The questions below are all on Hess’s Law. Each page has only one question on it, and the mark scheme for each question is on the page that immediately follows. A printable copy of the questions can be found here.
Question 17
This question is about the extraction of metals.
(a) Manganese can be extracted from Mn2O3 by reduction with carbon monoxide at high temperature.
(i) Use the standard enthalpy of formation data from the table and the equation for the extraction of manganese to calculate a value for the standard enthalpy change of this extraction.
Mn2O3(s) + 3CO(g) –> 2Mn(s) + 3CO2(g)
(3)
(ii) State why the value for the standard enthalpy of formation of Mn(s) is zero.
(1)
(b) Titanium is extracted in industry from titanium(IV) oxide in a two-stage process.
(i) Write an equation for the first stage of this extraction in which titanium(IV) oxide is converted into titanium(IV) chloride.
(2)
(ii) Write an equation for the second stage of this extraction in which titanium(IV) chloride is converted into titanium.
(2)
(c) Chromium is extracted in industry from chromite (FeCr2O4).
(i) In the first stage of this extraction, the FeCr2O4 is converted into Na2CrO4 Balance the equation for this reaction.
………FeCr2O4 + ………Na2CO3 + ………O2 –> ………Na2CrO4 + 2Fe2O3 + 8CO2
(1)
(ii) In the final stage, chromium is extracted from Cr2O3 by reduction with aluminium.
Write an equation for this reaction.
(1)
(Total 10 marks)
Q17.
(a) (i) M1 (could be scored by a correct mathematical expression which must have all ΔH symbols and the ⅀ or SUM)
Correct answer gains full marks
Credit 1 mark ONLY if –122 (kJ mol−1)
M1 ΔH = ⅀ΔHf (products) – ⅀ΔHf (reactants)
OR a correct cycle of balanced equations
M2 ΔH = 3(−394) − 3(−111) − (−971)
(This also scores M1)
M3 = (+) 122(kJ mol−1)
Award 1 mark ONLY for –122
For other incorrect or incomplete answers, proceed as follows
- check for an arithmetic error (AE), which is either a transposition error or an incorrect multiplication; this would score 2 marks (M1 and M2)
- If no AE, check for correct method; this requires either a correct cycle of balanced equations OR a clear statement of M1 which could be in words and scores M1 only
3
(ii) By definition
Ignore reference to “standard state”
OR
Because it is an element / elemental
1
(b) (i) TiO2 + 2Cl 2 + 2C –> TiCl4 + 2CO
Allow multiples
OR
TiO2 + 2Cl 2 + C –> TiCl4CO2
Ignore state symbols
M1 use of Cl2 and C
M2 a correct balanced equation
2
(ii) TiCl4 + 4Na –> Ti + 4NaCl
Allow multiples
OR
TiCl4 + 2Mg –> Ti + 2MgCl2
Ignore state symbols
M1 use of Na OR Mg
M2 a correct balanced equation
2
(c) (i) 4 FeCr2O4 + 8 Na2CO3 + 7O2 –> 8Na2CrO4 + 2Fe2O3 + 8CO2
Allow multiples
Ignore state symbols
1
(ii) Cr2O3 + 2Al –> Al2O3 + 2Cr
Allow multiples
Ignore state symbols
1
[10]
Q18.
A student used Hess’s Law to determine a value for the enthalpy change that occurs when anhydrous copper(II) sulfate is hydrated. This enthalpy change was labelled ΔHexp by the student in a scheme of reactions.
(a) State Hess’s Law.
(1)
(b) Write a mathematical expression to show how ΔHexp, ΔH1 and ΔH2 are related to each other by Hess’s Law.
(1)
(c) Use the mathematical expression that you have written in part (b), and the data book values for the two enthalpy changes ΔH1 and ΔH2 shown, to calculate a value
for ΔHexp
ΔH1 = −156 kJ mol−1
ΔH2 = +12 kJ mol−1
(1)
(d) The student added 0.0210 mol of pure anhydrous copper(II) sulfate to 25.0 cm3 of deionised water in an open polystyrene cup. An exothermic reaction occurred and the temperature of the water increased by 14.0 °C.
