Thermochemistry Unit
Section 16.2
Practice Problem #15:
a. H2O(g)
H2(s) + 1/2O2(g)  H2O(g) + 241.8 KJ
b. CaCl2(s)
Ca(s) + Cl2(g)  CaCl2(s) + 795.4 KJ
c. CH4(g)
C(s) + 2H2(g)  CH4(g) + 74.6 KJ
d. C6H6(l)
6C(s) + 3H2(g) + 49 KJ  C6H6(l)
e. Show the standard molar enthalpy of parts c) and d) using
another method
C(s) + 2H2(g)  CH4(g)
ΔHof = -74.6 KJ
6C(s) + 3H2(g)  C6H6(l)
ΔHof = +49 KJ
Practice Problem #16:
Draw an enthalpy diagram to represent the standard molar enthalpy of
formation of sodium chloride.
NaCl(s)
ΔHof = -411.1 KJ/mol
Na(s) + 1/2Cl2(g)  NaCl(s) + 411.1 KJ
Enthalpy, H
Na(s) + 1/2Cl2(g)
reactants
ΔH = -411.1 KJ
NaCl(s)
products
Exothermic
Practice Problem #17:
a. Ethane, C2H6(g)
C2H6(g) + 7/2O2(g)  2CO2(g) + 3H20(l) + 1250.9 KJ
b. Propane, C3H8(g)
C3H8(g) + 5O2(g)  3CO2(g) + 4H20(l) + 2323.7 KJ
c. Butane, C4H10(g)
C4H10(g) + 13/2O2(g)  4CO2(g) + 5H20(l) + 3003.0 KJ
c. Pentane, C5H12(l)
C5H12(l) + 8O2(g)  5CO2(g) + 6H20(l) + 3682.3 KJ
Practice Problem #18:
Draw an enthalpy diagram to represent the standard molar enthalpy of
combustion of heptane, C7H16(l) (use Table 16.3).
C7H16(l) + 11O2(g)  7CO2(g) + 8H20(l) + 5040.9 KJ
Exothermic reaction, so the enthalpy of reactants is higher
than the enthalpy of the products.
Enthalpy, H
C7H6(s) + 11O2(g)
reactants
ΔHcomb = - 5040.9 KJ
7CO2(g) + 8H20(l)
products
Sample Problem (page 644):
Methane is the main component of natural gas. Natural gas undergoes
combustion to provide energy for heating homes and cooking food.
a) How much heat is released when 500.00 g of methane forms from
the elements?
q=?
q = nΔHof
m = 500.0 g
nmethane = m/M
ΔHof = -74.6 KJ/mol
= (500.00 g) / (16.05 g/mol)
= 31.15 mol
q = nΔHof = (31.15 mol)(-74.6 KJ/mol) = -2323.99 KJ
b) How much heat is released when 50.00 g of methane undergoes
complete combustion?
q=?
q = nΔHocomb
m = 50.0 g
nmethane = m/M
ΔHocomb = -965.1 KJ/mol
= (500.00 g) / (16.05 g/mol)
= 3.115 mol
q = nΔHocomb = (3.115 mol)(-965.1 KJ/mol) = -3006.29 KJ
Practice Problem #19:
a) Hydrogen gas and oxygen gas react to form 0.534 g of liquid water.
How much heat is released to the surroundings?
q=?
q = nΔHof
m = 0.534 g
nwater = m/M
ΔHof = -285.8 KJ/mol
= (0.534 g) / (18.02 g/mol)
= 0.0296 mol
q = nΔHof = (0.0296 mol)(-285.8 KJ/mol) = -8.47 KJ
b) Hydrogen gas and oxygen gas react to form 0.534 g of gaseous
water. How much heat is released to the surroundings?
q=?
q = nΔHof
m = 0.534 g
nwater = m/M
ΔHof = -241.8 KJ/mol
= (0.534 g) / (18.02 g/mol)
= 0.0296 mol
q = nΔHof = (0.0296 mol)(-241.8 KJ/mol) = -7.16 KJ
Practice Problem #21:
a) Determine the heat released by the combustion of 56.78 g of
pentane, C5H12(l)
q=?
q = nΔHocomb
m = 56.78 g
nwater = m/M
ΔHocomb = -3682.3 KJ/mol
= (56.78 g) / (72.17 g/mol)
Mpentane = 72.17 g/mol
= 0.787 mol
q = nΔHocomb = (0.787 mol)(-3682.3 KJ/mol) = -2897.06 KJ
b) Determine the heat released by the combustion of 56.78 g of
pentane, C5H12(l)
q=?
q = nΔHocomb
m = 1.36 Kg = 1360 g
nwater = m/M
ΔHocomb = -5720.2 KJ/mol
= (1360 g) / (114.26 g/mol)
Mpentane = 114.26 g/mol
= 11.90 mol
q = nΔHocomb = (11.90 mol)(-5720.2 KJ/mol) = -68070.38 KJ
= -6.81 X 104 KJ
c) Determine the heat released by the combustion of 2.344 X 104 g of
hexane, C6H14(l)
q=?
