpy change, in kJ, for the reaction below?
2S(s) + 3O2(g)  2SO3(g)
A.
–100
B.
–400
C.
–500
D.
–800
6
36.
37.
Which reaction has the largest positive value of SӨ?
A.
CO2(g) + 3H2(g)  CH3OH(g) + H2O(g)
B.
2Al(s) + 3S(s)  Al2S3(s)
C.
CH4(g) + H2O(g)  3H2(g) + CO(g)
D.
2S(s) + 3O2(g)  2SO3(g)
Approximate values of the average bond enthalpies, in kJ mol–1, of three substances are:
H–H
430
F–F
155
H–F
565
What is the enthalpy change, in kJ, for this reaction?
2HF  H2 + F2
A.
38.
+545
B.
+2
–20
C.
D.
–545
The standard enthalpy change of formation values of two oxides of phosphorus are:
P4(s) + 3O2(g)  P4O6(s)
HӨf= –1600 kJ mol–1
P4(s) + 5O2(g)  P4O10(s)
HӨf= –3000 kJ mol–1
What is the enthalpy change, in kJ mol–1, for the reaction below?
P4O6(s) + 2O2(g)  P4O10(s)
A.
39.
40.
41.
+4600
B.
+1400
C.
–1400
D.
–4600
Which is a correct equation to represent the lattice enthalpy of magnesium sulfide?
A.
MgS(s)  Mg(s) + S(s)
B.
MgS(s)  Mg(g) + S(g)
C.
MgS(s)  Mg+(g) + S–(g)
D.
MgS(s)  Mg2+(g) + S2–(g)
Which equation represents a change with a negative value for S?
A.
2H2(g) + O2(g)  2H2O(g)
B.
H2O(s)  H2O(g)
C.
H2(g) + Cl2(g)  2HCl(g)
D.
2NH3(g)  N2(g) + 3H2(g)
The expression for the standard free energy change of a reaction is given by
GӨ = HӨ – TSӨ
What are the signs for HӨ and SӨ for a reaction that is spontaneous at all temperatures?
HӨ
SӨ
A.
+
–
B.
–
+
C.
+
+
D.
–
–
7
42.
43.
Which statement is correct for an endothermic reaction?
A.
The products are more stable than the reactants and H is positive.
B.
The products are less stable than the reactants and H is negative.
C.
The reactants are more stable than the products and H is positive.
D.
The reactants are less stable than the products and H is negative.
Which statement is correct about the reaction shown?
2SO2(g) + O2(g)  2SO3(g)
44.
A.
196 kJ of energy are released for every mole of SO2(g) reacted.
B.
196 kJ of energy are absorbed for every mole of SO2(g) reacted.
C.
98 kJ of energy are released for every mole of SO2(g) reacted.
D.
98 kJ of energy are absorbed for every mole of SO2(g) reacted.
Which statements are correct for all exothermic reactions?
A.
45.
46.
H = –196 kJ
I.
The enthalpy of the products is less than the enthalpy of the reactants.
II.
The sign of H is negative.
III.
The reaction is rapid at room temperature.
I and II only
B.
I and III onl
C. II and III only
D.
I, II and III
Which are characteristics of ions in an ionic compound with a large lattice enthalpy value?
A.
Large ionic radius and high ionic charge
B.
Small ionic radius and low ionic charge
C.
Large ionic radius and low ionic charge
D.
Small ionic radius and high ionic charge
Consider the specific heat capacity of the following metals.
Metal
Specific heat capacity / J kg–1 K–1
Cu
385
Ag
234
Au
130
Pt
134
Which metal will show the greatest temperature increase if 50 J of heat is supplied to a 0.001 kg
sample of each metal at the same initial temperature?
A.
47.
Cu
B.
Ag
C.
Au
D.
Pt
Consider the following reactions.
8
HӨ = 395 kJ mol1
S(s) + 1 12 O2(g)  SO3(g)
SO2(s) +
1
2
HӨ = 98 kJ mol1
O2(g)  SO3(g)
What is the HӨ value (in kJ mol–1) for the following reaction?
S(s) + O2(g)  SO2(g)
A.
48.
–297
B.
+297
– 493
C.
D.
+493
The following reaction is spontaneous only at temperatures above 850C.
CaCO3(s)  CaO(s) + CO2(g)
Which combination is correct for this reaction at 1000C?
49.
50.
G
H
S
A.
–
–
–
B.
+
+
+
C.
–
+
+
D.
+
–
–
Which statement is correct for an endothermic reaction?
A.
Bonds in the products are stronger than the bonds in the reactants.
B.
Bonds in the reactants are stronger than the bonds in the products.
C.
The enthalpy of the products is less than that of the reactants.
D.
The reaction is spontaneous at low temperatures but becomes non-spontaneous at high
temperatures.
Consider the following information.
Compound
C6H6(l)
CO2(g)
H2O(l)
HfӨ / kJ mol–1
+49
+394
–286
C6H6(l) + 7 12 O2(g)  6CO2(g) + 3H2O(l)
Which expression gives the correct value of the standard enthalpy change of combustion for
benzene (l), in kJ mol–1?
51.
52.
A.
12(394) + (286) 2(49)
B.
12(394) + 6(286) 2(49)
C.
6(394) + 3(286)  (49)
D.
6(394) + 3(286)  (49)
Which equation represents the lattice enthalpy of magnesium oxide?
1
2
A.
Mg(s) +
C.
Mg2+(g) +
O2(g)  MgO(s)
1
2
O2(g)  MgO(s)
B.
Mg2+(g) + O2–(g)  MgO(g)
D.
Mg2+(g) + O2(g)  MgO(s)
According to the enthalpy level diagram below, what is the sign for H and what term is used to
9
refer to the reaction?
H
reactants
products
reaction progress
53.
H
reaction
A.
positive
endothermic
B.
negative
exothermic
C.
positive
exothermic
D.
negative
endothermic
When 40 joules of heat are added to a sample of solid H2O at –16.0°C the temperature increases
to –8.0°C. What is the mass of the solid H2O sample? [Specific heat capacity of H2O(s) = 2.0 J
g–1K–1]
A.
