1. Chemsheets AS006 (Electron arrangement)
12/09/2018
CALORIMETRY

2. Introduction to Thermodynamics
10.2
Thermodynamics is the study of the interconversion of heat and other kinds of energy.
In thermodynamics, there are three types
of systems:
An open system can exchange mass and
energy with the surroundings.
A closed system allows the transfer
of energy but not mass.
An isolated system does not exchange
either mass or energy with its
surroundings.

3. Calorimetry
Fuse school ( calorimeters and heats of combustion )

4. PPT - AS Calorimetry 1 g of water
12/09/2018
1 ºC hotter
1 g of water
Energy required =
4.18 J
2 ºC hotter
1 g of water
Energy required =
2 x 4.18 J
= 8.36 J

5. Energy required = 4.18 J Energy required = 5 x 4.18 J 1 ºC hotter
1 g of water
Energy required =
4.18 J
1 ºC hotter
5 g of water
Energy required =
5 x 4.18 J
= 20.9 J

6. Energy required = 4.18 J Energy required = 3 x 10 x 4.18 1 ºC hotter
1 g of water
Energy required =
4.18 J
10 ºC hotter
3 g of water
Energy required =
3 x 10 x 4.18
= J

7. Energy required = 0.39 J Energy required = 10 x 5 x 0.39 1 ºC hotter
1 g of copper
Energy required =
0.39 J
5 ºC hotter
10 g of copper
Energy required =
10 x 5 x 0.39
= 19.5 J

8. energy needed to make 1 g of substance 1ºC hotter
q = m c ∆T
mass heated (m) x
energy needed to make 1 g of substance 1ºC hotter
temperature rise (∆T)

9. Calculate the heat to take 255 g of water from 25.2 C to 90.5 C.
Strategy Get the givens
c = J/g∙°C, m = 255 g, ΔT = 90.5°C – 25.2°C = 65.3°C.
Formula
q=mc ΔT
Solution
q = 255 g × J/g∙°C x 65.3°C
q= J
q= J / 1000 J/kJ
q= kJ (3 figs) the heat is going INTO the water, so it is +ve for the water
This is heat that was GIVEN to the water, but FROM where.
A chemical must have given it’s heat away so it is –VE for the chemical – 96.7kJ
NOW, molar heat

10. 2 moles of ethanol burned
Energy given out =
2734 kJ
Energy given out by 1 mole =
2734 kJ 2
= 1367 kJ
∆H =
–1367 kJ mol-1

11. q
moles in reaction
∆H =

12. Experimental Determination of Enthalpy Changes by Calorimetry
Calorimeter = a container used for measuring the temperature change of solution
Determination of Enthalpy Change of Neutralization

13. Determination of Enthalpy Change of Combustion
Experimental Determination of Enthalpy Changes by Calorimetry
Determination of Enthalpy Change of Combustion

14. Cal Problems worked 4 min https://www.youtube.com/watch?v=Hs5x0-IU2F4

15. Excess powdered zinc was added to 100ml of 0
Excess powdered zinc was added to 100ml of 0.2 mol/L copper (II) sulphate solution. A temperature rise of 10oC was recorded. Find the MOLAR enthalpy change for the reaction.
H = m x c x T
H = 100g x 4.18 KJ kg-1 K-1 x 10 oC
1000
= KJ
This is for the no of moles of CuSO4 used in the experiment
No of moles of CuSO4 = 0.2 x 100 = 0.02 moles
H = = 209 KJ mol-1
0.02
The reaction is exothermic so we need to put in a negative sign
Hr = KJ mol-1 of CuSO4

16. Combustion
To find the heat of combustion of a substance a known mass of the substance is burned, the heat released transferred to water and the enthalpy change found as before
In an experiment to find the heat of combustion of ethanol the following results were obtained
Initial mass of lamp + ethanol = g
Final mass of lamp = g
Final temperature of water = oC
Initial temperature of water = oC
Mass of the water = g
What are the products of complete combustion of ethanol?
What mass of ethanol was burnt? How many moles is this?
What quantity of heat was transferred to the water?
Find Hc of ethanol
Identify any sources of error
Is ethanol a good fuel?

