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 1
This question is about 1-chloropropane.
(a) Define the term standard enthalpy of formation.
(2)
(b) The equation for a reaction used to manufacture 1-chloropropane is
3 CH3CH2CH2OH(l) + PCl3(l) → 3 CH3CH2CH2Cl(l) + H3PO3(s)
The enthalpy change for this reaction, ΔH, is –114 kJ mol–1
The table contains some standard enthalpy of formation data.
Calculate a value for the standard enthalpy of formation of propan-1-ol using the enthalpy change for the reaction and data from the table.
(3)
(c) 1-chloropropane can also be produced by the reaction between propane and chlorine in the presence of ultraviolet light.
State why ultraviolet light is needed for this reaction to occur.
Give an equation for each propagation step in the formation of 1-chloropropane from propane.
(3)
(d) The C–Cl bond in 1-chloropropane is polar because carbon and chlorine have different electronegativities.
Define the term electronegativity.
(1)
(e) Ammonia reacts with 1-chloropropane to form propylamine.
Name and outline the mechanism for this reaction.
(5)
(Total 14 marks)
Q1.
(a) M1 The enthalpy / heat energy change when 1 mol (of a substance / compound / product) is formed from its (constituent) elements
M1 energy change is not sufficient – must refer to enthalpy change or heat energy change
1
M2 with (all) reactants and products / all substances in standard states
M2 or with (all) reactants and products / substances in normal states under standard conditions / 100 kPa and any specified temperature (usually 298 K)
Ignore reference to 1 atmosphere
If enthalpy of combustion given rather than formation, then mark M1 and M2 independently, and M2 could score.
1
(b) M1 ∆H = [sum ∆fH products] – [sum ∆fH reactants]
or –114 = [3(–130) – 972] – [3X – 339]
or 3X = 3(–130) – 972 + 339 + 114
–303 scores 3 marks (+303 scores 2 marks)
–909 scores 2 marks (+909 scores 1 mark)
ignore units
1
M2 3X = –909
M2 No ECF from M1 (except +909 or arithmetic error)
1
M3 X = –303 (kJ mol–1)
M3 ECF from M2, ie M3 ÷ 3
1
(c) M1 provides energy to break (covalent) bond in chlorine / Cl2 or to form chlorine free radicals
1
M2 CH3CH2CH3 + •Cl → •CH2CH2CH3 + HCl
1
M3 •CH2CH2CH3 + Cl2 → ClCH2CH2CH3 + •Cl
M2 and M3:
- must show structure of •CH2CH2CH3 in at least one of the equations to score both marks (dot must be on or around the end CH2 group), but only penalise •C3H7 once across both equations if both equations otherwise correct
- on this occasion, molecular formula of propane can be allowed for M2
- on this occasion, molecular formula of 1-chloropropane can be allowed for M3
- penalise absence of radical dots once
- allow equations in either order
1
(d) the ability/power of atom to attract/withdraw the 2/pair of electrons in a covalent bond
allow nucleus in place of atom
1
(e) M1 nucleophilic substitution
1
M2 curly arrow from lone pair on N of NH3 to the correct C atom
Penalise M2 if negative charge on ammonia
1
M3 must show the movement of a pair of electrons from the C−Cl bond to the Cl atom; mark M3 independently provided it is from their original molecule
Penalise M3 for formal charge on C and/or Cl of C–Cl or incorrect partial charges on C–Cl; ignore other partial charges on uncharged atoms
Penalise M3 for any additional arrow(s) to/from the Cl to/from anything else
1
M4 is for the structure of the alkylammonium ion, which could be a condensed formula;
a positive charge must be shown on, or close to, the N atom
1
M5 is for an arrow from the N−H bond to the N atom
The second molecule of NH3 is not essential for M5, but penalise M5 if used incorrectly (but only penalise once in M2 and M5 for negative charge on ammonia)
1
SN1 mechanism alternative (loss of Cl first followed by attack by NH3) :
M2 curly arrow from C–Cl bond to the Cl
M3 curly arrow from lone pair of NH3 to correct C on the correct carbocation
[14]
Question 2
Two reactions of iron with oxygen are shown.
| Fe(s) + 1/2 O2(g) → FeO(s) | ΔH = – 272 kJ mol–1 |
| 2 Fe(s) + 3/2 O2(g) → Fe2O3(s) | ΔH = – 822 kJ mol–1 |
What is the enthalpy change, in kJ mol–1, for this reaction?