54.
2.5 g
B.
5.0 g
C.
10 g
D.
160 g
The HӨ values for the formation of two oxides of nitrogen are given below.
1
2
HӨ = –57 kJ mol–1
N2(g) + O2(g)  NO2(g)
HӨ = +9 kJ mol–1
N2(g) + 2O2(g)  N2O4(g)
Use these values to calculate HӨ for the following reaction (in kJ):
2NO2(g)  N2O4(g)
A.
55.
56.
–105
B.
– 48
C.
+66
D.
+123
The HӨ and SӨ values for a reaction are both negative. What will happen to the spontaneity
of this reaction as the temperature is increased?
A.
The reaction will become more spontaneous as the temperature is increased.
B.
The reaction will become less spontaneous as the temperature is increased.
C.
The reaction will remain spontaneous at all temperatures.
D.
The reaction will remain non-spontaneous at any temperature.
How much energy, in joules, is required to increase the temperature of 2.0 g of aluminium from
25 to 30°C? (Specific heat of Al = 0.90 J g–1 K–1).
A.
0.36
B.
4.5
C.
9.0
D.
54
10
57.
58.
Which combination is correct for a chemical reaction that absorbs heat from the surroundings?
Type of reaction
ΔH at constant pressure
A.
Exothermic
Positive
B.
Exothermic
Negative
C.
Endothermic
Positive
D.
Endothermic
Negative
Using the equations below:
C(s) + O2(g) → CO2(g)
∆Hο = –394 kJ mol–1
Mn(s) + O2(g) → MnO2(s)
∆Hο = –520 kJ mol–1
What is ∆H, in kJ, for the following reaction?
MnO2(s) + C(s) → Mn(s) + CO2(g)
A.
59.
60.
61.
62.
63.
914
B.
126
–126
C.
D.
–914
Which reaction has the most negative ∆Hο value?
A.
LiF(s) → Li+(g) + F–(g)
B.
Li+(g) + F–(g) → LiF(s)
C.
NaCl(s) → Na+(g) + Cl–(g)
D.
Na+(g) + Cl–(g) → NaCl(s)
Which equation represents the electron affinity of calcium?
A.
Ca(g) →Ca+(g) + e–
B.
Ca(g) →Ca–(g) + e–
C.
Ca(g) + e– → Ca–(g)
D.
Ca+(g) + e– → Ca(g)
Which reaction causes a decrease in the entropy of the system?
A.
CaCO3(s) → CaO(s) + CO2(g)
B.
2H2(g) + O2(g) → 2H2O(l)
C.
2C(s) + O2(g) → 2CO(g)
D.
2SO3(g) → 2SO2(g) + O2(g)
What are the signs of ∆Hο and ∆Sο for a reaction that is non-spontaneous at low temperature but
spontaneous at high temperature?
Hο
Sο
A.
–
–
B.
+
–
C.
–
+
D.
+
+
Given the following data:
C(s) + F2(g) → CF4(g); ∆H1 = –680 kJ mol–1
F2(g) → 2F(g); ∆H2 = +158 kJ mol–1
C(s) → C(g); ∆H3 = +715 kJ mol–1
11
calculate the average bond enthalpy (in kJ mol–1) for the C––F bond.
64.
For the process: C6H6(l)  C6H6(s)
the standard entropy and enthalpy changes are:
∆Hο = –9.83kJ mol–1 and ∆Sο = –35.2J K–1 mol–1.
Predict and explain the effect of an increase in temperature on the spontaneity of the process.
65.
Explain in terms of Gο, why a reaction for which both Hο andSο values are positive can
sometimes be spontaneous and sometimes not.
66.
Consider the following reaction. N2(g) +3H2(g) → 2NH3(g)
(i)
Use values from Table 10 in the Data Booklet to calculate the enthalpy change, Hο, for
this reaction.
(ii)
The magnitude of the entropy change, S, at 27°C for the reaction is 62.7 J K–1 mol–1.
State, with a reason, the sign of S. (2)
(iii)
Calculate G for the reaction at 27°C and determine whether this reaction is spontaneous
at this temperature.
67.
Explain in terms of Gο, why a reaction for which bothHο and Sο are positive is sometimes
spontaneous and sometimes not.
68.
Consider the following reaction.
N2(g) + 3H2(g) → 2NH3(g)
(i)
Using the average bond enthalpy values in Table 10 of the Data Booklet, calculate the
standard enthalpy change for this reaction. (4)
(ii)
The absolute entropy values, S, at 300 K for N2(g), H3(g) and NH2(g) are 193, 131 and
192 JK–1 mol–1 respectively. Calculate Sο for the reaction and explain the sign of Sο.
iii)
Calculate Gο for the reaction at 300 K.
(iv) If the ammonia was produced as a liquid and not as a gas, state and explain the effect this
would have on the value of Hο for the reaction.
(2)
69.
Define the term standard enthalpy of formation, and write the equation for the standard enthalpy
of formation of ethanol.
70.
Two reactions occurring in the manufacture of sulfuric acid are shown below:
reaction I
S(s) +O2(g)  SO2(g)
reaction II
SO2(g) +
(i)
1
2
O2(g)
HӨ = –297 kJ
SO3(g)
HӨ = –92 kJ
State the name of the term HӨ. State, with a reason, whether reaction I would be
accompanied by a decrease or increase in temperature.
(3)
(ii)
At room temperature sulfur trioxide, SO3, is a solid. Deduce, with a reason, whether the
HӨ value would be more negative or less negative if SO3(s) instead of SO3(g) were
12
formed in reaction II.
(2)
(iii)
Deduce the HӨ value of this reaction:
S(s) + 1 12 O2(g)  SO3(g)
(1)
71.
(i)
Define the term average bond enthalpy.
(3)
(ii)
Explain why Br2 is not suitable as an example to illustrate the term average bond
enthalpy.