(i) Use these data to calculate the enthalpy change, in kJ mol−1, for this reaction of copper(II) sulfate. This is the student value for ΔH1
In this experiment, you should assume that all of the heat released is used to raise the temperature of the 25.0 g of water. The specific heat capacity of water is 4.18 J K−1 g−1.
(3)
(ii) Suggest one reason why the student value for ΔH1 calculated in part (d)(i) is less accurate than the data book value given in part (c).
(1)
(e) Suggest one reason why the value for ΔHexp cannot be measured directly.
(1)
(Total 8 marks)
Q18.
(a) The enthalpy change / heat (energy) change (at constant pressure) in a reaction is independent of the route / path taken (and depends only on the initial and final states)
Ignore the use of ΔH for enthalpy
1
(b) ΔHexp + ΔH2 – ΔH1 = 0
Any correct mathematical statement that uses all three terms
OR
ΔHexp + ΔH2 = ΔH1 OR ΔH1 = ΔHexp + ΔH2
OR
ΔHexp= ΔH1 – ΔH2 OR ΔHexp = ΔH1 +( – ΔH2 )
1
(c) ΔHexp = ΔH1 – ΔH2
ΔHexp = −156 −12 = −168 (kJ mol−1)
Ignore units
Award the mark for the correct answer without any working
1
(d) (i) M1 q = m c ΔT OR calculation (25.0 x 4.18 x 14.0)
Award full marks for correct answer
M2 = 1463J OR 1.46 kJ (This also scores M1)
In M1, do not penalise incorrect cases in the formula
M3 must have both the correct value within the range specified and the minus sign
Penalise M3 ONLY if correct numerical value but sign is incorrect; e.g. +69.5 to +69.7 gains 2 marks (ignore +70 after correct answer)
For 0.0210 mol, therefore
ΔH1 = − 69.67 to − 69.52 (kJ mol-1)
OR ΔH1 = − 69.7 to − 69.5 (kJ mol−1)
Penalise M2 for arithmetic error but mark on
Accept answers to 3sf or 4sf in the range − 69.7 to − 69.5
ΔT = 287, score q = m c ΔT only
Ignore -70 after correct answer
If c = 4.81 (leads to 1684J ) penalise M2 ONLY and mark on for M3 = − 80.17 (range − 80.0 to − 80.2)
Ignore incorrect units
3
(ii) The idea of heat loss
NOT impurity
OR
Incomplete reaction (of the copper sulfate)
NOT incompetence
OR
Not all the copper sulfate has dissolved
NOT incomplete combustion
1
(e) Impossible to add / react the exact / precise amount of water
Not just “the reaction is incomplete”
OR
Very difficult to measure the temperature rise of a solid
OR
Difficult to prevent solid dissolving
OR
(Copper sulfate) solution will form
1
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Question 19
Methanol (CH3OH) is an important fuel that can be synthesised from carbon dioxide.
(a) The table shows some standard enthalpies of formation.
(i) Use these standard enthalpies of formation to calculate a value for the standard enthalpy change of this synthesis.
(3)
(ii) State why the standard enthalpy of formation for hydrogen gas is zero.
(1)
(b) State and explain what happens to the yield of methanol when the total pressure is increased in this synthesis.
(3)
(c) The hydrogen required for this synthesis is formed from methane and steam in a reversible reaction. The equation for this reaction is shown below.
CH4(g) + H2O(g) –> C0(g) + 3H2(g) ∆H = +206 kJ mol–1
State and explain what happens to the yield of hydrogen in this reaction when the temperature is increased.
(3)
(d) The methanol produced by this synthesis has been described as a carbon-neutral fuel.
(i) State the meaning of the term carbon-neutral.
(1)
(ii) Write an equation for the complete combustion of methanol.
(1)
(iii) The equation for the synthesis of methanol is shown below.
Use this equation and your answer to part (d)(ii) to deduce an equation to represent the overall chemical change that occurs when methanol behaves as a carbon-neutral fuel.
(1)
(e) A student carried out an experiment to determine the enthalpy change when a sample of methanol was burned.
