Section 5.5: Standard Enthalpies of Formation
Tutorial 1 Practice, page 323
1. Given: from Table 1, H f C2H2 (g)  228.2 kJ/mol , H f CO2 (g)  393.5 kJ/mol , and
H f H2O(l)  285.8 kJ/mol ; H f O2 (g)  0 kJ/mol
Required: H r  for the combustion of acetylene gas
Analysis: H r    nproducts  H products   nreactants  H reactants
Since O2(g) is in its standard state, the equation can be written as:
H r   [nCO2 (g)  H f CO2 (g)  nH2O(l)  H f H2O(l) ]  nC2H 2 (g)  H f C2H 2 (g)
Solution:
Step 1: Write a balanced chemical equation for the combustion of acetylene gas so that acetylene
gas has a coefficient of 1.
5
C2H2(g) + 2 O2(g)  2 CO2(g) + H2O(l)
Step 2: Substitute the appropriate values for standard enthalpy of formation into the equation and
solve.
H r   [nCO2 (g)  H f CO2 (g)  nH2O(l)  H f  H2O(l) ]  nC2 H2 (g)  H f C2 H2 (g)
 [2(393.5 kJ)  (285.8 kJ)]  (228.2 kJ)
 787.0 kJ  285.8 kJ  228.2 kJ
H r   1301.0 kJ
Statement: Since the reaction is for 1 mol of acetylene gas, the standard enthalpy of combustion
for acetylene gas is –1301.0 kJ/mol.
2. Given: from Question 1, H r  for combustion of acetylene gas = –1301 kJ/mol;
from Table 1, H f C3H8 (g)  104.7 kJ/mol , H f CO2 (g)  393.5 kJ/mol , and
H f H2O(l)  285.8 kJ/mol ; H f O2 (g)  0 kJ/mol
Required: H r  per 1.00 g acetylene gas and per 1.00 g propane gas
Analysis: H r    nproducts  H products   nreactants  H reactants
Since O2(g) is in its standard state, the equation for propane gas can be written as:
H r   [nCO2 (g)  H f CO2 (g)  nH2O(l)  H f H2O(l) ]  nC3H8 (g)  H f C3H8 (g)
Solution:
Step 1: Write a balanced chemical equation for the combustion of propane gas so that propane
gas has a coefficient of 1.
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
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Chapter 5: Thermochemistry
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Step 2: Substitute the appropriate values for standard enthalpy of formation into the equation and
solve.
H r   [nCO2 (g)  H f CO2 (g)  nH 2O(l)  H f  H 2O(l) ]  nC3H8 (g)  H f C3H 8 (g)
 [3( 393.5 kJ)  4( 285.8 kJ)]  ( 104.7 kJ)
 1180 kJ  1143.2 kJ  104.7 kJ
H r   2218.5 kJ
Step 3: Convert to enthalpy per gram of acetylene and enthalpy per gram of propane.
M C2H2 (g)  26.04 g/mol and M C3H8 (g)  44.11 g/mol
For acetylene:
H r per mole C2 H2 (g)
H r per gram C2 H2 (g) 
M C2 H2 (g)
1301 kJ 1 mol

26.04 g
1 mol
 50.0 kJ/g

H r per gram C2 H2 (g)
For propane:
H r per gram C3H8 (g) 
H r per mole C3H8 (g)
M C3H8 (g)
2218.5 kJ 1 mol