(1)
(iii)
Using values from Table 10 of the Data Booklet, calculate the enthalpy change for the
following reaction:
CH4(g) + Br2(g)  CH3Br(g) + HBr(g)
(3)
(iv)
Sketch an enthalpy level diagram for the reaction in part (iii).
(2)
(v)
Without carrying out a calculation, suggest, with a reason, how the enthalpy change for
the following reaction compares with that of the reaction in part (iii):
CH3Br(g) + Br2(g)  CH2Br2(g) + HBr(g)
(2)
72.
Throughout this question, use relevant information from the Data Booklet.
(a)
Define the term standard enthalpy change of formation, and illustrate your answer with
an equation, including state symbols, for the formation of nitric acid.
(4)
(b) Propyne undergoes complete combustion as follows:
C3H4(g) + 4O2(g)  3CO2(g) + 2H2O(l)
Calculate the enthalpy change of this reaction, given the following additional values:
HfӨ of CO2(g) = –394 kJ mol–1
HfӨ of H2O(l) = –286 kJ mol–1
(4)
(c)
Predict and explain whether the value of SӨ for the reaction in part (b) would be
negative, close to zero, or positive.
(3)
73.
(a)
Propyne reacts with hydrogen as follows:
C3H4(g) + 2H2(g)  C3H8(g)
HӨ = –287 kJ
Calculate the standard entropy change of this reaction, given the following additional
information:
SӨ of H2(g) = 131 J K–1 mol–1
(3)
13
(b) Calculate the standard free energy change at 298 K, GӨ, for the reaction in part (a). Use
your answer and relevant information from part (d). If you did not obtain an answer to
part (a), use SӨ = –360 J K–1 (this is not the correct value).
(3)
74.
(a)
The lattice enthalpy of an ionic compound can be calculated using a Born-Haber cycle.
Using lithium fluoride as the example, construct a Born-Haber cycle, labeling the cycle
with the formulas and state symbols of the species present at each stage.
(6)
(b)
Two values of the lattice enthalpies for each of the silver halides are quoted in the Data
Booklet. Discuss the bonding in silver fluoride and in silver iodide, with reference to
these values.
(2)
75.
But–1–ene gas, burns in oxygen to produce carbon dioxide and water vapour according to the
following equation.
C4H8 + 6O2  4CO2 + 4H2O
(a)
Use the data below to calculate the value of HӨ for the combustion of but-1-ene.
Bond
CC
C=C
CH
O=O
C=O
O–H
348
612
412
496
743
463
Average bond
enthalpy / kJ
mol–1
(3)
(b) State and explain whether the reaction above is endothermic or exothermic.
(1)
(Total 4 marks)
76.
Calculate the enthalpy change, H4 for the reaction
C + 2H2 +
1
2
O2  CH3OH
H4
using Hess’s Law and the following information.
CH3OH + 1 12 O2  CO2 + 2H2O
H1 = 676 kJ mol1
C + O2  CO2
H2 = 394 kJ mol1
H2 +
77.
1
2
H3 = 242 kJ mol1
O2  H2O
Hex-1-ene gas, C6H12, burns in oxygen to produce carbon dioxide and water vapour.
(a)
Write an equation to represent this reaction.
(b) Use the data below to calculate the values of HcӨ and ScӨ for the combustion of
hex-1-ene.
O2(g)
C6H12(g)
CO2(g)
H2O(g)
Standard enthalpy of
formation, HfӨ / kJ1 mol
0.0
–43
–394
–242
Entropy, SӨ / J K1 mol1
205
385
214
189
Substance
(i)
Value of HcӨ (2)
14
(ii)
(c)
(d)
78.
Value of ScӨ (2)
Calculate the standard free energy change for the combustion of hex-1-ene. (2)
State and explain whether or not the combustion of hex-1-ene is spontaneous at 25C. (1)
Calculate the enthalpy change, H4 for the reaction
C + 2H2 +
1
2
O2  CH3OH
H4
using Hess’s Law, and the following information.
CH3OH + 1 12 O2  CO2 + 2H2O
H1 = 676 kJ mol1
C + O2  CO2
H2 = 394 kJ mol1
H2 +
79.
1
2
H3 = 242 kJ mol1
O2  H2O
Methylamine can be manufactured by the following reaction.
CH3OH(g) + NH3(g)  CH3NH2(g) + H2O(g)
(a)
Define the term average bond enthalpy. (2)
(b) Use information from Table 10 of the Data Booklet to calculate the enthalpy change for
this reaction. (4)
80.
Methylamine can be manufactured by the following reaction.
CH3OH(g) + NH3(g)  CH3NH2(g) + H2O(g)
(a)
Define the term standard enthalpy change of formation. (2)
(b)
The values of standard enthalpy changes of formation for some compounds are shown in
the table.
Compound
HfӨ / kJ mol–1
NH3(g)
– 46
H2O(g)
– 242
Predict, with a reason, whether the value of HfӨ for H2O(l) is less than, greater than, or
equal to, the value of HfӨ for H2O(g). (2)
(c)
81.
(a)
Use information from the table in (b) and from Table 11 of the Data Booklet to calculate
the enthalpy change for the reaction used to manufacture methylamine. (3)
Define the term average bond enthalpy. (2)
(b) Use the information from Table 10 in the Data Booklet to calculate the enthalpy change
for the complete combustion of but-1-ene according to the following equation
C4H8(g)  4CO2(g) + 4H2O(g)
(c)
(d)
(3)
Predict, giving a reason, how the enthalpy change for the complete combustion of
but-2-ene would compare with that of but-1-ene based on average bond enthalpies.
The enthalpy level diagram for a certain reaction is shown below.
15
Enthalpy
HR
enthalpy of
reactants
HP
enthalpy of
products
Time
State and explain the relative stabilities of the reactants and products.
(2)
82.
(a)
Define the term standard enthalpy change of formation, HfӨ.
(2)
(b)
(i)
Use the information in the following table to calculate the enthalpy change for the
complete combustion of but-1-ene according to the following equation.