17. C2H5OH + 3O2  2CO H2O
q=mc∆t
H = x KJ kg-1 K-1 x 18.6K = 23.3KJ
1000
H lost = h gained
(so 23.3 KJ gained by calorimeter water was lost in burning)
Mass of ethanol used = 0.92g
0.92g = 0.92 g = mol
46g/mol
H=mc∆t / n
Hc = = KJ/mol
0.020

18. Errors
Heat lost to surroundings (air, can thermometer)
Errors in measuring temperature change (unavoidable error in reading thermometer)
Errors in measuring masses (unavoidable error in reading balance)
The enthalpy change of combustion is high ethanol is a good fuel

19. Chemistry in Action: Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l) DH = kJ/mol
Per gram Sugar = 2999 KJ / 180 g
= KJ/gram
1 cal = J
1 Cal = 1000 cal = 4184 J =4.184KJ
If 1.0 gram sugar is = KJ Or
15.7 KJ / KJ/Cal = 3.75 Cal
So sugar burned is approx 4Cal/g
Here 26g x 4 Cal/g = 104 Cal from sugar

20. Foods and Fuels Cheetos burned v Cereal
Foods
1 nutritional Calorie, 1 Cal = 1000 cal = 1 kcal.
Energy in us comes from carbohydrates and fats TED X (3 min, Calories)
Intestines: carbohydrates converted into glucose:
C6H12O6 + 6O2  6CO2 + 6H2O, DH = kJ
Fats break down as follows:
2C57H110O O2  114CO H2O, DH = -75,520 kJ
Fats contain more energy; are not water soluble, so are good for energy storage.

21. moles of propane = mass / Mr = 1.00 / 58.0 = 0.01724
1) In an experiment, 1.00 g of propanone (CH3COCH3) was completely burned in air. The heat evolved raised the temperature of 150 g of water from 18.8C to 64.3C. Use this data to calculate the enthalpy of combustion of propanone (the specific heat capacity of water is 4.18 J g-1 K-1).
q = mc∆T
m = 150
c = 4.18
∆T = 45.5
q = 150 x x = J
∆H = q / mol
moles of propane = mass / Mr
= 1.00 / =
∆H = –28.53 /
= kJ mol-1 (3 sig fig)

22. moles of propane = mass / Mr = 1.00 / 86.0 = 0.01163
2) In an experiment, 1.00 g of hexane (C6H14) was completely burned in air. The heat evolved raised the temperature of 200 g of water from K to K. Use this data to calculate the enthalpy of combustion of hexane (the specific heat capacity of water is 4.18 J g-1 K-1).
q = mc∆T
m = 200
c = 4.18
∆T = 51.6
q = 200 x x = J
∆H = q / mol
moles of propane = mass / Mr
= 1.00 / =
∆H = –43.14 /
= kJ mol-1 (3 sig fig)

23. moles of propane = mass / Mr = 1.56 / 60.0 = 0.02600
3) In an experiment, 1.56 g of propan-1-ol (CH3CH2CH2OH) was completely burned in air. The heat evolved raised the temperature of dm3 of water from K to K. Use this data to calculate the enthalpy of combustion of propan-1-ol (the specific heat capacity of water is 4.18 J g-1 K-1).
q = mc∆T
m = 250
c = 4.18
∆T = 47.3
q = 250 x x = J
∆H = q / mol
moles of propane = mass / Mr
= 1.56 / =
∆H = –49.43 /
= kJ mol-1 (3 sig fig)

24. 4). 25 cm3 of 2. 0 mol dm-3 nitric acid was added to 25 cm3 of 2
4) 25 cm3 of 2.0 mol dm-3 nitric acid was added to 25 cm3 of 2.0 mol dm-3 potassium hydroxide solution. The temperature rose by 13.7C. Calculate the enthalpy of neutralisation for this reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1.
q = mc∆T
m = 50
c = 4.18
∆T = 13.7
q = 50 x x = J
∆H = q / mol
Mol HNO3 = conc x vol
= 2.0 x 25/1000 =
Mol KOH = conc x vol
= 2.0 x 25/1000 =
HNO3 + KOH → NaNO3 + H2O
∆H = –2.863 / 0.050
= kJ mol-1 (3 sig fig)

