Calorimetry
Energetics Worksheet
Energetics Worksheet
1. 100cm3 of 0.5moldm-3 copper sulfate reacted with exactly 2.00g of magnesium. The temperature of the mixture increased from 18.0oC to 62.0oC.
a) Write a balanced symbol equation for the reaction.
CuSO4(aq) + Mg(s)
→
MgSO4(aq) + Cu(s)
b) Calculate the moles of copper sulfate and moles of magnesium and determine which one was in excess.
Moles of Copper Sulfate
0.5 × 0.100 = 0.050 mol
Moles of Magnesium
2.00 / 24.3 = 0.0823 mol
Conclusion
Reaction is 1:1. Magnesium is in excess. Copper sulfate is limiting.
c) Calculate the enthalpy change for the reaction.
Step 1: Calculate q
q = mcΔT = 100 × 4.18 × (62.0-18.0) = 18,392 J = 18.392 kJ
Step 2: Calculate ΔH
ΔH = -18.392 / 0.050 = -368 kJ mol-1
2. 150cm3 of 0.8moldm-3 KOH reacted with 150cm3 of 0.5moldm-3 H2SO4. The temperature of the mixture increased from 18oC to 23oC.
a) Write a balanced symbol equation for the reaction.
2KOH + H2SO4
→
K2SO4 + 2H2O
b) Define enthalpy of neutralisation.
The enthalpy change when one mole of water is produced from the reaction between an acid and an alkali under standard conditions.
c) Calculate the moles of KOH, H2SO4 and H2O produced.
Moles
KOH: 0.8 × 0.150 = 0.120 mol
H2SO4: 0.5 × 0.150 = 0.075 mol
H2SO4: 0.5 × 0.150 = 0.075 mol
Moles Water Produced
0.075 mol acid requires 0.150 mol KOH. We only have 0.120 mol KOH.
KOH is limiting. Ratio KOH:H2O is 1:1.
Water = 0.120 mol
KOH is limiting. Ratio KOH:H2O is 1:1.
Water = 0.120 mol
d) Calculate the enthalpy change of neutralisation for KOH and H2SO4.
Step 1: Calculate q
Total Vol = 300cm3. ΔT = 5oC.
q = 300 × 4.18 × 5 = 6270 J = 6.27 kJ
q = 300 × 4.18 × 5 = 6270 J = 6.27 kJ
Step 2: Calculate ΔHn
ΔHn = -6.27 / 0.120 = -52.3 kJ mol-1
3. Propan-1-ol was burnt in a spirit burner and used to heat up 200cm3 of water in a calorimeter.
| Mass of burner before | 15.61 g |
|---|---|
| Mass of burner after | 14.08 g |
| Initial temp | 20.0 oC |
| Final temp | 62.0 oC |
a) Write a balanced symbol equation for the reaction.
C3H7OH + 4.5O2
→
3CO2 + 4H2O
b) Calculate enthalpy change of combustion of propan-1-ol.
Step 1: Calculate q
q = 200 × 4.18 × (62.0 – 20.0) = 35,112 J = 35.112 kJ
Step 2: Calculate Moles
Mass burned = 15.61 – 14.08 = 1.53 g
Mr = 60.0
Moles = 1.53 / 60.0 = 0.0255 mol
Mr = 60.0
Moles = 1.53 / 60.0 = 0.0255 mol
Step 3: Calculate ΔHc
ΔH = -35.112 / 0.0255 = -1377 kJ mol-1
c) Suggest two reasons why the value is different to that from the enthalpy of combustion for propan-1-ol in a data book.
- Heat loss to the surroundings (beaker, air, tripod).
- Incomplete combustion (formation of soot).
- Evaporation of fuel from the wick.
- Non-standard conditions.