C4H8(g) + 6O2(g)  4CO2(g) + 4H2O(g)
Compound
C4H8(g)
CO2(g)
H2O(g)
HfӨ / kJ mol–1
+1
– 394
– 242
(3)
(ii)
Deduce, giving a reason, whether the reactants or the products are more stable.
(2)
(iii)
83.
84.
Predict, giving a reason, how the enthalpy change for the complete combustion of
but-2-ene would compare with that of but-1-ene based on average bond enthalpies.
(1)
The reaction between ethene and hydrogen gas is exothermic.
(i)
Write an equation for this reaction. (1)
(ii)
Deduce the relative stabilities and energies of the reactants and products. (2)
(iii)
Explain, by referring to the bonds in the molecules, why the reaction is exothermic. (2)
(i)
Define the term standard enthalpy change of formation, HfӨ. (2)
(ii)
Construct a simple enthalpy cycle and calculate the value of HfӨ (C2H5OH(l)) given the
following data.
Compound
HfӨ / kJ mol–1
H2O(l)
–286
CO2(g)
–394
C2H5OH(l)
ΔHӨcomb/ kJ mol–1
–1371
(5)
16
85.
(i)
Define the term average bond enthalpy.
(2)
(ii)
The equation for the reaction of ethyne and hydrogen is:
C2H2(g) + 2H2(g)  C2H6(g)
Use information from Table 10 of the Data Booklet to calculate the change in enthalpy
for the reaction.
(2)
(iii)
State and explain the trend in the bond enthalpies of the C–Cl, C–Br and C–I bonds.
(2)
86.
Consider the following reaction:
N2(g) + 3H2(g)
(i)
2NH3(g)
Suggest why this reaction is important for humanity. (1)
(ii) Using the average bond enthalpy values in Table 10 of the Data Booklet, calculate the
standard enthalpy change for this reaction. (4)
(iii) The absolute entropy values, S, at 238 K for N2(g), H2(g) and NH3(g) are 192, 131 and
193 J K–1 mol–1 respectively. Calculate ∆Sο for the reaction and explain the sign of ∆Sο.
(2)
(iv) Calculate ∆Gο for the reaction at 238 K. State and explain whether the reaction is
spontaneous. (3)
(v)
87.
(i)
If ammonia was produced as a liquid and not as a gas, state and explain the effect this
would have on the value of ∆Hο for the reaction. (2)
Define the terms lattice enthalpy and electron affinity. (2)
(ii) Use the data in the following table and from the data booklet to construct the Born-Haber
cycle for sodium chloride, NaCl, and determine the lattice enthalpy of NaCl(s). (4)
Na(s) +
1
Cl2(g) → NaCl(g)
2
Na(s) → Na(g)
∆Hο = –411 kJ mol–1
∆Hο = +108 kJ mol–1
(iii) Describe the structure of sodium chloride. (2)
88.
Number of
molecules
Ea
Energy
The diagram shows the distribution of energy for the molecules in a sample of gas at a given
temperature, T1.
(a)
In the diagram Ea represents the activation energy for a reaction. Define this term.
17
(b) On the diagram above draw another curve to show the energy distribution for the same
gas at a higher temperature. Label the curve T2. (2)
89.
(c)
With reference to your diagram, state and explain what happens to the rate of a reaction
when the temperature is increased. (2)
(a)
Define the term average bond enthalpy, illustrating your answer with an equation for
methane, CH4. (3)
(b) The equation for the reaction between methane and chlorine is
CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
Use the values from Table 10 of the Data Booklet to calculate the enthalpy change for
this reaction.
(3)
(c)
Explain why no reaction takes place between methane and chlorine at room
temperature unless the reactants are sparked, exposed to UV light or heated. (2)
(d) Draw an enthalpy level diagram for this reaction.
90.
The equation for the decomposition of calcium carbonate is given below.
CaCO3(s) → CaO(s) + CO2(g)
At 500 K, ∆H for this reaction is +177 kJ mol–1 and ∆S is 161 J K–1 mol–1.
(a)
Explain why ∆H for the reaction above cannot be described as ∆Hfο.
(b) State the meaning of the term ∆S.
(c)
91.
Calculate the value of ∆G at 500 K and determine, giving a reason, whether or not the
reaction will be spontaneous.
In aqueous solution, potassium hydroxide and hydrochloric acid react as follows.
KOH(aq) + HCl(aq) → KCl(aq)+ H2O(l)
The data below is from an experiment to determine the enthalpy change of this reaction.
50.0 cm3 of a 0.500 mol dm–3 solution of KOH was mixed rapidly in a glass beaker with
50.0 cm3 of a 0.500 mol dm–3 solution of HCl.
Initial temperature of each solution = 19.6°C
Final temperature of the mixture = 23.1°C
(a)
(b)
(c)
State, with a reason, whether the reaction is exothermic or endothermic. (1)
Explain why the solutions were mixed rapidly. (1)
Calculate the enthalpy change of this reaction in kJ mol–1. Assume that the specific heat
capacity of the solution is the same as that of water. (4)
(d) Identify the major source of error in the experimental procedure described above.
Explain how it could be minimized. (2)
(e)
The experiment was repeated but with an HCl concentration of 0.510 mol dm–3 instead
18
of 0.500 mol dm–3. State and explain what the temperature change would be. (2)
92.
The standard enthalpy change for the combustion of phenol, C6H5OH(s), is –3050 kJ mol–1
at 298 K.
(a)
Write an equation for the complete combustion of phenol (1)
(b) The standard enthalpy changes of formation of carbon dioxide, CO2(g), and of water,
H2O(l), are –394 kJ mol–1 and –286 kJ mol–1 respectively.
Calculate the standard enthalpy change of formation of phenol, C6H5OH(s).
(c)
(3)
The standard entropy change of formation, ∆Sο, of phenol, C6H5OH(s) at
298 K is –385 J K–1 mol –1. Calculate the standard free energy change of formation,
∆Gο, of phenol at 298 K. (3)
(d) Determine whether the reaction is spontaneous at 298 K, and give a reason. (2)
(e)
93.
Predict the effect, if any, of an increase in temperature on the spontaneity of this reaction.