2 FeO(s) + 1/2 O2(g) → Fe2O3(s)
(Total 1 mark)
Question 3
This question is about enthalpy changes.
(a) A student determined the enthalpy of combustion of cyclohexane (C6H12).
The student
- placed a pure sample of cyclohexane in a spirit burner
- placed the spirit burner under a beaker containing 50.0 g of water and ignited the cyclohexane
- extinguished the flame after a few minutes.
The results for the experiment are shown in Table 1.
Table 1
The student determined from this experiment that the enthalpy of combustion of cyclohexane is –1216 kJ mol–1
Use the data to calculate the final temperature of the water in this experiment.
The specific heat capacity of water = 4.18 J K–1 g–1
The relative molecular mass (Mr) of cyclohexane = 84.0
(4)
(b) A data book value for the enthalpy of combustion of cyclohexane is –3920 kJ mol–1
The student concluded that the temperature rise recorded in the experiment was smaller than it should have been.
Suggest a practical reason for this.
(1)
(c) Table 2 gives some values of standard enthalpies of combustion (∆cH⦵).
Table 2
Use the data in Table 2 to calculate the enthalpy change for the reaction represented by this equation
6 C(s) + 6 H2(g) → C6H12(l)
(3)
(Total 8 marks)
Q2.
B
–278
[1]
Q3.
(a)
M1 
Correct answer scores 4 marks
1
M2 heat released = 1216 x 1000 x 0.0075 (= 9120) (J)
[or 1216 x 0.0075 = (9.12) (kJ)]
0.0075 scores M1 with or without working
9120 or 9.12 scores M1 and M2 with or without working
1
M3 
Allow ECF at each stage
correct M3 scores M1 and M2
1
M4 final temperature = 19.1 + M3 = 62.7 or 63 (°C)
1
Alternative M3/4
M3 9120 = 50 × 4.18 × (Final T – 19.1)
M4 Final T = 62.7 or 63 (°C)
Ignore negative sign for q in M2 and/or ΔT in M3, but penalise if used as a temperature fall in M4 (if alternative method used for M3/4 and negative value for q is used, allow M3 for expression with negative q value but do not allow M4)
(temperatures to at least 2sf)
If candidates use a value in kJ rather than J to find ΔT / final T then they lose M3, but ECF to M4 [e.g. 9.12 rather than 9120 giving ΔT = 0.0436 and final temperature = 19.1(436) – this would give 3 marks]
If candidates use 0.63 g for m in M3, they will get ΔT = 3.46 and final temperature = 22.56 – this would give 3 marks]
Cannot score M2 using moles = 1
(b) thermal energy / heat loss or
or idea of heat being transferred to calorimeter
incomplete combustion or
allow idea that it is not under standard conditions
evaporation
allow no lid / poor/no insulation
1
(c) M1 6 × (–394), 6 × (–286) and –3920
1
M2 (ΔH =) [6 × (–394)] + [6 × (–286)] + 3920
(or (ΔH =) [–2364)] + [–1716)] + 3920)
(or (ΔH =) –4080 + 3920)
1
M3 = –160 (kJ mol–1)
1
–160 scores 3 marks; +160 scores 2 marks
–8000 scores 2 marks; +8000 scores 1 mark
–1876 scores 2 marks; +1876 scores 1 mark
M1 is for correct coefficients, i.e. 6 × ΔcH H2 & 6 × ΔcH C & 1 x ΔcH C6H12 (ignore whether + or –)
ECF from M1 to M2/3 for incorrect coefficients / arithmetic error / transposition
ECF from M2 to M3 for use of products – reactants
Ignore any cycle
[8]
Question 4
Nitrogen dioxide is produced from ammonia and air as shown in these equations
4 NH3(g) + 5 O2 (g) → 4 NO(g) + 6 H2O(g) ΔH = –909 kJ mol–1
2 NO(g) + O2 (g) → 2 NO2 (g) ΔH = –115 kJ mol–1
What is the enthalpy change (in kJ mol–1) for the following reaction?