q = nΔHocomb
m = 2.344 X 104 g
nwater = m/M
ΔHocomb = -4361.6 KJ/mol
= (2.344 X 104 g) / (86.20 g/mol)
Mhexane = 86.20 g/mol
= 271.93 mol
q = nΔHocomb = (271.93 mol)(-4361.6 KJ/mol) = -1 186 031.37 KJ
= -1.186 X 106 KJ
Practice Problem #23:
What mass of methanol, CH3OH(l), is formed from its elements if
2.34 X 104 kJ of energy is released during the process?
m=?
q = nΔHof
q = -2.34 X 104 kJ
ΔHof = -238.6 KJ/mol
Mmethanol = 32.05 g/mol
n = q / ΔHof
=(-2.34 X 104 kJ)/(-238.6 KJ/mol)
= 98.07 mol
mmethanol= (n)(M)
= (98.07 mol)(32.05 g/mol)
= 3143.21 g
Practice Problem #24:
An ice cube with a mass of 8.2 g is placed in some lemonade. The ice
cube melts completely. How much heat does the ice cube absorb from
the lemonade as it melts?
q=?
q = nΔHofus
mice cube = 8.2 g
nwater = m/M
ΔHomelt = ΔHofus= 6.02 KJ/mol
= (8.2g) / (18.02 g/mol)
Mice cube = 18.02 g/mol
= 0.455 mol
q = nΔHofus = (0.455 mol)(6.02 KJ/mol) = 2.74 KJ
Practice Problem #25:
A teacup contains 0.100 kg of water at its freezing point. The water
freezes solid.
a) How much heat is released to its surroundings?
q=?
q = nΔHofus
mwater = 0.100 kg = 100 g
nwater = m/M
ΔHofreezing = ΔHofus= -6.02 KJ/mol
= (100 g) / (18.02 g/mol)
Mwater = 18.02 g/mol
= 5.55 mol
q = nΔHofus= (5.55 mol)(-6.02 KJ/mol) = -33.41 KJ
qfreezing = -33.41 KJ
b) qmelting = 33.41 KJ
Practice Problem #26:
A sample of mercury vaporizes. The mercury is at its boiling point and
has a mass of 0.325 g. How much heat is absorbed or released to the
surroundings?
q=?
q = nΔHovap
mmercury = 0.325 g
nwater = m/M
ΔHovap= 59 KJ/mol
= (0.325 g) / (200.59 g/mol)
Mmercury = 200.59 g/mol
= 0.00162 mol
q = nΔHovap = (0.00162 mol)(59 KJ/mol) = 0.0956 KJ
This is then an endothermic reaction since heat energy is absorbed.
Practice Problem #27:
The molar enthalpy of solution for sodium chloride, NaCl, is 3.9 kJ/mol.
a) Write a thermochemical reaction to represent the dissolution of
sodium chloride?
Dissolution:
Solid state  Liquid state
NaCl(s) + 3.9 kJ  NaCl(aq)
b) Suppose you dissolve 25.3 g of sodium chloride in a glass of water
at room temperature. How much heat is absorbed or released by the
process?
q=?
q = nΔHosol
mNaCl = 25.3 g
nNaCl= m/M
ΔHosol= 3.9 KJ/mol
= (25.3 g) / (58.44 g/mol)
MNaCl = 58.44 g/mol
= 0.433 mol
q = nΔHosol= (0.433 mol)(3.9 KJ/mol) = 1.69 KJ
This is then an endothermic reaction since heat energy is absorbed.
c) Do you expect the glass containing the salt solution to feel warm or
cool? Explain your answer.
Answer:
Since heat is absorbed to dissolve the salt, heat is removed from the
glass and it will feel cold.
Practice Problem #28:
What mass of diethyl ether, C4H10O, can be vaporized by adding
80.7 kJ of heat?
q = +80.7 kJ
q = nΔHovap
mdiethyl ether = ?
nNaCl= q / ΔHovap
ΔHovap= 29 KJ/mol
= (+80.7 kJ) / (29 kJ/mol)
Mdiethyl ether= 74.14 g/mol
= 2.78 mol
m= nM= (2.78 mol)(74.14 g/mol) = 206.08 g
Practice Problem #29:
3.97 X 104 J of heat is required to vaporize 100 g of benzene, C6H6.
What is the molar enthalpy of vaporisation of benzene?
q = +3.97 X 104 J
q = nΔHovap
mbenzene= 100 g
ΔHovap= q/n
ΔHovap= ?
= (+3.97 X 104 kJ)/(1.28 mol)
Mbenzene= 78.12 g/mol
= 31 015.63 J/mol
n= m/M= (100 g) / (78.12 g/mol) = 1.28 mol