(2)
The data below is from an experiment used to measure the enthalpy change for the combustion
of 1 mole of sucrose (common table sugar), C12H22O11(s). The time-temperature data was taken
from a data-logging software programme.
Mass of sample of sucrose, m = 0.4385 g
Heat capacity of the system, Csystem = 10.114 kJ K–1
(a)
Calculate ΔT, for the water, surrounding the chamber in the calorimeter. (1)
19
(b)
(c)
Determine the amount, in moles, of sucrose. (1)
(i)
Calculate the enthalpy change for the combustion of 1 mole of sucrose. (1)
(ii) Using Table 12 of the Data Booklet, calculate the percentage experimental error
based on the data used in this experiment. (1)
(d)
A hypothesis is suggested that TNT, 2-methyl-1,3,5-trinitrobenzene, is a powerful
explosive because it has:
• a large enthalpy of combustion
• a high reaction rate
• a large volume of gas generated upon combustion
Use your answer in part (c)(i) and the following data to evaluate this hypothesis:
Equation for combustion
Relative
rate of
combustion
Sucrose
C12H22O11(s) + 12O2(g)  12CO2(g) + 11H2O(g)
Low
TNT
2C7H5N3O6(s)  7CO(g) + 7C(s) + 5H2O(g) + 3N2(g)
High
Enthalpy of
combustion
/ kJ mol–1
3406
(3)
94.
(a)
Use the information from Table 10 of the Data Booklet to calculate the enthalpy change
for the complete combustion of but-1-ene, according to the following equation.
C4H8(g) + 6O2(g) → 4CO2(g) + 4H2O(g)
(3)
20
ANSWERS
95.
D
96.
C
97.
B
98.
D
99.
D
100. D
101. D
102. B
103. A
104. A
105. C
106. B
107. B
108.
A
109. B
110.
B
111. C
112.
B
113. C
114.
D
115. C
116.
B
117. A
118. A
119. D
120. D
121. A
122.
123. C
124. B
125. A
126. C
127. C
128. A
129. D
130. C
131. A
132. C
133. D
134. A
135. B
136. C
137. C
138. A
139. D
140. C
141. A
142. C
143. B
144. C
145. D
146. B
147. A
148. D
149. B
150. C
151. C
152. B
153. B
154. C
155. B
156. D
D
157. C(s) + 2F2(g) → CF4(g)
∆H1 = –680 kJ;
4F(g) → 2F2(g)
∆H2 = 2(–158) kJ;
C(g) → C(s)
∆H3 = –715 kJ;
Accept reverse equations with +∆H values.
C(g) + 4F(g) → CF4(g)
∆H = –1711 kJ,
so average bond enthalpy = – 1711
4
= –428 kJ mol–1;
4
Accept + or – sign.
21
Lots of ways to do this! The correct answer is very different
from the value in the Data Booklet, so award [4] for final
answer with/without sign units not needed, but deduct [1] if
incorrect units. Accept answer in range of 427 to 428 without
penalty for sig figs.
If final answer is not correct use following;
Award [1] for evidence of cycle or enthalpy diagram or adding
of equations.
Award [1] for 2F2 (g) → 4F(g) 2×158 seen.
Award [1] for dividing 1711 or other value by 4.
[4]
158. (∆Gο = ∆H – T∆Sο)
as T increases, –T∆Sο becomes larger/more positive;
∆G increases/becomes more positive/less negative;
process becomes less spontaneous/reverse reaction favoured;
3
[3]
ο
ο
159. a reaction is spontaneous when ∆G is negative/non-spontaneous when ∆G is positive;
at high T, ∆Gο is negative;
(because) T∆Sο is greater than ∆Hο;
at low T, ∆Gο is positive because T∆Sο is smaller than ∆Hο/OWTTE;
4
[4]
160. (i)
selection of all the correct bonds or values from Data Booklet;
∆H = (N≡≡N) + 3(H—H) – 6(N—H) / 944 + 3(436) – 6(388);
= –76 (kJ);
Allow ECF for one error (wrong bond energy/wrong
coefficient/reverse reaction) but not for two errors
(so –611, –857, +76, +1088 all score 2 out of 3).
(ii) negative;
decrease in the number of gas molecules/OWTTE;
(iii)
∆G = ∆H – T∆S
∆G = –76.0 – 300 (–0.0627);
Award [1] for 300 K.
Award [1] for conversion of units J to kJ or vice versa.
Allow ECF from (i) from ∆H.
Allow ECF from (ii) for sign of ∆S.
3
2
3
= –57.2 (kJ mol–1) is spontaneous/or non-spontaneous if positive value
obtained;
[3 max]
[8]
161. a reaction is spontaneous when ∆Gο is negative;
at high T, ∆Gο is negative;
–T∆S ο is larger/greater than ∆Hο;
at low T, ∆Gο is positive because –T∆Sο is smaller than ∆Hο/OWTTE;
4
[4]
162. (i)
∆H = (sum of energies of bonds broken) – (sum of energies of bonds formed);
Can be implied by working.
Correct substitution of values and numbers of bonds broken;
Correct substitution of values and numbers of bonds made;
(∆H = (N≡≡N) + 3(H—H) – 6(N—H) = 944 + 3(436) – 6(388) =) –76 (kJ);
4
Allow ECF.
Do not penalize for SF or units.
(ii) ∆Sο = (sum of entropies of products) – (sum of entropies of reactants);
3
22
Can be implied by working.
(= 2×192 – (193 + 3×131) =) –202(J K–1 mol–1);
four molecules make two molecules/fewer molecules of gas;
(iii)
(∆Gο =∆Hο – T∆Sο = –76.0 – 300(–0.202)) = – 15.4 (kJ mol –1);
Do not penalize for SF.
1
(iv)
∆Hο becomes more negative;
heat released when gas → liquid;
2
[10]
163. enthalpy change associated with the formation of one mole of a
compound/substance; from its elements;
in their standard states/under standard conditions;
2C(s) + 3H2(g) +
1
2
O2(g) → C2H5OH(l);
5
Award [1] for formulas and coefficients, [1] for state symbols.
