Mixed Calculation Questions
Energetics Worksheet
Practice Questions
1. Propanone undergoes complete combustion when reacted with excess oxygen.
a) Write an equation for the combustion of propanone.
CH3COCH3(g) + 4O2(g)
→
3CO2(g) + 3H2O(g)
b) Given that the enthalpy of formation of propanone, water and carbon dioxide are -248, -286, and -394 kJmol-1 respectively, calculate the enthalpy change of combustion for propanone.
Formula
ΔHc = ΣΔHf(Products) – ΣΔHf(Reactants)
Calculation
[3(-394) + 3(-286)] – [-248]
[-1182 – 858] + 248 = -2040 + 248
[-1182 – 858] + 248 = -2040 + 248
Answer
-1792 kJ mol-1
2. Propane can be broken down to form ethene and methane.
a) Given the enthalpy of combustion for propane, ethene and methane are -2044, -1411 and -890 kJ mol-1 respectively, calculate the enthalpy change of the reaction.
C3H8
→
C2H4 + CH4
Formula
ΔHr = ΣΔHc(Reactants) – ΣΔHc(Products)
Calculation
[-2044] – [-1411 + (-890)]
-2044 – [-2301] = -2044 + 2301
-2044 – [-2301] = -2044 + 2301
Answer
+257 kJ mol-1
b) Calculate the enthalpy change of the cracking of propane using the bond enthalpies below: C-H: 413 kJ mol-1, C-C: 347 kJ mol-1, C=C: 612 kJ mol-1.
Bonds Broken (Propane)
2(C-C) + 8(C-H)
(Note: 8 C-H bonds are conserved, so we can focus on C-C vs C=C changes)
2(347) = 694 kJ (Breaking 2 C-C)
(Note: 8 C-H bonds are conserved, so we can focus on C-C vs C=C changes)
2(347) = 694 kJ (Breaking 2 C-C)
Bonds Formed (Ethene + Methane)
1(C=C)
612 kJ
612 kJ
Answer
ΔH = 3998 – 3916 = +82 kJ mol-1
c) Explain why the answers to question 2a and 2b are different.
2b is based on mean bond enthalpies, which are an average across many molecules, not just the specific bonds in this reaction.
2a is based on experimental calorimetry data, which can have errors due to heat loss, evaporation, or incomplete combustion.
3. A scientist wants to determine the enthalpy change for the conversion of sodium hydrogen carbonate to sodium carbonate. She cannot determine the enthalpy change directly, so instead she does two reactions. In the first reaction, the scientist reacts 6.20g of sodium hydrogen carbonate with 100cm3 of 0.500 moldm-3 HCl and the temperature changed from 20.0oC to 18.5oC. In the second reaction, the scientist reacts 5.40g of sodium carbonate with 100cm3 of 0.500 moldm-3 HCl and the temperature changed from 20.0oC to 24.0oC.
a) Draw a Hess diagram for the two reactions and the conversion of sodium hydrogen carbonate into sodium carbonate.
b) Calculate the enthalpy change of neutralisation for each reaction and therefore calculate the enthalpy change of reaction for the conversion of sodium hydrogen carbonate into sodium carbonate.
Rxn 1: Limiting Reagent & Energy
Mol NaHCO3 = 6.20/84.0 = 0.0738 mol
Mol HCl = 0.100 × 0.500 = 0.0500 mol (Limiting)
q = 100 × 4.18 × 1.5 = 627 J (0.627 kJ)
Mol HCl = 0.100 × 0.500 = 0.0500 mol (Limiting)
q = 100 × 4.18 × 1.5 = 627 J (0.627 kJ)
ΔH1 (Based on Limiting)
ΔH1 = +0.627 / 0.0500 = +12.54 kJ mol-1
Rxn 2: Limiting Reagent & Energy
Mol Na2CO3 = 5.40/106.0 = 0.0509 mol
Mol HCl = 0.100 × 0.500 = 0.0500 mol (Limiting)
Stoichiometry (2 HCl : 1 Na2CO3) → 0.0500 mol HCl reacts with 0.0250 mol Na2CO3
q = 100 × 4.18 × 4.0 = 1672 J (1.672 kJ)
Mol HCl = 0.100 × 0.500 = 0.0500 mol (Limiting)
Stoichiometry (2 HCl : 1 Na2CO3) → 0.0500 mol HCl reacts with 0.0250 mol Na2CO3
q = 100 × 4.18 × 4.0 = 1672 J (1.672 kJ)
ΔH2 (Based on Reacting Moles of solid)
ΔH2 = -1.672 / 0.0250 = -66.88 kJ mol-1
Overall ΔHr
Target: 2NaHCO3 → Na2CO3 + H2O + CO2
ΔHr = 2(ΔH1) – ΔH2
= 2(+12.54) – (-66.88) = 25.08 + 66.88 = +92.0 kJ mol-1
ΔHr = 2(ΔH1) – ΔH2
= 2(+12.54) – (-66.88) = 25.08 + 66.88 = +92.0 kJ mol-1
4. A student determines the enthalpy change for the reaction between calcium carbonate and sulfuric acid. The student follows the method: measure out 75 cm3 of 1.00 mol dm–3 sulfuric acid using a measuring cylinder and pour the acid into a 100 cm3 conical flask; weigh out 3.00 g of solid calcium carbonate on a weighing boat and tip it into the acid; record the maximum temperature reached using a thermometer.
