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Thermochemistry Unit Section 16.2. Practice Problem #15:. a . H 2 O (g). H 2(s) + 1/2O 2(g)  H 2 O (g) + 241.8 KJ. b . CaCl 2(s). Ca (s) + Cl 2(g)  CaCl 2(s) + 795.4 KJ. c. CH 4(g). C (s) + 2H 2(g)  CH 4(g) + 74.6 KJ. d . C 6 H 6(l). 6C (s) + 3H 2(g) + 49 KJ  C 6 H 6(l).

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slide2

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

slide3

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

slide4

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

reactants

Enthalpy, H

ΔH = -411.1 KJ

NaCl(s)

products

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

slide5

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

slide6

C7H6(s) + 11O2(g)

reactants

Enthalpy, H

ΔHcomb= - 5040.9 KJ

7CO2(g) + 8H20(l)

products

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.

slide7

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 =nΔHof

q= ?

nmethane=m/M

m = 500.0 g

= (500.00 g) / (16.05 g/mol)

ΔHof= -74.6 KJ/mol

= 31.15 mol

q =nΔHof = (31.15 mol)(-74.6 KJ/mol) = -2323.99 KJ

slide8

b) How much heat is released when 50.00 g of methane undergoes

complete combustion?

q =nΔHocomb

q= ?

nmethane=m/M

m = 50.0 g

= (500.00 g) / (16.05 g/mol)

ΔHocomb= -965.1 KJ/mol

= 3.115 mol

q =nΔHocomb = (3.115 mol)(-965.1 KJ/mol) = -3006.29 KJ

slide9

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 =nΔHof

q= ?

nwater =m/M

m = 0.534 g

= (0.534 g) / (18.02 g/mol)

ΔHof= -285.8 KJ/mol

= 0.0296 mol

q =nΔHof = (0.0296 mol)(-285.8 KJ/mol) = -8.47 KJ

slide10

b) Hydrogen gas and oxygen gas react to form 0.534 g of gaseous water. How much heat is released to the surroundings?

q =nΔHof

q= ?

nwater =m/M

m = 0.534 g

= (0.534 g) / (18.02 g/mol)

ΔHof= -241.8 KJ/mol

= 0.0296 mol

q =nΔHof = (0.0296 mol)(-241.8 KJ/mol) = -7.16 KJ

slide11

nwater =m/M

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

= 0.787 mol

  • Practice Problem #21:
  • Determine the heat released by the combustion of 56.78 g of
  • pentane, C5H12(l)

q= ?

q =nΔHocomb

m = 56.78 g

ΔHocomb= -3682.3 KJ/mol

Mpentane= 72.17 g/mol

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

slide12

q= ?

q =nΔHocomb

nwater =m/M

m = 1.36 Kg = 1360 g

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

ΔHocomb= -5720.2 KJ/mol

= 11.90 mol

Mpentane= 114.26 g/mol

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

= -6.81 X 104 KJ

b) Determine the heat released by the combustion of 56.78 g of

pentane, C5H12(l)

slide13

q= ?

q =nΔHocomb

nwater =m/M

m = 2.344 X 104 g

= (2.344 X 104g) / (86.20 g/mol)

ΔHocomb= -4361.6 KJ/mol

= 271.93 mol

Mhexane= 86.20 g/mol

q =nΔHocomb = (271.93 mol)(-4361.6 KJ/mol) = -1 186 031.37 KJ

= -1.186 X 106 KJ

c) Determine the heat released by the combustion of 2.344 X 104 g of

hexane, C6H14(l)

slide14

mmethanol=(n)(M)

= (98.07 mol)(32.05 g/mol)

= 3143.21 g

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 #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

n =q / ΔHof

=(-2.34 X 104 kJ)/(-238.6 KJ/mol)

= 98.07 mol

ΔHof= -238.6 KJ/mol

Mmethanol= 32.05 g/mol

slide15

nwater =m/M

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

= 0.455 mol

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

ΔHomelt= ΔHofus=6.02 KJ/mol

Mice cube= 18.02 g/mol

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

slide16

nwater =m/M

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

= 5.55 mol

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

ΔHofreezing= ΔHofus=-6.02 KJ/mol

Mwater= 18.02 g/mol

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

b)qmelting=33.41 KJ

qfreezing=-33.41 KJ

slide17

nwater =m/M

= (0.325 g) / (200.59 g/mol)

= 0.00162 mol

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

ΔHovap=59 KJ/mol

Mmercury= 200.59 g/mol

q =nΔHovap = (0.00162 mol)(59 KJ/mol) = 0.0956 KJ

This is then an endothermic reaction since heat energy is absorbed.

slide18

Practice Problem #27:

  • The molar enthalpy of solution for sodium chloride, NaCl, is 3.9 kJ/mol.
  • Write a thermochemical reaction to represent the dissolution of
  • sodium chloride?

Dissolution:

Solid state  Liquid state

NaCl(s) + 3.9 kJ  NaCl(aq)

slide19

nNaCl=m/M

= (25.3 g) / (58.44 g/mol)

= 0.433 mol

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

ΔHosol=3.9 KJ/mol

MNaCl= 58.44 g/mol

q =nΔHosol= (0.433 mol)(3.9 KJ/mol) = 1.69 KJ

This is then an endothermic reaction since heat energy is absorbed.

slide20

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.

slide21

nNaCl=q / ΔHovap

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

= 2.78 mol

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= ?

ΔHovap=29 KJ/mol

Mdiethyl ether= 74.14 g/mol

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

slide22

ΔHovap=q/n

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

= 31 015.63 J/mol

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=?

Mbenzene= 78.12 g/mol

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

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