The student found that the temperature of 140 g of water increased by 7.5 °C when 0.011 mol of methanol was burned in air and the heat produced was used to warm the water.
Use the student’s results to calculate a value, in kJ mol–1, for the enthalpy change when one mole of methanol was burned.
(The specific heat capacity of water is 4.18 J K–1 g–1).
(3)
(Total 16 marks)
Q19.
(a) (i) M1 (could be scored by a correct mathematical expression which must have
all ∆Hsymbols and the ∑ or SUM)
M1 ΔHr = ΣΔHf (products) – ΣΔHf (reactants)
OR a correct cycle of balanced equations with 1C, 3H2 and 1O2
M2 ΔHr = – 201 + (– 242) – (– 394)
ΔHr = – 201 – 242 + 394
ΔHr = – 443 + 394
(This also scores M1)
M3 = – 49 (kJ mol–1)
(Award 1 mark ONLY for + 49)
Correct answer gains full marks
Credit 1 mark ONLY for + 49 (kJ mol–1)
For other incorrect or incomplete answers, proceed as follows
• check for an arithmetic error (AE), which is either
a transposition error or an incorrect multiplication;
this would score 2 marks (M1 and M2)
• If no AE, check for a correct method; this requires either
correct cycle of balanced equations with 1C, 3H2 and 1O2
OR a clear statement of M1 which could be in words and
scores only M1
3
(ii) It is an element / elemental
Ignore reference to “standard state”
OR
By definition
1
(b) M1 (The yield) increases / goes up / gets more
If M1 is given as “decreases” / “no effect” / “no change” then CE= 0 for clip, but mark on only M2 and M3 from a blank M1
M2 There are more moles / molecules (of gas) on the left / of reactants
OR fewer moles / molecules (of gas) on the right
/ products
OR there are 4 moles /molecules (of gas) on the left and 2 moles / molecules on the right.
OR (equilibrium) shifts / moves to the side with less moles / molecules
Ignore “volumes”, “particles” “atoms” and “species” for M2
M3: Can only score M3 if M2 is correct
The (position of) equilibrium shifts / moves (from left to right) to oppose the increase
in pressure
For M3, not simply “to oppose the change”
For M3 credit the equilibrium shifts / moves (to right) to lower / decrease the pressure
(There must be a specific reference to the change that is opposed)
3
(c) M1 Yield increases goes up
M2 The (forward) reaction / to the right is endothermic OR takes in/ absorbs
heat
OR
The reverse reaction / to the left is exothermic OR gives out / releases heat
If M1 is given as “decrease” / “no effect” / “no change” then CE= 0 for clip, but mark on only M2 and M3 from a blank M1
Can only score M3 if M2 is correct
M3 The (position of) equilibrium shifts / moves (from left to right) to oppose the increase
in temperature (QoL)
For M3, not simply “to oppose the change”
For M3, credit the (position of) equilibrium shifts / moves (QoL)
to absorb the heat OR
to cool the reaction OR
to lower the temperature
(There must be a specific reference to the change that is opposed)
3
(d) (i) An activity which has no net / overall (annual) carbon emissions to the
atmosphere
OR
An activity which has no net / overall (annual) greenhouse gas emissions
to the atmosphere.
OR
There is no change in the total amount / level of carbon dioxide /CO2 carbon /greenhouse gas present in the atmosphere.
The idea that the carbon /CO2 given out equals the carbon /CO2 that was taken in from the atmosphere
1
(ii) CH3OH + 1½ O2 CO2 + 2H2O
Ignore state symbols
Accept multiples
1
(iii) 3H2 + 1½ O2 –> 3H2O
Ignore state symbols
OR
Accept multiples
2H2 + O2 –> 2H2O
Extra species must be crossed through
1
(e) M1 q = m c ∆T
Award full marks for correct answer
Ignore the case for each letter
OR q = 140 × 4.18 × 7.5
M2 = 4389 (J) OR 4.389 (kJ) OR 4.39 (kJ) OR 4.4 (kJ)(also scores M1)
M3 Using 0.0110 mol
therefore ∆H = – 399 (kJmol–1 )
OR – 400
Penalise M3 ONLY if correct numerical answer but sign is incorrect; +399 gains 2 marks
Penalise M2 for arithmetic error and mark on
In M1, do not penalise incorrect cases in the formula
If ∆T = 280.5; score q = m c ∆T only
If c = 4.81 (leads to 5050.5) penalise M2 ONLY and mark on for M3 = – 459
+399 or +400 gains 2 marks
Ignore incorrect units
3
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Question 20
(a) Iron is extracted from iron(III) oxide using carbon at a high temperature.
