1. Calorimetry

2. The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.
C = m x s
Heat (q) absorbed or released:
q = m x s x Dt
q = C x Dt
Dt = tfinal - tinitial

3. How much heat is given off when an 869 g iron bar cools from 94oC to 5oC?
s of Fe = J/g • oC
Dt = tfinal – tinitial = 5oC – 94oC = -89oC
q = msDt
= 869 g x J/g • oC x –89oC
= -34,000 J

4. Problem 6.33
A 6.22 kg piece of copper metal is heated from 20.5 ºC and ºC. Calculate the heat absorbed (in kJ) by the metal.

6. Heating Curve for Water
Ice
Ice + water
Water
Water + steam
ΔHvap = 40.6 kJ/mol, ΔHfus = 6.02 kJ/mol, s = J/g˚C

7. Calculate the energy to heat 17. 9g of ice from 0
Calculate the energy to heat 17.9g of ice from 0.00˚C to liquid water at 100.˚C. (ΔHvap = 40.6 kJ/mol, ΔHfus = 6.02 kJ/mol, s = J/g˚C)

8. Constant-Volume Calorimetry
qsys = qcal+ qrxn
qsys = 0
qrxn = - qcal
qcal = m x s x Dt
qcal = Ccal x Dt
Reaction at Constant V
DH = qrxn
DH ~ qrxn
No heat enters or leaves!

9. Dt = tfinal – tinitial = 25.95oC – 20.28oC = 5.67oC
A quantity of g of naphthalene (C10H8), a pungent smelling substance used in moth repellents, was burned in a constant-volume bomb calorimeter. Consequently the temperature of the water rose from 20.28ºC to 25.95ºC. If the heat capacity of the bomb plus water was kJ/ºC, calculate the heat of combustion of naphthalene on a molar basis; that is find the molar heat of combustion.
qcal = Ccal Dt
Ccal= kJ/oC
Dt = tfinal – tinitial = 25.95oC – 20.28oC = 5.67oC
qcal = Ccal Dt
= kJ/oC x 5.67oC
= kJ
qrxn
moles =
128.2 g
mole x
Dhcomb kJ
1.435 g =
= x 103 kJ/mol

10. Problem 6.37
A g sample of solid magnesium is burned in a constant volume bomb calorimeter that has a heat capacity of 3024 J/ºC. The temperature increases by 1.126ºC. Calculate the heat given off by the burning Mg in kJ/mol.

11. Constant-Pressure Calorimetry
qsys = qcal + qrxn
qsys = 0
qrxn = - (qcal)
qcal = m x s x Dt
qcal = Ccal x Dt
Reaction at Constant P
DH = qrxn
No heat enters or leaves!

12. A lead (Pb) pellet having a mass of 26. 47 g at 89
A lead (Pb) pellet having a mass of g at 89.98ºC was placed in a constant pressure calorimeter of negligible heat capacity containing mL of water. The water temperature rose from 22.50ºC to 23.17ºC. What is the specific heat of the lead pellet?
qPb + qwater = 0
qPb = - qwater
qwater = ms∆t
Dt = tfinal – tinitial = 23.17oC – 22.50oC = 0.67oC
qwater = ms∆t
= 100.0g x J/goC x 0.67oC
= J

13. The heat lost be the pellet is equal to the heat gained by the water, so
qPb = J
Dt = tfinal – tinitial = 23.17oC – 89.98oC = oC
qPb = ms∆t
J= 26.47g x s x oC
s = J/goC

14. Problem 6.82
A 44.0 g sample of an unknown metal at 99.0ºC was placed in a constant pressure calorimeter containing 80.0 g of water at 24.0ºC. The final temperature of the system was found to be 28.4ºC. Calculate the specific heat capacity of the metal. (the specific heat of water is J/gºC)

15. Chemistry in Action: Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l) DH = kJ/mol
1 cal = J
1 Cal = 1000 cal = 4184 J
Substance
DHcombustion (kJ/g)
Apple -2
Beef -8
Beer
-1.5
Gasoline
-34

16. Enthalpy of Formations

17. Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of formation (DH0) as a reference point for all enthalpy expressions. f
Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. f
The standard enthalpy of formation of any element in its most stable form is zero.
DH0 (C, graphite) = 0 f
DH0 (O2) = 0 f
DH0 (C, diamond) = 1.90 kJ/mol f
DH0 (O3) = 142 kJ/mol f

19. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn
aA + bB cC + dD
DH0
rxn
dDH0 (D) f
cDH0 (C) = [ + ] -
bDH0 (B)
aDH0 (A)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

20. C (graphite) + 1/2O2 (g) CO (g)
CO (g) + 1/2O2 (g) CO2 (g)
C (graphite) + O2 (g) CO2 (g)

21. Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ/mol
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ/mol
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ/mol
rxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
rxn
C(graphite) + O2 (g) CO2 (g) DH0 = kJ/mol
2S(rhombic) + 2O2 (g) SO2 (g) DH0 = kJ/mol x 2
rxn +
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJ/mol
rxn
C(graphite) + 2S(rhombic) CS2 (l)
DH0 = (2x-296.1) = 86.3 kJ/mol
rxn

22. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[ 12x– x–187.6 ] – [ 2x49.04 ] = kJ
-5946 kJ
2 mol
= kJ/mol C6H6

23. Chemistry in Action: Bombardier Beetle Defense
C6H4(OH)2 (aq) + H2O2 (aq) C6H4O2 (aq) + 2H2O (l) DH0 = ?
C6H4(OH)2 (aq) C6H4O2 (aq) + H2 (g) DH0 = 177 kJ/mol
H2O2 (aq) H2O (l) + ½O2 (g) DH0 = kJ/mol
H2 (g) + ½ O2 (g) H2O (l) DH0 = -286 kJ/mol
DH0 = – 286 = -204 kJ/mol
Exothermic!

24. Problem 6.62
From the following data C(graphite) + O2(g) CO2(g) ∆Hrxnº= kJ/mol H2(g) + ½O2(g) H2O(l) ∆Hrxnº= kJ/mol 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) ∆Hrxnº= kJ/mol calculate the enthalpy change for the reaction 2C(graphite) + 3H2(g) C2H6(g)

25. The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be used for melting ice?
Which substance(s) could be used for a cold pack?

26. The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol

27. The energy required to completely separate one mole of a solid ionic compound into gaseous ions is called lattice energy (U).
The energy change associated with the hydration process is called the heat of hydration, ∆Hhyd
∆Hsoln = U + ∆Hhyd
U = 788 kJ/mol +
∆Hhydr = -784 kJ/mol
∆Hsoln = 4 kJ/mol
DH0 = (-784) = 4 kJ/mol
rxn

28. Lattice energy (kJ/mol)
Problem 6.133
From the following data, calculate the heat of solution for KI:
NaCl
NaI
KCl KI
Lattice energy (kJ/mol)
788
686
699
632
Heat of Solution
(kJ/mol)
4.0
-5.1
17.2 ?