[5]
164. (a)
(i)
(ii)
(iii)
standard enthalpy (change) of reaction;
(temperature) increase;
reaction is exothermic/sign of H is negative;
3
more (negative);
heat given out when gas changes to solid/solid has less enthalpy than
gas/OWTTE;
2
–389 kJ;
1
[6]
165. (i)
the energy needed to break one bond;
(in a molecule in the) gaseous state;
value averaged using those from similar compounds;
3
(ii)
it is an element/no other species with just a Br-Br bond/OWTTE;
1
(iii)
(sum bonds broken =) 412 + 193 = 605;
(sum bonds formed =) 276 + 366 = 642;
(H =) –37 kJ;
Award [3] for correct final answer.
Award [2] for “+ 37”.
Accept answer based on breaking and making extra C-H bonds.
3
(iv)
CH4 + Br2
Enthalpy
;
CH3Br + HBr
2
Award [1] for enthalpy label and two horizontal lines, [1] for
reactants higher than products.
ECF from sign in (iii), ignore any higher energy level involving
atoms.
(v)
(about) the same/similar;
same (number and type of) bonds being broken and formed;
2
[11]
166. (a)
the enthalpy/energy/heat change for the formation of one mole of a
compound/substance from its elements;
in their standard states/under standard conditions/at 298 K and 1 atm;
1
2
H 2 (g)  12 N 2 (g)  1 12 O 2 (g)  HNO3 (l) ;
4
23
Award [1] for correctly balanced equation, [1] for all state
symbols correct.
Do not award equation mark if 2HNO3 formed.
(b)
Hr = ∑HfӨ (products)  ∑HfӨ (reactants)/suitable cycle;
= 3(  394) + 2(  286)  185;
Award [1] for correct coefficients of CO2 and H2O values, [1]
for correct value for C3H4 from Data Booklet.
= 1939 or 1940 kJ;
Ignore units.
Award [4] for correct final answer.
Award [3] for +1939 or  1569.
(c)
4
negative;
decrease in disorder/increase in order;
5 mol of gas  3 mol of gas/reduction in number of gas moles;
Award [1] for answer of close to zero based on use of H2O(g).
3
[11]
167. (a)
S = SӨ (products)  SӨ (reactants)/suitable cycle;
= 270  248  2131;
=  240 (J K1);
Units not needed for mark, but penalize incorrect units.
Award [3] for correct final answer.
(b)
3
ΔGӨ =  287  (2980.240);
Award [1] for correct substitution of values and [1] for
conversion of units.
= 215 kJ;
3
Units needed for mark.
Apply ECF from  360 kJ or incorrect answer from (a).
[6]
168. (a)
6
Li +(g) + e- +F(g)
Li +(g) + e- + 21 F2(g)
Li + (g) + F- (g)
Li(g) + 12 F2(g)
Li(s) + 12 F2(g)
LiF(s)
Award [6] for completely correct cycle, with endothermic
processes in any order.
Deduct [1] for each line in which species symbol and/or state
symbol is incorrect or missing.
24
Penalize missing electrons once only.
(b)
bonding in AgF more ionic than in AgI/bonding in AgI more covalent than
in AgF;
Accept AgF is ionic and AgI is covalent.
values closer/in better agreement in AgF/big(ger) difference in values for
AgI/OWTTE;
2
[8]
169. (a)
(Amount of energy required to break bonds of reactants)
8×412 + 2×348 + 612 + 6×496/7580 (kJ mol1);
(Amount of energy released during bond formation)
4×2×743 + 4×2×463/9648 (kJ mol1);
H = 2068 (kJ or kJ mol1);
3
ECF from above answers.
Correct answer scores [3].
Award [2] for (+)2068.
If any other units apply 1(U), but only once per paper.
(b) exothermic and HӨ is negative/energy is released;
Apply ECF to sign of answer in part (a).
Do not mark if no answer to (a).
1
[4]
170. 1×H1/676;
1×H2/–394;
2×H3/– 484;
H4 = 202 (kJ mol1 );
4
Accept alternative methods.
Correct answers score [4].
Award [3] for (+)202 or (+)40 (kJ/kJ mol1).
1(U) if units incorrect (ignore if absent).
[4]
171. (a)
(b)
C6H12 + 9O2  6CO2 + 6H2O;
(i)
(HӨ = ∑HfӨproducts  ∑HfӨreactants)
HӨ = (6×394 + 6×242)  (43);
HӨc = 3773/3.8×103 (kJ mol1);
Accept 2, 3 or 4 sig. fig..
1
2
Award [1] for + 3773/+ 3.8 ×103 (kJ mol1).
Allow ECF from (a) only if coefficients used.
(ii)
SӨ = (SpӨ  SrӨ) = (6×189 + 6×214)  (385 + 9×205);
SӨ =188 (J K1 mol1 );
2
Accept only 3 sig. fig..
Award [1] for –188.
Allow ECF from (a) only if coefficients used.
(c)
(GӨc = HӨc  TSӨc) = 3800  (298×0.188);
=  3900 kJ mol1.
Accept  3800 to  3900.
Accept 2, 3 or 4 sig. fig.
Allow ECF from (b).
2
25
Units needed for second mark.
(d)
spontaneous and GӨ negative;
Allow ECF from (c).
1
[8]
172.  1×H1/676;
1×H2/ 394;
2×H3/ 484;
H4 = 202 (kJ mol1);
Accept alternative methods.
Correct answers score [4].
4
Award [3] for (+)202 or (+)40 (kJ/kJ mol1).
[4]
173. (a)
(b)
energy needed to break (1 mol of) a bond in a gaseous molecule;
averaged over similar compounds;
2
bonds broken identified as CO and NH;
bonds formed identified as CN and OH;
H = 748  768 (kJ);
=  20 kJ/kJ mol1 (units needed for this mark);
If wrong bonds identified apply ECF to 3rd and 4th marks.
Accept answer based on breaking and making all bonds.
Award [4] for correct final answer.
Award max [3] if only one bond missed.