4. 200cm3 of 0.5moldm-3 hydrochloric acid and 2.4g of calcium were reacted in a calorimeter. Mixing occurred at 5.0 minutes.
| Time / mins | 0.0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4.0 | 4.5 |
|---|---|---|---|---|---|---|---|---|---|---|
| Temp / oC | 15.5 | 15.4 | 15.3 | 15.3 | 15.3 | 15.2 | 15.2 | 15.2 | 15.1 | 15.1 |
| Time / mins | 5.5 | 6.0 | 6.5 | 7.0 | 7.5 | 8.0 | 8.5 | 9.0 | 9.5 | 10.0 |
| Temp / oC | 47.5 | 46.7 | 46.0 | 45.4 | 44.8 | 44.2 | 43.6 | 43.0 | 42.5 | 42.0 |
| Time / mins | 10.5 | 11.0 | 11.5 | 12.0 | 12.5 | 13.0 | 13.5 | 14.0 | 14.5 | 15.0 |
| Temp / oC | 41.5 | 41.0 | 40.6 | 40.2 | 39.8 | 39.4 | 39.0 | 38.6 | 38.3 | 38.0 |
a) Write the equation for the reaction.
Ca + 2HCl
→
CaCl2 + H2
b) Draw a graph and use it to determine the temperature change during the reaction.
By extrapolating the lines to 5 minutes:
- Initial Temp ≈ 15.0 oC
- Final Temp (Theoretical Max) ≈ 48.3 oC
- ΔT = 33.3 oC
c) Calculate the moles of each reactant.
Moles HCl
0.5 × 0.200 = 0.100 mol
Moles Ca
2.4 / 40.1 = 0.0599 mol
d) Calculate the enthalpy change of the reaction.
Step 1: Calculate q
q = 200 × 4.18 × 33.3 = 27,839 J = 27.84 kJ
Step 2: Identify Limiting Reagent
Reaction: Ca + 2HCl. 0.0599 mol Ca needs 0.1198 mol HCl. We have 0.100 mol HCl. HCl is limiting.
Step 3: Calculate ΔH
Reaction moles = 0.100 / 2 = 0.050 mol
ΔH = -27.84 / 0.050 = -557 kJ mol-1
ΔH = -27.84 / 0.050 = -557 kJ mol-1
e) Suggest a possible reason for why the value you calculated is different to the value found in a data-book.
Heat loss to surroundings is the main factor, though extrapolation minimizes this error. Specific heat capacity of the solution is assumed to be that of water, which might be inaccurate.
5. 80cm3 of 0.6moldm-3 citric acid (H3C6H5O7) and 5.8g of sodium hydrogen carbonate react in a calorimeter. The temperature decreases from 20.0oC to 15.5oC.
a) Write the symbol equation for the reaction.
H3C6H5O7 + 3NaHCO3
→
Na3Cit + 3CO2 + 3H2O
b) Calculate the enthalpy change of neutralisation for the reaction.
Step 1: Calculate q
ΔT = -4.5 oC (Endothermic)
q = 80 × 4.18 × 4.5 = 1,505 J = 1.505 kJ
q = 80 × 4.18 × 4.5 = 1,505 J = 1.505 kJ
Step 2: Calculate Moles
Acid: 0.6 × 0.080 = 0.048 mol
NaHCO3: 5.8 / 84.0 = 0.069 mol
NaHCO3 is limiting.
NaHCO3: 5.8 / 84.0 = 0.069 mol
NaHCO3 is limiting.
Step 3: Calculate ΔHneut
Water formed = 0.069 mol
ΔH = +1.505 / 0.069 = +21.8 kJ mol-1
ΔH = +1.505 / 0.069 = +21.8 kJ mol-1
6.
50cm3 of 0.2moldm-3 sulfuric acid and 50cm3 of 0.2moldm-3 NaOH are reacted in a calorimetry experiment. Both solutions start at 20oC and the ΔHn is -83.6kJmol-1. What is the final temperature of the mixture assuming that no heat is lost in the experiment?
Step 1: Moles Reacting
NaOH = 0.2 × 0.050 = 0.010 mol (Limiting)
Water produced = 0.010 mol
Water produced = 0.010 mol
Step 2: Energy Released
q = 0.010 × 83,600 = 836 J
Step 3: Final Temperature
ΔT = q / mc = 836 / (100 × 4.18) = 2.0 oC
Final Temp = 20.0 + 2.0 = 22.0 oC
Final Temp = 20.0 + 2.0 = 22.0 oC