Thermochemistry unit section 16 2
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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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Thermochemistry Unit Section 16.2

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Thermochemistry unit section 16 2

Thermochemistry Unit Section 16.2


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


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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.


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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)


Thermochemistry unit section 16 2

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)


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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.


Thermochemistry unit section 16 2

  • 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)


Thermochemistry unit section 16 2

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.


Thermochemistry unit section 16 2

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.


Thermochemistry unit section 16 2

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


Thermochemistry unit section 16 2

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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