25. 5) 50 cm3 of 0.10 mol dm-3 silver nitrate solution was put in a calorimeter and 0.2 g of zinc powder added. The temperature of the solution rose by 4.3C. Work out which reagent was in excess and then calculate the enthalpy change for the reaction (per mole Limiting reagent . Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. Ignore the heat capacity of the metals.
q = mc∆T
m = 50
c = 4.18
∆T = 4.3
q = 50 x x = J
∆H = q / mol
Mol AgNO3 = conc x vol
= 0.1 x 50/1000 =
Mol Zn = mass / Mr
= 0.2 / = XS
2 AgNO3 + Zn → ZnNO3 + 2 Ag
∆H = – kJ / AgNO3
= kJ mol-1 (3 sig fig)

26. EXTRA PROBLEMS

27. moles of propane = mass / Mr = 0.600 / 44.0 = 0.01364
1) In an experiment, g of propane (C3H8) was completely burned in air. The heat evolved raised the temperature of 100 g of water by 64.9C. Use this data to calculate the enthalpy of combustion of propane (the specific heat capacity of water is 4.18 J g-1 K-1).
q = mc∆T
m = 100
c = 4.18
∆T = 64.9
q = 100 x x = J
∆H = q / mol
moles of propane = mass / Mr
= / =
∆H = –27.13 /
= kJ mol-1 (3 sig fig)

28. 2). 50 cm3 of 1. 0 mol dm-3 hydrochloric acid was added to 50 cm3 of 1
2) 50 cm3 of 1.0 mol dm-3 hydrochloric acid was added to 50 cm3 of 1.0 mol dm-3 sodium hydroxide solution. The temperature rose by 6.8C. Calculate the enthalpy of neutralisation for this reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1.
q = mc∆T
m = 100
c = 4.18
∆T = 6.8
q = 100 x x = J
∆H = q / mol
Mol HCl = conc x vol
= 1.0 x 50/1000 =
Mol NaOH = conc x vol
= 1.0 x 50/1000 =
HCl + NaOH → NaCl + H2O
∆H = –2.842 / 0.050
= -57 kJ mol-1 (2 sig fig)

29. 3) 100 cm3 of mol dm-3 copper sulphate solution was put in a calorimeter and 2.00 g of magnesium powder added. The temperature of the solution rose by 25.1C. Work out which reagent was in excess and then calculate the enthalpy change for the reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. Ignore the heat capacity of the metals.
q = mc∆T
m = 100
c = 4.18
∆T = 25.1
q = 100 x x = J
∆H = q / mol
Mol CuSO4 = conc x vol
= x 100/1000 =
Mol Mg = mass / Mr
= 2.00 / = XS
CuSO4 + Mg → MgSO4 + Cu
∆H = –10.49 / 0.020
= kJ mol-1 (3 sig fig)

30. 4). 50 cm3 of 2. 0 mol dm-3 hydrochloric acid was added to 50 cm3 of 2
4) 50 cm3 of 2.0 mol dm-3 hydrochloric acid was added to 50 cm3 of 2.0 mol dm-3 ammonia solution. The temperature rose by 12.4C. Calculate the enthalpy of neutralisation for this reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1.
q = mc∆T
m = 100
c = 4.18
∆T = 12.4
q = 100 x x = J
∆H = q / mol
Mol HCl = conc x vol
= 2.0 x 50/1000 =
Mol NH3 = conc x vol
= 2.0 x 50/1000 =
HCl + NH3 → NH4Cl
∆H = –5.183 / 0.100
= kJ mol-1 (3 sig fig)

31. 2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O
5) 50 cm3 of 1.0 mol dm-3 nitric acid was added to 20 cm3 of 1.0 mol dm-3 barium hydroxide solution. The temperature rose by 7.9C. Calculate the enthalpy of neutralisation for this reaction (per mole of nitric acid reacting). Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1.
q = mc∆T
m = 70
c = 4.18
∆T = 7.9
q = 70 x x = J
∆H = q / mol
Mol HNO3 = conc x vol
= 1.0 x 50/1000 = XS
Mol Ba(OH)2 = conc x vol
= 1.0 x 20/1000 =
2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O
∆H = –2.312 / 0.040
= kJ mol-1 (3 sig fig)