4
Answer of 20 or +20 kJ (mol1 ) scores [3].
174. (a)
enthalpy/energy change for the formation of 1 mol of a compound from
its elements;
Do not accept heat needed to form 1 mol…
in their standard states/under standard conditions/at 298 K and 1 atm;
(b) greater value/more negative value;
energy given out when steam condenses/turns to water;
(c)
2
2
HӨ = ∑HfӨ (products)  ∑HfӨ (reactants)/suitable cycle;
= (28242)(20146);
= 23 kJ/kJ mol1;
Units needed for 3rd mark.
Correct final answer scores [3].
3
23 or +23 kJ/kJ mol1 scores [2].
If 239 used instead of 201 for CH3OH, award [2] for +15 kJ.
[7]
175. (a)
(b)
amount of energy needed to break one mole of (covalent) bonds;
in the gaseous state;
average calculated from a range of compounds;
Award [1] each for any two points above.
2
bonds broken: 161 + 2×348 + 8×412 + 6×496/7580 kJ mol1;
bonds made: 8×743 + 8×463/9648 kJ mol1;
(bonds broken  bonds made =) H = 2068(kJ mol1);
Award [3] for the correct answer.
3
26
Allow full ECF  1 mistake equals 1 penalty.
Allow kJ but not other wrong units.
(c)
same/equal, because the same bonds are being broken and formed;
1
(d) products more stable than reactants;
bonds are stronger in products than reactants/HP < HR/enthalpy/stored
energy of products less than reactants;
2
[8]
176. (a)
(b)
the enthalpy change when one mole of compound is formed from its elements
in their (standard state);
at (standard conditions of) 298 K/25C and 101 325 Pa/1 atm;
2
HP = (4×242 + 4×394) kJ mol1;
(i)
HR = 1 kJ mol1;
HӨ = (∑HӨ p∑HӨR) = 2545 /2.55×103/ 2550 (kJ mol1);
3
Allow ECF.
(ii)
(iii)
products more stable than reactants;
bonds are stronger in products than reactants/Hp < HR/enthalpy/stored
energy of products less than reactants;
2
same/equal, because the same bonds are being broken and formed;
1
[8]
177. (a)
(i)
C2H4(g) + H2(g)  C2H6(g);
State symbols not required for mark
(ii)
products more stable than reactants/reactants less stable than products;
products lower in energy/reactants higher in energy;
(iii)
1
2
(overall) bonds in reactants weaker/(overall) bonds in product stronger
/all bonds in product are  bonds/weaker  bond broken and a
(stronger)  bond formed;
less energy needed to break weaker bonds/more energy produced
to make stronger bonds (thus reaction is exothermic)/OWTTE;
OR
bond breaking is endothermic/requires energy and bond making is
exothermic/releases energy;
stronger bonds in product mean process is exothermic overall;
2
[5]
178. (i)
change in energy for the formation of (1 mol) of a substance from its
elements; under standard conditions/1 atm pressure or 101 kPa and
298 K/25C;
2
(ii)

C 2 H 5 OH(l) + 3O 2 (g)

Hcomb

Hf (C 2 H 5OH(l))
2CO 2 (g) + 3H 2O(l)
H f (CO2 (g))
2C(s) + 2O2(g)
+

H f (H 2O(l))
3H 2 (g) + 1.5 O2 (g)
States not required.
Correct cycle showing:
27
HcombӨ
HfӨ (C2H5OH(l));
2HfӨ (CO2(g)) and 3HfӨ (H2O(l));
(HfӨ (C2H5OH(l)) = (2HfӨ (CO2(g)) + 3HfӨ (H2O(l))  HcombӨ
= 2(394) + 3(286) + 1371;
= 275 kJ mol1;
If values are substituted for symbols in the enthalpy cycle
diagram to give correct answer, award last [2] marks.
If no enthalpy cycle drawn but equation written and Hess’s
Law applied or calculated as follows, then [3 max]
5
(Hr = ∑Hf (products)  ∑Hf (reactants))
1371 = (394×2) + (286×3) Hf (ethanol);
Hf (ethanol) = 788  858 + 1371;
=  275(kJ mol1);
Award [2] for correct answer without enthalpy cycle and
without working and [1] for 275 or + 275.
[7]
179. (i)
(ii)
energy required to break (a mole of) bonds in the gaseous state
/energy given out when (a mole of) bonds are made in the
gaseous state;
average value from a number of similar compounds;
2
(HӨreaction = (∑BEbreak  BEmake))
= [(837) + 2(436)]  [(348 + 4(412)];
=  287(kJ/kJ mol1);
Award [1 max] for 287 or + 287.
(iii)
2
(BE): CCl > CBr > CI/CX bond becomes weaker;
halogen size/radius increases/bonding electrons further away from
the nucleus/bonds become longer;
2
[6]
180. (i)
fertilizers/increasing crop yields;
production of explosives for mining;
1 max
(ii) H = (sum of energies of bonds broken) – (sum of energies of bonds formed);
Can be implied by working.
correct substitution of values and numbers of bonds broken;
correct substitution of values and numbers of bonds made;
(H = (NN) + 3(H–H) – 6(N–H) = 944 + 3(436) – 6(388) =) –76.0 (kJ);
Allow ECF.
Do not penalize for sig. fig. or units.
Award [4] for correct final answer.
(iii) (Sο[2×193] – [192 + 3×131]) = –199 (J K–1 mol–1);
Allow ECF.
four gaseous molecules generating two gaseous
molecules/fewer molecules of gas;
(iv) (Gο = Hο – TSο = –76.0 – 298(–0.199)) = –16.7 (kJ);
Spontaneous;
G is negative;
Do not penalize for SF.