4 NH3(g) + 7 O2 (g) → 4 NO2 (g) + 6 H2O(g)
(Total 1 mark)
Question 5
Chlorine is used to decrease the numbers of microorganisms in water.
When chlorine is added to water, there is a redox reaction, as shown by the equation
Cl2 + H2O ⇌ HClO + HCl
(a) Deduce the oxidation state of chlorine in HClO and the oxidation state of chlorine in HCl
(1)
(b) Give two half-equations to show the oxidation and reduction processes that occur in this redox reaction.
(2)
(c) Chlorine is reacted with cold, aqueous sodium hydroxide in the manufacture of bleach.
Give an equation for this reaction between chlorine and sodium hydroxide.
(1)
(d) The concentration of ClO– ions in bleach solution can be found by reaction with iodide ions.
The overall equation for this reaction is shown.
ClO– + 2I– + 2H+ → I2 + Cl– + H2O
A sample of bleach solution was found to contain ClO– ions with a concentration of 0.0109 mol dm–3
Potassium iodide is added to a 20.0 cm3 portion of this bleach solution.
Calculate the mass, in mg, of potassium iodide needed to react with all of the
ClO– ions in the sample of bleach.
Give your answer to the appropriate number of significant figures.
Give one observation during this reaction.
(4)
(e) Potassium chlorate(VII), KClO4, is used in fireworks. When potassium chlorate(VII) decomposes, it produces potassium chloride and oxygen.
Give an equation for the decomposition of potassium chlorate(VII).
Use the data in the table to calculate the enthalpy change for this reaction.
(2)
(Total 10 marks)
Q4.
D
[1]
Q5.
(a) Two correct Cl ox states: HClO = +1 HCl = -1
1
(b) Oxidation:
Cl2 + 2H2O → 2HClO + 2H+ + 2e–
Accept – 2e– on the other side
Allow Cl2 + 2H2O → 2ClO– + 2e–
1
Reduction:
Cl2 + 2H+ + 2e– → 2HCl
Allow Cl2 + 2e– → 2Cl–
If both equations correct but incorrect order, allow 1
Ignore state symbols
1
(c) 2NaOH + Cl2 → NaCl + NaClO + H2O
Allow
2OH- + Cl2 → Cl– + ClO– + H2O
Allow NaOCl
Ignore state symbols
1
(d) mol ClO– = conc × vol = 0.0109 × 0.02
= 0.000218 / 2.18 × 10–4 mol
1
mol KI = 0.000218 × 2 = 0.000436 mol
M2 = M1 x2
If incorrect ratio, M2 & M3 = 0
1
mass KI = Mr ×mol = 166.0 × 0.000436
= 0.072376 g
= 72.4 (mg)
M3 = M1 × 2 × 166.0 × 1000
Must be to 3 sig figs
1
black solid/ppt appears/forms (in a colourless solution)
or (colourless solution) turns brown (solution)
Not purple. Not red. Not brown ppt/solid
Ignore grey.
1
(e) KClO4 → KCl + 2O2
Ignore state symbols
Allow multiples
1
∆H = – 436 – – 434 = – 2 kJ mol–1
Must be negative
Mark independently
Allow consequential for multiples
1
[10]
Question 6
What is the enthalpy of formation of buta-1,3-diene, C4H6(g)?
(Total 1 mark)
Question 7
This question is about energetics.
(a) Write an equation, including state symbols, for the reaction with an enthalpy change equal to the enthalpy of formation for iron(III) oxide.
(1)
(b) Table 1 contains some standard enthalpy of formation data.
Fe2O3(s) + 3CO(g) ⟶ 2Fe(s) + 3CO2(g) ΔH = −19 kJ mol−1
Use these data and the equation for the reaction of iron(III) oxide with carbon monoxide to calculate a value for the standard enthalpy of formation for carbon dioxide.
Show your working.
(3)
(c) Some enthalpy data are given in Table 2.
Use the data from Table 2 to calculate the bond enthalpy for N−H in ammonia.
(3)
(d) Give one reason why the bond enthalpy that you calculated in part (c) is different from the mean bond enthalpy quoted in a data book (388 kJ mol−1).
(1)
(Total 8 marks)
Q6.
A
[1]
Q7.