44.11 g
1 mol
 50.3 kJ/g

H r per gram C3H8 (g)
Statement: The enthalpy of combustion of 1.00 g of acetylene gas is –50.0 kJ and that of 1.00 g
of propane gas is –50.3 kJ. For 1.00 g of the gas, propane releases 0.3 kJ more energy than
acetylene.
Section 5.5 Questions, page 324
1. Under SATP conditions, magnesium, Mg, is in solid state and bromine, Br2, is in liquid state.
So, (b) and (d) are in standard states.
2. (a) Given: from Table 1, H f H2O(l)  285.8 kJ/mol ; H f H2 (g)  0 kJ/mol ;
H f O2 (g)  0 kJ/mol
Required: H r 
Analysis: H r    nproducts  H products   nreactants  H reactants
Since H2(g) and O2(g) are in their standard states, the equation can be written as:
H r   nH2O(l)  H f H2O(l)
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Chapter 5: Thermochemistry
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Solution:
From the equation: 2 H2(g) + O2(g)  2 H2O(l),
nH2O(l)  2
H r   nH 2O(l)  H f  H 2O(l)
 2( 285.8 kJ)
H r   571.6 kJ
Statement: H r  for the reaction is –571.6 kJ.
(b) Given: from Table 1, H f CO2 (g)  393.5 kJ/mol ; H f C(s)  0 kJ/mol ;
H f O2 (g)  0 kJ/mol
Required: H r 
Analysis: H r    nproducts  H products   nreactants  H reactants
Since C(s) and O2(g) are in their standard states, the equation can be written as:
H r   nCO2 (g)  H f CO2 (g)
Solution:
From the equation: C(s) + O2(g)  CO2(g),
nCO2 (g)  1
H r   nCO2 (g)  H f CO2 (g)
H r   393.5 kJ
Statement: H r  for the reaction is –393.5 kJ.
(c) Given: from Table 1, H f H2O(l)  285.8 kJ/mol ; H f H2 (g)  0 kJ/mol ;
H f O2 (g)  0 kJ/mol
Required: H r 
Analysis: H r    nproducts  H products   nreactants  H reactants
Since H2(g) and O2(g) are in their standard states, the equation can be written as:
H r   nH2O(l)  H f H2O(l)
Solution:
From the equation: 2 H2O(l)  2 H2(g) + O2(g),
nH2O(l)  2
H r   nH 2O(l)  H f H 2O(l)
 2(285.8 kJ)
H r   571.6 kJ
Statement: H r  for the reaction is 571.6 kJ.
(d) Given: from Table 1, H f C2H5OH(l)  235.2 kJ/mol , H f CO2 (g)  393.5 kJ/mol , and
H f H2O(l)  285.8 kJ/mol ; H f O2 (g)  0 kJ/mol
Required: H r 
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Chapter 5: Thermochemistry
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Analysis: H r    nproducts  H products   nreactants  H reactants
Since O2(g) is in its standard state, the equation can be written as:
H r   [nCO2 (g)  H f CO2 (g)  nH2O(l)  H f H2O(l) ]  nC2H5OH(l)  H f C2H5OH(l)
Solution:
From the equation: C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l),
nC2H5OH(l)  1 , nCO2 (g)  2 , and nH2O(l)  3
H r   [nCO2 (g)  H f CO2 (g)  nH 2O(l)  H f  H 2O(l) ]  nC2H 5OH(l)  H f C2H 5OH(l)
 [2( 393.5 kJ)  3( 285.8 kJ)]  ( 235.2 kJ)
 787.0 kJ  857.4 kJ  235.2 kJ
H r   1409.2 kJ
Statement: H r  for the reaction is –1409.2 kJ.
3. Given: H c   1.09  104 kJ
Required: H r  for the required reaction
Analysis: rules from Hess’s law
Solution:
Step 1: Compare the equations to change equation (1) to equation (2).
(1) 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(l) H c   1.09  104 kJ
25
(2) 8 CO2(g) + 9 H2O(l)  C8H18(l) +
O2(g)
2
Step 2: Reverse (1) so that reactants and products are on the same side as (2). Reverse the sign of
H c  .
4
(3) 16 CO2(g) + 18 H2O(l)  2 C8H18(l) + 25 O2(g) H r   1.09  10 kJ
Step 3: Equation (3) has 2 mol of C8H18(l) on the product side, while equation (2) has 1 mol.
1
Multiply equation (3) and its H r  by the factor .
2
1
4
H r    1.09  10 kJ
2
H r   5.45  103 kJ
25
3
8 CO2(g) + 9 H2O(l)  C8H18(l) +
O2(g) H r   5.45  10 kJ
2
Statement: H r  for the required reaction is 5.45  103 kJ.
4. Given: from Table 1, H f NH4ClO4 (s)  295.8 kJ/mol , H f Al2O3 (s)  1675.7 kJ/mol ,
H f AlCl3 (s)  704.2 kJ/mol , H f NO(g)  90.2 kJ/mol , and H f H2O(g)  241.8 kJ/mol ;
H f Al(s)  0 kJ/mol
Required: H r 
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Chapter 5: Thermochemistry
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Analysis: H r    nproducts  H products   nreactants  H reactants
Since Al(s) is in its standard state, the equation can be written as:
H r   [nAl2O3 (s)  H f Al2O3 (s)  nAlCl3 (s)  H f  AlCl3 (s)  nNO(g)  H f  NO(g) 
nH 2O(g)  H f H 2O(g) ]  nNH 4ClO4 (s)  H f  NH 4ClO4 (s)
Solution:
From the equation:
3 Al(s) + 3 NH4ClO4(s)  Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g),