32. 6) 25 cm3 of 1.00 mol dm-3 copper sulphate solution was put in a calorimeter and 6.0 g of zinc powder added. The temperature of the solution rose by 50.6C. Work out which reagent was in excess and then calculate the enthalpy change for the reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1. Ignore the heat capacity of the metals.
q = mc∆T
m = 25
c = 4.18
∆T = 50.6
q = 25 x x = J
∆H = q / mol
Mol CuSO4 = conc x vol
= 1.0 x 25/1000 =
Mol Zn = mass / Mr
= 6.0 / = XS
CuSO4 + Zn → ZnSO4 + Cu
∆H = –5.288 / 0.025
= kJ mol-1 (3 sig fig)

33. HCl + NaHCO3 → NaCl + H2O + CO2
7) 3.53 g of sodium hydrogencarbonate was added to 30.0 cm3 of 2.0 mol dm-3 hydrochloric acid. The temperature fell by 10.3 K. Work out which reagent was in excess and then calculate the enthalpy change for the reaction. Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1.
q = mc∆T
m = 30
c = 4.18
∆T = 10.3
q = 30 x x = J
∆H = q / mol
Mol HCl = conc x vol
= 2.0 x 30/1000 = XS
Mol NaHCO3 = mass / Mr
= 3.53 / =
HCl + NaHCO3 → NaCl + H2O + CO2
∆H = /
= kJ mol-1 (3 sig fig)

34. ∆H = q / mol q = ∆H x mol Mol CH3OH = mass / Mr = 2.00 / 32.0 = 0.0625
8) A calorimeter was calibrated by burning 2.00 g of methanol (CH3OH) whose enthalpy of combustion is -715 kJ mol-1. The temperature of the calorimeter rose from 19.6C to 52.4C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol g of propan-2-ol CH3CH(OH)CH3 raised the temperature by from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol.
∆H = q / mol
q = ∆H x mol
Mol CH3OH = mass / Mr
= 2.00 / =
q = ∆H x mol
= 715 x = kJ
q = mc∆T
mc = q/∆T
= / = kJ K-1

35. q = mc∆T mc = 1.363 ∆T = 36.4 q = 1.363 x 36.4 = 49.61 kJ ∆H = q / mol
9) A calorimeter was calibrated by burning 2.00 g of methanol (CH3OH) whose enthalpy of combustion is -715 kJ mol-1. The temperature of the calorimeter rose from 19.6C to 52.4C. The same calorimeter was used to measure the enthalpy of combustion of propan-2-ol g of propan-2-ol CH3CH(OH)CH3 raised the temperature by from 19.8C to 56.2C. Calculate the heat capacity of the calorimeter and then the enthalpy of combustion of propan-2-ol.
q = mc∆T
mc = 1.363
∆T = 36.4
q = x = kJ
∆H = q / mol
Mol C3H7OH = mass / Mr
= 1.50 / =
∆H = –49.61 /
= kJ mol-1 (3 sig fig)

36. PRACTICE enthalpy change of solution ΔHS
Enthalpy change of solution ΔHS – is when 1 mole of substance is dissolved
Place a clean, dry polystyrene cup into an outer plastic cup to increase insulation and stability.
Use the measuring cylinder to transfer 25 cm3 of water into the polystyrene cup.
Place a thermometer into the cup through a cardboard lid and record this temperature as an initial temperature
Weight 1 g NH4Cl of and put it into polystyrene cup.
Measure the temperature of the mixture in the cup and record the greatest temperature change that you observed as a final temperature;
Rinse the plastic cup with water and shake to dry.
Repeat same steps with NaOH (keep this solution for next experiment)
Calculate the enthalpy change for both using the equation q = mcΔT
NH4Cl
NaOH
Mass (g)
Volume of water is measured (cm3)
Initial temperature (K)
Final temperature (K)
ΔH ( kJmol-1 )

37. PRACTICE enthalpy change of neutralisation ΔHN
The standard enthalpy change of neutralisation is the enthalpy change when
solutions of an acid and an alkali react together under standard conditions to produce 1 mole of water.
Place a thermometer into the cup with NaOH through a cardboard lid and record this temperature as an initial temperature;
Use the measuring cylinder to transfer 25cm3 of HCl into the polystyrene cup.
Measure the temperature of the mixture in the cup immediately record the greatest temperature change that you observed as a final temperature;
Calculate the enthalpy change for reaction using the equation q = mcΔT