a) Describe how the student could improve their method to improve the accuracy of their result.
- Use a polystyrene cup rather than a conical flask for insulation.
- Use a pipette rather than a measuring cylinder.
- Reweigh the weighing boat to determine exact mass added.
- Plot a cooling curve (measure temp before and after mixing) and extrapolate to find ΔT.
b) Assuming that the balance was a 2dp balance and that the thermometer and measuring cylinder were accurate to 0.5oC and 0.5cm3, calculate the percentage error of the experiment. Assume that the temperature increased by 12oC.
Breakdown
Mass (2 readings): (0.01 / 3.00) × 100 = 0.33%
Temp (2 readings): (1.0 / 12.0) × 100 = 8.33%
Vol (1 reading): (0.5 / 75.0) × 100 = 0.67%
Total = 9.33%
Temp (2 readings): (1.0 / 12.0) × 100 = 8.33%
Vol (1 reading): (0.5 / 75.0) × 100 = 0.67%
Total = 9.33%
5. A student wants to determine the enthalpy change for the hydrogenation of cyclohexene to cyclohexane. He cannot determine this enthalpy directly from an experiment so he does two experiments instead. In the first experiment he combusts cyclohexene, causing 500cm3 of water to heat up from 20.0oC to 45.1oC. The spirit Burner weighed 21.50g prior to burning and 20.28g afterwards. In the second experiment, the student combusts cyclohexane, causing 500cm3 of water to heat up from 20.0oC to 89.6oC. The spirit Burner weighed 18.50g prior to burning and 15.30g afterwards.
a) State three sources of error in the experiments that the student conducted.
- Incomplete combustion of the fuel.
- Heat loss to surroundings/beaker.
- Evaporation of the fuel (cyclohexene/cyclohexane).
b) Calculate the enthalpy change for the combustion of the two fuels and use these results, along with the enthalpy change of combustion of hydrogen (-241kJmol-1) to calculate the enthalpy change of the hydrogenation of cyclohexene.
Exp 1: Combustion of Cyclohexene (C6H10)
Energy: q = 500 × 4.18 × (45.1-20.0) = 52,459 J = 52.46 kJ
Moles: Mass burned = 21.50 – 20.28 = 1.22g
Mr(C6H10) = 82.0. n = 1.22/82.0 = 0.01488 mol
ΔHc(ene) = -52.46 / 0.01488 = -3526 kJ mol-1
Moles: Mass burned = 21.50 – 20.28 = 1.22g
Mr(C6H10) = 82.0. n = 1.22/82.0 = 0.01488 mol
ΔHc(ene) = -52.46 / 0.01488 = -3526 kJ mol-1
Exp 2: Combustion of Cyclohexane (C6H12)
Energy: q = 500 × 4.18 × (89.6-20.0) = 145,464 J = 145.46 kJ
Moles: Mass burned = 18.50 – 15.30 = 3.20g
Mr(C6H12) = 84.0. n = 3.20/84.0 = 0.03810 mol
ΔHc(ane) = -145.46 / 0.03810 = -3818 kJ mol-1
Moles: Mass burned = 18.50 – 15.30 = 3.20g
Mr(C6H12) = 84.0. n = 3.20/84.0 = 0.03810 mol
ΔHc(ane) = -145.46 / 0.03810 = -3818 kJ mol-1
Hess’s Law Step
Reaction: C6H10 + H2 → C6H12
ΔHhyd = ΣΔHc(Reactants) – ΔHc(Products)
= [ΔHc(ene) + ΔHc(H2)] – ΔHc(ane)
= [-3526 + (-241)] – [-3818]
= -3767 + 3818 = +51 kJ mol-1
ΔHhyd = ΣΔHc(Reactants) – ΔHc(Products)
= [ΔHc(ene) + ΔHc(H2)] – ΔHc(ane)
= [-3526 + (-241)] – [-3818]
= -3767 + 3818 = +51 kJ mol-1