(i) State the type of reaction that iron(III) oxide undergoes in this extraction.
(1)
(ii) Write a half-equation for the reaction of the iron(III) ions in this extraction.
(1)
(b) At a high temperature, carbon undergoes combustion when it reacts with oxygen.
(i) Suggest why it is not possible to measure the enthalpy change directly for the following combustion reaction.
C(s,graphite) + 1/2O2(g) –> CO(g)
(1)
(ii) State Hess’s Law.
(1)
(iii) State the meaning of the term standard enthalpy of combustion.
(3)
(c) Use the standard enthalpies of formation in the table below and the equation to calculate a value for the standard enthalpy change for the extraction of iron using carbon monoxide.
(3)
(d) (i) Write an equation for the reaction that represents the standard enthalpy of formation of carbon dioxide.
(1)
(ii) State why the value quoted in part (c) for the standard enthalpy of formation of CO2(g) is the same as the value for the standard enthalpy of combustion of carbon.
(1)
(Total 12 marks)
Q20.
(a) (i) reduction OR reduced OR redox OR reduction–oxidation
Not “oxidation” alone
1
(ii)
Ignore state symbols
Do not penalise absence of charge on electron
Credit Fe3+ Fe – 3e–
Credit multiples
1
(b) (i) Because (one of the following)
CO is not the only product OR
Reference to “incomplete combustion to form CO” does not answer the question
(Some) complete combustion (also)occurs OR
CO2 is (also) formed
Further oxidation occurs
1
(ii) The enthalpy change / heat (energy) change at constant pressure in a reaction
is independent of the route / path taken (and depends only on the initial and
final states)
1
(iii) M1 The enthalpy change / heat change at constant pressure when 1 mol
of a compound / substance / element
For M1, credit correct reference to molecule/s or atom/s
M2 is burned completely / undergoes complete combustion in (excess)
oxygen
M3 with all reactants and products / all substances in standard states
For M3
Ignore reference to 1 atmosphere
OR all reactants and products / all substances in normal / specified states
under standard conditions / 100 kPa / 1 bar and specified T / 298 K
3
(c) M1 (could be scored by a correct mathematical expression which must
have all ∆H symbols and the ∑)
Correct answer gains full marks
Credit 1 mark ONLY for –1 (kJ mol–1)
M1 ∆Hr = ∑∆Hf (products) – ∑∆Hf (reactants)
Credit 1 mark ONLY for – 27 (kJ mol–1) i.e. assuming value for Fe(l) = 0
OR correct cycle of balanced equations with 2Fe, 3C and 3O2
M2 ∆Hr = 2(+14) + 3(– 394) – (– 822) – 3(–111)
= 28 –1182 + 822 + 333
(This also scores M1)
M3 = (+) 1 (kJ mol–1)
(Award 1 mark ONLY for – 1)
(Award 1 mark ONLY for – 27)
For other incorrect or incomplete answers, proceed as follows
• check for an arithmetic error (AE), which is either a transposition error or an incorrect multiplication; this would score 2 marks (M1 and M2)
• If no AE, check for a correct method; this requires either a correct cycle with 2Fe, 3C and 3O2 OR a clear statement of M1 which could be in words and scores only M1
3
(d) (i) C(s) + O2(g) CO2(g)
State symbols essential
Possible to include C(s, graphite)
1
(ii) These two enthalpy changes are for the same reaction / same equation /
same reactants and products
Penalise reference to CO2 being produced by a different route
OR
They both make one mole of carbon dioxide only from carbon and oxygen
(or this idea clearly implied)
“both form CO2” is not sufficient (since other products might occur e.g.CO)
OR
The same number and same type of bonds are broken and formed
1
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Q21.
