Thermochemistry Unit Thermochemistry Unit Section 16.2Section 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 methodC(s) + 2H2(g) CH4(g) ΔHo

f = -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 Exothermic

Na(s) + 1/2Cl2(g) NaCl(s) + 411.1 KJ

Na(s) + 1/2Cl2(g)

NaCl(s)

ΔH = -411.1 KJ

reactants

products

Enth

alpy

, H

Practice Problem #17:

a. Ethane, C2H6(g)

C2H6(g) + 7/2O2(g) 2CO2(g) + 3H20(l) + 1250.9 KJ

C3H8(g) + 5O2(g) 3CO2(g) + 4H20(l) + 2323.7 KJ

b. Propane, C3H8(g)

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).

Exothermic reaction, so the enthalpy of reactants is higher than the enthalpy of the products.

C7H16(l) + 11O2(g) 7CO2(g) + 8H20(l) + 5040.9 KJ

C7H6(s) + 11O2(g)

ΔHcomb = - 5040.9 KJ

reactants

products

Enth

alpy

, H

7CO2(g) + 8H20(l)

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 = ?

m = 500.0 g

ΔHof = -74.6 KJ/mol

q = nΔHof

nmethane = m/M

= (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 = ?

m = 50.0 g

ΔHocomb = -965.1 KJ/mol

q = nΔHocomb

nmethane = m/M

= (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 = ?

m = 0.534 g

ΔHof = -285.8 KJ/mol

q = nΔHof

nwater = m/M

= (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 = ?

m = 0.534 g

ΔHof = -241.8 KJ/mol

q = nΔHof

nwater = m/M

= (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 = ?

m = 56.78 g

ΔHocomb = -3682.3 KJ/mol

nwater = m/M

= (56.78 g) / (72.17 g/mol)

= 0.787 mol

q = nΔHocomb = (0.787 mol)(-3682.3 KJ/mol) = -2897.06 KJ

Mpentane = 72.17 g/mol

q = nΔHocomb

b) Determine the heat released by the combustion of 56.78 g of pentane, C5H12(l)

q = ?

m = 1.36 Kg = 1360 g

ΔHocomb = -5720.2 KJ/mol

nwater = m/M

= (1360 g) / (114.26 g/mol)

= 11.90 mol

q = nΔHocomb = (11.90 mol)(-5720.2 KJ/mol) = -68070.38 KJ

= -6.81 X 104 KJ

Mpentane = 114.26 g/mol

q = nΔHocomb

c) Determine the heat released by the combustion of 2.344 X 104 g of hexane, C6H14(l)

q = ?

m = 2.344 X 104 g

ΔHocomb = -4361.6 KJ/mol

nwater = m/M

= (2.344 X 104 g) / (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

Mhexane = 86.20 g/mol

q = nΔHocomb

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 = -2.34 X 104 kJ

ΔHof = -238.6 KJ/mol

mmethanol= (n)(M)

= (98.07 mol)(32.05 g/mol)

= 3143.21 g

n = q / ΔHof

=(-2.34 X 104 kJ)/(-238.6 KJ/mol) = 98.07 mol Mmethanol = 32.05 g/mol

q = nΔHof

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?

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 = ?

mice cube = 8.2 g

ΔHomelt = ΔHo

fus= 6.02 KJ/molnwater = m/M

= (8.2g) / (18.02 g/mol)

= 0.455 mol

q = nΔHofus = (0.455 mol)(6.02 KJ/mol) = 2.74 KJ

Mice cube = 18.02 g/mol

q = nΔHofus

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 = ?

mwater = 0.100 kg = 100 g

ΔHofreezing = ΔHo

fus= -6.02 KJ/molnwater = m/M

= (100 g) / (18.02 g/mol)

= 5.55 mol

q = nΔHofus= (5.55 mol)(-6.02 KJ/mol) = -33.41 KJ

Mwater = 18.02 g/mol

q = nΔHofus

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 = ?

mmercury = 0.325 g

ΔHovap= 59 KJ/mol

nwater = m/M

= (0.325 g) / (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.

Mmercury = 200.59 g/mol

q = nΔHovap

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 = ?

mNaCl = 25.3 g

ΔHosol= 3.9 KJ/mol

nNaCl= m/M

= (25.3 g) / (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.

MNaCl = 58.44 g/mol

q = nΔHosol

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

mdiethyl ether = ?

ΔHovap= 29 KJ/mol

nNaCl= q / ΔHovap

= (+80.7 kJ) / (29 kJ/mol)

= 2.78 mol

m= nM= (2.78 mol)(74.14 g/mol) = 206.08 g

Mdiethyl ether= 74.14 g/mol

q = nΔHovap

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

mbenzene= 100 g

ΔHovap= ?

ΔHovap= q/n

= (+3.97 X 104 kJ)/(1.28 mol)

= 31 015.63 J/mol

n= m/M= (100 g) / (78.12 g/mol) = 1.28 mol

Mbenzene= 78.12 g/mol

q = nΔHovap