nNH4ClO4 (s)  3 , nAl2O3 (s)  1 , nAlCl3 (s)  1 , nNO(g)  3 , and nH2O(g)  6
H r   [nAl2O3 (s)  H f  Al2O3 (s)  nAlCl3 (s)  H f  AlCl3 (s)  nNO(g)  H f  NO(g) 
nH 2O(g)  H f  H 2O(g) ]  nNH 4ClO4 (s)  H f  NH 4ClO4 (s)
 [( 1675.7 kJ)  ( 704.2 kJ)  3(90.2 kJ)  6( 241.8 kJ)]  3( 295.8 kJ)
 [  1675.7 kJ  704.2 kJ  270.6 kJ  1450.8 kJ]  887.4 kJ
H r   2672.7 kJ
Statement: H r  for the reaction of a mixture of aluminum and ammonium perchlorate is –
2672.7 kJ.
5. Given: from Table 1, H f NH3 (g)  45.9 kJ/mol and H f HF(g)  273.3 kJ/mol ;
H r   1196 kJ/mol ; H f N2 (g)  0 kJ/mol ; H f Cl2 (g)  0 kJ/mol
Required: H f ClF3 (g)
Analysis: H r    nproducts  H products   nreactants  H reactants
Since N2(g) and Cl2(g) are in their standard states, the equation can be written as:
H r   nHF(g)  H f HF(g)  [nClF3 (g)  H f ClF3 (g)  nNH3 (g)  H f NH3 (g) ]
Solution:
From the equation: 2 ClF3(g) + 2 NH3(g)  N2(g) + 6 HF(g) + Cl2(g),
nClF3 (g)  2 , nNH3 (g)  2 , and nHF(g)  6
H r   nHF(g)  H f  HF(g)  [nClF3 (g)  H f ClF3 (g)  nNH 3 (g)  H f  NH 3 (g) ]
1196 kJ  6( 273.3 kJ)  [2 mol  H f ClF3 (g)  2( 45.9 kJ)]
1196 kJ  1693.8  2 mol  H f ClF3 (g)  91.8 kJ
2 mol  H f ClF3 (g)  352 kJ
352 kJ
2 mol
 176 kJ/mol
H f ClF3 (g) 
H f ClF3 (g)
Statement: H f  for ClF3(g) is –176 kJ/mol.
6. Given: from Table 1, H f H2O(g)  241.8 kJ/mol and H f CO2 (g)  393.5 kJ/mol ;
H f N2 (g)  0 kJ/mol ; H f N2 H3CH3 (l)  53 kJ/mol ; H f N2O4 (l)  20 kJ/mol
Required: H r 
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Chapter 5: Thermochemistry
5.5-5
Analysis: H r    nproducts  H products   nreactants  H reactants
Since N2(g) is in its standard state, the equation can be written as:
H r   [nCO2 (g)  H f CO2 (g)  nH 2O(g)  H f  H 2O(g) ] 
[nN 2H3CH3 (l)  H f  N 2H 3CH 3 (l)  nN 2O4 (l)  H f N 2O4 (l) ]
Solution:
From the equation: 4 N2H3CH3(l) + 5 N2O4(l)  12 H2O(g) + 9 N2(g) + 4 CO2(g),
nN2H3CH3 (l)  4 , nN2O4 (l)  5 , nH2O(g)  12 , and nCO2 (g)  4
H r   [nCO2 (g)  H f CO2 (g)  nH 2O(g)  H f  H 2O(g) ] 
[nN 2 H 3CH 3 (l)  H f  N 2H 3CH 3 (l)  nN 2O4 (l)  H f N 2O4 (l) ]
 [4( 393.5 kJ)  12( 241.8 kJ)]  [4(53 kJ)  5( 20 kJ)]
 [1574.0 kJ  2901.6 kJ]  [212 kJ  100 kJ]
H r   4587.6 kJ
Statement: H r  for the required reaction is –4587.6 kJ.
7. Given: from Table 1, H f CO2 (g)  393.5 kJ/mol and H f H2O(l)  285.8 kJ/mol ;
H c   1411.1 kJ/mol ; H f O2 (g)  0 kJ/mol
Required: H f C2H4 (g)
Analysis: H r    nproducts  H products   nreactants  H reactants
Since O2(g) is in its standard state, the equation can be written as:
H c   [nCO2 (g)  H f CO2 (g)  nH2O(l)  H f H2O(l) ]  nC2H4 (g)  H f C2H 4 (g)
Solution:
Step 1: Write a balanced chemical equation for the combustion of ethene gas so that ethene gas
has a coefficient of 1.
C2H4(g) + 3 O2(g)  2 CO2(g) + 2 H2O(l)
Step 2: Substitute the appropriate values for standard enthalpy into the equation and solve.
H c   [nCO2 (g)  H f CO2 (g)  nH 2O(l)  H f H 2O(l) ]  nC2H 4 (g)  H f C2H 4 (g)
1411.1 kJ  [2(393.5 kJ)  2(285.8 kJ)]  H f C2 H 4 (g)
1411.1 kJ  [787 kJ  571.6 kJ]  1 mol  H f C2 H 4 (g)
1 mol  H f C2 H 4 (g)  1385.6 kJ  1411.1 kJ
H f C2 H 4 (g)  52.5 kJ/mol
Statement: H f  for ethene gas is 52.5 kJ/mol.
8. Given: from Table 1, H f C2H5OH(l)  235.2 kJ/mol , H f CO2 (g)  393.5 kJ/mol , and
H f H2O(l)  285.8 kJ/mol ; H f O2 (g)  0 kJ/mol
Required: H r  per gram of liquid ethanol
Analysis: H r    nproducts  H products   nreactants  H reactants
Since O2(g) is in its standard state, the equation can be written as:
H r   [nCO2 (g)  H f CO2 (g)  nH2O(l)  H f H2O(l) ]  nC2H5OH(l)  H f C2H5OH(l)
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Chapter 5: Thermochemistry
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Solution:
Step 1: Write a balanced chemical equation for the combustion of liquid ethanol.
C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
Step 2: Substitute the appropriate values for standard enthalpy of formation into the equation and
solve.
H r   [nCO2 (g)  H f CO2 (g)  nH 2O(l)  H f  H 2O(l) ]  nC2H 5OH(l)  H f C2H 5OH(l)
 [2( 393.5 kJ)  3( 285.8 kJ)]  ( 235.2 kJ)
 787.0 kJ  857.4 kJ  235.2 kJ
H r   1409.2 kJ
Step 3: Convert to enthalpy per gram of liquid ethanol.
M C2H5OH(l)  46.08 g/mol
H r per gram C2 H5OH(l) 
H r per mole C2 H5OH(l)
M C2 H5OH(l)
1409.2 kJ 1 mol