This question is about the extraction of titanium from titanium(IV) oxide by a two-stage
process.
The first stage in the process produces titanium(IV) chloride. In the second stage,
titanium(IV) chloride is converted into titanium.
The enthalpy change for the second stage can be determined using Hess’s Law.
(a) Give one reason why titanium is not extracted directly from titanium(IV) oxide using carbon.
(1)
(b) Give the meaning of the term enthalpy change.
(1)
(c) State Hess’s Law.
(1)
(d) Define the term standard enthalpy of formation.
(3)
(e) The following standard enthalpy of formation data refer to the second stage in the extraction of titanium.
(i) State why the value for the standard enthalpy of formation of Na(I) is not zero.
(1)
(ii) Use data from the table to calculate a value for the standard enthalpy change of the following reaction.
TiCl4(g) + 4Na(I) –> 4NaCl(s) + Ti(s)
(3)
(iii) State the role of sodium in this reaction.
(1)
(Total 11 marks)
Q21.
(a) One from
- Ti is not produced
- TiC / carbide is produced OR titanium reacts with carbon
- Product is brittle
- Product is a poor engineering material
Penalise “titanium carbonate”
Ignore “impure titanium”
Credit “titanium is brittle”
1
(b) Heat (energy) change at constant pressure
QoL
1
(c) The enthalpy change in a reaction is independent of
the route taken (and depends only on the initial and final states)
Credit “heat change at constant pressure” as an alternative to “enthalpy change”
1
(d) M1 The enthalpy change / heat change at constant pressure
when 1 mol of a compound / substance / product
For M1, credit correct reference to molecule/s or atom/s
M2 is formed from its (constituent) elements
M3 with all reactants and products / all substances in
standard states
OR all reactants and products / all substances in normal
states under standard conditions / 100 kPa / 1 bar and any
specified T (usually 298 K)
Ignore reference to 1 atmosphere
3
(e) (i) Na / it is not in its standard state / normal state under
standard conditions
OR
Standard state / normal state under standard conditions
for Na is solid / (s)
QoL
Ignore “sodium is a liquid or sodium is not a solid”
1
(ii) M1 ∆Hr = ∑∆Hf (products) – ∑∆Hf (reactants)
M2 ∆Hr = 4(−411) − (−720) − 4(+3)
= −1644 + 720 − 12
(This also scores M1)
M3 = −936 (kJ mol−1)
Correct answer gains full marks
Credit 1 mark for + 936 (kJ mol−1)
Credit 1 mark for – 924 (kJ mol−1)
i.e. assuming value for Na(l) = 0
For other incorrect or incomplete answers, proceed as follows
• check for an arithmetic error (AE), which is either a
transposition error or an incorrect multiplication; this would
score 2 marks (M1 and M2)
• If no AE, check for a correct method; this requires either
a correct cycle with 2Cl2 and 4Na OR a clear complete
statement of M1 which could be in words and scores
only M1
3
(iii) Reducing agent
Ignore “reduces titanium”
OR reductant OR reduces TiCl4
OR electron donor
1
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Q22.
Hess’s Law is used to calculate the enthalpy change in reactions for which it is difficult to determine a value experimentally.
(a) State the meaning of the term enthalpy change.
(1)
(b) State Hess’s Law.
(1)
(c) Consider the following table of data and the scheme of reactions.
Use the data in the table, the scheme of reactions and Hess’s Law to calculate a value for ∆Hr
(3)
(Total 5 marks)
Q22.
(a) Heat (energy) change at constant pressure
Ignore references to standard conditions, but credit specified pressure.
1
(b) The enthalpy change/heat (energy) change (at constant pressure) in a
reaction is independent of the route/path taken (and depends only
on the initial and final states)
1
(c) ΔH + 963 = –75 – 432 OR ΔH + 963 = – 507 (M1)
ΔH = –75 – 432 – 963 (M1 and M2)
ΔH = –1470 (kJ mol–1)
Award 1 mark for + 1470
Award full marks for correct answer
Ignore units.
Ignore numbers on the cycle
M1 and M2 can score for an arithmetic error
3
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