(a) 2Fe(s) + 2/3 O2(g) ⟶ Fe2O3(s) ONLY
Don’t allow multiples. States must be shown
1
(b) M1 Correct cycle or equation
If M1 and M2 not awarded then M3 can be awarded for their M2 divided by 3
1
M2 (3 × ΔfHCO2) = −19 + (−822) + 3(−111) − 0
(3 × ΔfHCO2) = −1174
1
M3 ΔfHCO2 = −391 kJ mol−1
−317 for 1 mark
+391 for 1 mark
1
Allow 2 sig fig or more
(c) M1 Correct Hess’s law cycle or equation
If M1 and M2 not awarded then M3 can be awarded for their M2 divided by 6
1
M2 (6(N−H)) = 944 + 3(+436) + 92
(6(N−H)) = 2344
−391 for 1 mark
1
M3 N−H = (+)391 kJ mol−1
1
Allow 2 sig fig or more
(d) Data book value derived from (a number of) different compounds (not just different NH3 molecules)
1
[8]
Question 8
This question is about enthalpy changes.
(a) Write an equation, including state symbols, to show the reaction taking place when the standard enthalpy of combustion for ethanol is measured.
(2)
(b) State the name given to the enthalpy change represented by the following chemical equation.
Explain why this enthalpy change would be difficult to determine directly.
(2)
(c) Standard enthalpies of combustion for carbon and carbon monoxide are −393 kJ mol−1 and −283 kJ mol−1, respectively.
Use these data to calculate the enthalpy change for the reaction in part (b).
(2)
(d) Use the following data to calculate a value for the Xe–F bond enthalpy in XeF4
(3)
(e) Suggest a reason why the value calculated in part (d) differs from the mean Xe–F bond enthalpy quoted in a data source.
(1)
(Total 10 marks)
Q8.
(a) C2H5OH(I) + 3O2(g) ⟶ 2CO2(g) + 3H2O(I)
1 mark for correct formulae and balancing
1
1 mark for all correct state symbols
1
(b) (Standard) enthalpy of formation
1
Difficult to prevent C reacting with O2 to form some CO2
1
(c) ΔH = ΣΔHc reactants − ΣΔHc products or a correct cycle
1
OR ΔH = −393−(−283)
ΔH = −110 (kJ mol−1)
1
(d) Correctly drawn Hess’s law cycle
1
4 (Xe–F) = 252 + (2 × 158) = 568
Xe–F = 568 / 4
1
Xe–F = 142 (kJ mol−1))
1
(e) Mean bond enthalpy found by taking an average for Xe–F in a range of compounds
1
[10]
Question 9
Hydrogen can be manufactured by the reaction of methane with steam. An equilibrium is established as shown by the equation.
CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g)
(a) Use Le Chatelier’s principle to predict the effect on the equilibrium yield of hydrogen if the overall pressure is increased.
Explain your answer.
(3)
(b) Explain why the equilibrium yield of hydrogen is unchanged if a catalyst is used in the reaction.
(2)
(c) The table shows the standard enthalpy of formation and the standard entropy for each substance in this equilibrium reaction.
Use data from the table to calculate the standard enthalpy change for this equilibrium reaction.
Standard enthalpy change _____________ kJ mol−1
(2)
Use your answer from part (c) and the entropy data from the table above to calculate the minimum temperature, in °C, needed for this reaction to be feasible.
Give your answer to the appropriate number of significant figures.
(If you did not complete part (c) you should assume a value of 120 kJ mol−1 for the standard enthalpy change. This is not the correct value).
Minimum temperature _________ °C
(5)
(Total 12 marks)
Q9.
(a) Decrease
1
Increasing pressure moves equilibrium to the side of least moles i.e. backward reaction
1
To oppose the increase in pressure or to decrease the pressure
1
(b) A catalyst speeds up the rate of the forward and backward reaction
1
By the same amount
1
(c) ΔH = −111 − (−75 − 242)
1
206 (kJ mol−1)
1
(d) ΔS = 3 × 131 + 198 − (186 + 189) = 216 J K−1 mol−1
1
ΔG = ΔH − TΔS
1
0 = 206 − T 
1
T = 953.7 or 954 K
1
T = 681 (°C)
If the value given in the question is used then the answer is 283 (°C)
1
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