(v)
heat released when gas → liquid;
Hο becomes more negative;
4
2
3
2
28
[12]
181. (i)
lattice enthalpy for a particular ionic compound is defined as ΔH for the
process, MX(s) → M+(g) + X–(g);
Accept definition for exothermic process
electron affinity is the energy change that occurs when an electron is added
to a gaseous atom or ion;
(ii)
2
H f = –411 kJ mol –1
Na(s)
+
+108 kJ mol –1
1
2
Cl 2 (g)
+121 kJ mol –1
Cl(g)
Na(g)
+494 kJ mol –1
Na+ (g)
NaCl(s)
–364 kJ mol –1
+
–
Cl (g)
lattice enthalpy = –[(–411) – (+108) – (+494) – (+121) – (–364)]
= 770 (kJ mol–1)
Award [2] for all correct formulas in correct positions on cycle
diagram.
1 incorrect or missing label award [1].
Award [1] for all correct values in correct positions on cycle
diagram.
calculation of lattice enthalpy of NaCl(s) = 770 (kJ mol–1);
Allow ECF.
Accept alternative method e.g. energy level diagram.
(iii) lattice/network/regular structure;
each chloride ion is surrounded by six sodium ions and each sodium ion is
surrounded by six chloride ions/6:6 coordination;
4
2
[8]
182. (a)
(b)
(c)
activation energy = minimum energy required for a reaction to occur;
curve moved to the right;
peak lower,
Deduct [1] if shaded area smaller at T2 or if T2 line touches the
x-axis
rate increased;
as more molecules with energy  Ea;
1
2
2
[5]
183. (a)
(b)
energy for the conversion of a gaseous molecule into (gaseous) atoms;
(average values) obtained from a number of similar bonds/compounds/OWTTE;
CH4(g) → C(g) + 4H(g);
State symbols needed.
(bond breaking) = 1890/654;
(bond formation) = 2005/769;
enthalpy = –115(kJ mol–1)
Allow ECF from bond breaking and forming.
Award [3] for correct final answer.
Penalize [1] for correct answer with wrong sign.
3
3
29
(c)
molecules have insufficient energy to react (at room temperature)/
wrong collision geometry/unsuccessful collisions;
extra energy needed to overcome the activation energy/Ea for the reaction;
2
(d)
Ea
energy
reactants
products
reaction path
exothermic shown;
activation energy/Ea shown;
Allow ECF from (b).
2
[10]
(cannot be ο as) conditions are not standard/at 500 K/OWTTE;
(cannot be f as) not formation from elements/is decomposition/OWTTE;
2
(b)
change in entropy/degree of (dis)order (of system);
1
(c)
∆G = 177000 – (500×161) = +96500;
reaction is not spontaneous;
∆G is positive;
Allow ECF from calculation for last two marks.
184. (a)
3
[6]
185. (a)
(b)
(c)
exothermic because temperature rises/heat is released;
1
to make any heat loss as small as possible/so that all the heat will be
given out very quickly;
Do not accept “to produce a faster reaction”.
1
heat released = mass×specific heat capacity×temp increase/q = mc∆T =/
100×4.18×3.5;
= 1463 J/1.463 kJ; (allow 1.47 kJ if specific heat = 4.2)
amount of KOH/HCl used = 0.500×0.050 = 0.025 mol;
∆H = (1.463÷0.025) = –58.5 (kJ mol–1); (minus sign needed for mark)
Use ECF for values of q and amount used.
Award [4] for correct final answer.
Final answer of 58.5 or +58.5 scores [3].
Accept 2,3 or 4 significant figures.
4
(d) heat loss (to the surroundings);
insulate the reaction vessel/use a lid/draw a temperature versus time graph;
(e)
3.5°C/temperature change would be the same;
amount of base reacted would be the same/excess acid would not react/
KOH is the limiting reagent;
2
2
[10]
186. (a)
(b)
C6H5OH + 7O2 → 6CO2 + 3H2O;
Ignore state symbols.
1
∆Hrο =Σ∆Hfο products – Σ∆Hfο reactants;
–3050 = (6(–394) +3 (–286) – (∆Hfο phenol + O));
30
∆Hfο phenol =–172 kJ mol–1;
Award [3] for correct final answer.
Apply –1 (U) if appropriate.
3
Award [2 max] for ∆Hfο phenol = +172 kJ mol –1.
(c)
appropriate conversion of units;
∆G = –172 – 298(– 0.385)
= –57.3 kJ mol–1/–57 300 J mol–1;
Award [3] for correct final answer.
Accept answers in range –57.0 to –57.3 kJ mol–1.
Accept 3 sig. fig. only.
Allow ECF from (b).
Apply –1 (U) if appropriate.
3
(d) spontaneous;
since ∆G is negative;
Allow ECF from (c).
(e)
2
reaction becomes less spontaneous;
∆G becomes less negative/more positive;
Accept a suitable calculation.
Allow ECF from (c).
2
ΔT = 23.70 – 23.03 = 0.67 (°C/K);
1
(b)
 0.4385g

 = 1.281×10–3;
n  
1 
 342.34g mol 
1
(c)
(i)
187. (a)
(ii)
ΔHc = (C ΔT)/n =
 [(10.114 kJ K 1 )(0.67 K)]
(1.28110 3 mol)
Use ECF for values of T and n.
= –5.3×103 kJ mol–1;
1
 (5.3 103 )  (5.6 103 ) 
Percentage experimental error = 
 100 = 5.4%;
(5.6 103 )


Use ECF for values of ΔHc.
(d) enthalpy change of combustion of sucrose > TNT, and therefore not important;
rate of reaction for TNT is greater than that of sucrose, so this is valid;
amount of gas generated (in mol) for sucrose > than that of TNT
(according to the given equation), so this is not important;
1
3
[7]
188. (a)
The amount of energy needed to break 1 mole of (covalent) bonds;
in the gaseous state;
average calculated from a range of compounds;
Award [1] each for any two points above.
(b) Bonds broken
(612) + (2×348) + (8×412) + (6×496)/7580 (kJ mol–1);
Bonds made
(8×743) + (8×463) / 9648 (kJ mol–1);
H = –2068 (kJ mol–1);
Award [3] for the correct answer.
Allow full ECF.
Allow kJ but no other incorrect units.
2 max
3
31
Even if the first two marks are lost, the candidate can score [1]
for a clear correct subtraction for H.
32