46.08 g
1 mol
 30.58 kJ/g

H r per gram C2 H5OH(l)
Statement: The standard enthalpy of combustion per gram of liquid ethanol is –30.58 kJ/g.
9. (a) Given: from Table 1, H f CH3OH(l)  239.1 kJ/mol , H f CO2 (g)  393.5 kJ/mol , and
H f H2O(l)  285.8 kJ/mol ; H f O2 (g)  0 kJ/mol
Required: H r  per gram of liquid methanol
Analysis: H r    nproducts  H products   nreactants  H reactants
Since O2(g) is in its standard state, the equation can be written as:
H r   [nCO2 (g)  H f CO2 (g)  nH2O(l)  H f H2O(l) ]  nCH3OH(l)  H f CH3OH(l)
Solution:
Step 1: Write a balanced chemical equation for the combustion of liquid ethanol.
3
CH3OH(l) + O2(g)  CO2(g) + 2 H2O(l)
2
Step 2: Substitute the appropriate values for standard enthalpy of formation into the equation and
solve.
H r   [nCO2 (g)  H f CO2 (g)  nH 2O(l)  H f  H 2O(l) ]  nCH 3OH(l)  H f CH 3OH(l)
 [( 393.5 kJ)  2( 285.8 kJ)]  ( 239.1 kJ)
 393.5 kJ  571.6 kJ  239.1 kJ
H r   726.0 kJ
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Chapter 5: Thermochemistry
5.5-7
Step 3: Convert to enthalpy per gram of liquid ethanol.
M CH3OH(l)  32.05 g/mol
H r per gram CH3OH(l) 
H r per mole CH3OH(l)
M CH3OH(l)
726.0 kJ 1 mol

32.05 g
1 mol
 22.65 kJ/g

H r per gram CH3OH(l)
Statement: The standard enthalpy of combustion per gram of liquid methanol is –22.65 kJ/g.
(b) The standard enthalpy of combustion per gram of liquid ethanol is –30.58 kJ and that of
liquid methanol is –22.65 kJ. For 1 gram of the liquid fuel, liquid ethanol releases 7.93 kJ more
energy than liquid methanol.
(c) The fuel that would be the most convenient source of energy for a vehicle is ethanol because
for the mass of fuel burned, liquid ethanol releases 7.93 kJ/g more energy than methanol. That
means a lighter mass of fuel would be needed to produce the same amount of energy.
10. (a) A themochemical equation for the melting of gallium is:
Ga(s)  Ga(l) ΔH = 5.59 kJ/mol
or
Ga(s) + 5.59 kJ  Ga(l)
(b) Since the standard state of gallium is solid, H f Ga(s)  0 kJ/mol .
From the thermochemical equation, the enthalpy of formation of liquid gallium can be
calculated.
H r    nGa(l)  H Ga(l)   nGa(s)  H Ga(s)
5.59 kJ/mol  H f Ga(l)
The enthalpy of formation of liquid gallium is H f Ga(l)  5.59 kJ/mol , which is different from
that of solid gallium. The enthalpy of formation of liquid gallium is greater than that of solid
gallium because gallium is liquid at a temperature higher than 25 °C, so has more enthalpy than
solid gallium does.
Copyright © 2012 Nelson Education Ltd.
Chapter 5: Thermochemistry
5.5-8