Enthalpy of Reaction (∆Hrxn)
1
a)
2 CO(g) + O2(g)  2 CO2(g)
∆Hrx = (2 mol)(∆Hf CO2) – (2 mol)(∆Hf CO) – (1 mol)(∆Hf O2)
∆Hrx = (2 mol)(–393.5 kJ/mol) – (2 mol)(–110.5 kJ/mol) – (1 mol)(0 kJ/mol)
∆Hrx = –566.0 kJ
b)
Cu(s) + 2 H2SO4(aq)  CuSO4(s) + SO2(g) + 2 H2O(l)
∆Hrx = (1 mol)(∆Hf CuSO4) + (1 mol)(∆Hf SO2) + (2 mol)(∆Hf H2O(l)) – (1 mol)(∆Hf Cu) – (2 mol)(∆Hf H2SO4)
∆Hrx = (1 mol)(–771.4 kJ/mol) +(1 mol)(–296.8 kJ/mol) + (2 mol)(–285.8 kJ/mol) – (1 mol)(0 kJ/mol) – (2 mol)(–814.0 kJ/mol)
∆Hrx = –11.8 kJ
c)
N2(g) + 3 H2(g)  2 NH3(g)
∆Hrx = (2 mol)(∆Hf NH3) – (1 mol)(∆Hf N2) – (3 mol)(∆Hf H2)
∆Hrx = (2 mol)(–46.1 kJ/mol) – (1 mol)(0 kJ/mol) – (3 mol)(0 kJ/mol)
∆Hrx = –92.2 kJ
d)
Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
∆Hrx = (2 mol)(∆Hf Fe) + (3 mol)(∆Hf CO2) – (1 mol)(∆Hf Fe2O3) – (3 mol)(∆Hf CO)
∆Hrx = (2 mol)(0 kJ/mol) +(3 mol)(–393.5 kJ/mol) – (1 mol)(–824.2 kJ/mol) – (3 mol)(–110.5 kJ/mol)
∆Hrx = –24.8 kJ
e)
2 SO2(g) + O2(g)  2 SO3(g)
∆Hrx = (2 mol)(∆Hf SO3) – (2 mol)(∆Hf SO2) – (1 mol)(∆Hf O2)
∆Hrx = (2 mol)(–395.7 kJ/mol) – (2 mol)(–296.8 kJ/mol) – (1 mol)(0 kJ/mol)
∆Hrx = –197.8 kJ
f)
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(l)
∆Hrx = (4 mol)(∆Hf NO) + (6 mol)(∆Hf H2O(l)) – (4 mol)(∆Hf NH3) – (5 mol)(∆Hf O2)
∆Hrx = (4 mol)(90.2 kJ/mol) +(6 mol)(–285.8 kJ/mol) – (4 mol)(–46.1 kJ/mol) – (5 mol)(0 kJ/mol)
∆Hrx = –1169.6 kJ
L. Farrell - Enthalpy of Reaction (∆Hrx) and Hess’ Law – Answers – Page 1 of 9
g)
2 NO(g) + O2(g)  2 NO2(g)
∆Hrx = (2 mol)(∆Hf NO2) – (2 mol)(∆Hf NO) – (1 mol)(∆Hf O2)
∆Hrx = (2 mol)(33.2kJ/mol) – (2 mol)(90.2kJ/mol) – (1 mol)(0 kJ/mol)
∆Hrx = –114.0 kJ
h)
Ca(s) + 2 H2O(l)  Ca(OH)2(s) + H2(g)
∆Hrx = (1 mol)(∆Hf Ca(OH)2) + (1 mol)(∆Hf H2) – (1 mol)(∆Hf Ca) – (2 mol)(∆Hf H2O(l))
∆Hrx = (1 mol)(–986.1 kJ/mol) +(1 mol)(0 kJ/mol) – (1 mol)(0 kJ/mol) – (2 mol)(–285.8 kJ/mol)
∆Hrx = –414.5 kJ
2) Calculate the amount of heat released when 15.0 g of methanol burns.
CH3OH(l) + 3 O2(g)  CO2(g) + 2H2O(l)
2
∆Hrxn = (1 mol)( CO2(g)) + (2 mol)( 2H2O(l)) – (1 mol)( CH3OH(l)) – ( 3 mol)(O2(g))
2
∆Hrxn = (1 mol)(-393.5 kJ/mol) + (2 mol)(-285.8 kJ/mol) – (1 mol)(-238.6 kJ/mol) - ( 3 mol)(0 kJ/mol)
2
∆Hrxn = -726.5 kJ
n=
15.0 g
mass
=
= 0.468 mol methanol
molar mass 32.04216 g mol
0.468 mol
x
=
1 mol
− 726.5 kJ
x = −340. kJ
3) Calculate the mass of ethanol that must be burned in order to produce 20.0 kJ.
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l)
∆Hrxn = (2 mol)(CO2(g)) + (3 mol)(H2O(l)) – (1 mol)(C2H5OH(l)) – (3 mol)(O2(g))
∆Hrxn = (2 mol)(-393.5 kJ/mol) + (3 mol)(-285.8 kJ/mol) – (1 mol)(-277.1 kJ/mol) – (3 mol)(0 kJ/mol)
∆Hrxn = -1367.3 kJ
1 mol −1367.3 kJ
=
x
− 20.0 kJ
x = 0.0146 mol
mass = (mol)(mol mass ) = (0.0146 mol)(46.06904 g mol) = 0.674 g
ethanol
L. Farrell - Enthalpy of Reaction (∆Hrx) and Hess’ Law – Answers – Page 2 of 9
4) One teaspoon of sugar has a mass of 5.00g.
a) Calculate the energy change when a person consumes 2.00 teaspoons of sugar.
C12H22O11(s) + 12O2(g)  12CO2(g) + 11H2O(l)
∆Hrxn = (12 mol)( CO2(g)) + (11 mol)( H2O(l)) – (1 mol)(C12H22O11(s)) – (12 mol)(O2(g))
∆Hrxn = (12 mol)(-393.5 kJ/mol) + (11 mol)(-285.8 kJ/mol) – (1 mol)(-2225.5 kJ/mol) – (12 mol)(0 kJ/mol)
∆Hrxn = -5640.3 kJ
n=
10.0 g
mass
=
= 0.0292 mol
mol mass 342.30008 g mol
1 mol
− 5640.3 kJ
=
0.0292 mol
x
x = −164.8 kJ
b) Energy obtained from food is often reported in food calories, or Calories
1 Calorie (1 Cal) = 1 kilocalorie (1 kcal) = 1000 calories (1000 cal) = 4.18 kJ
Calculate the number of food calories contained in 2.00 teaspoons of sugar.
1 Cal 4.18 kJ
=
x
165 kJ
x = 39.4 Cal
c) A 150.0 lb person walking at a moderate speed will burn 40.0 Cal while walking for 15.0 minutes.
How long would this person have to walk to burn the Calories in 2.00 teaspoons of sugar?
15.0 min 40.0 Cal
=
x
39.4 Cal
x = 14.8 minutes
5) Which sugar compound is more stable, glucose or sucrose? Explain.
Glucose
Sucrose
∆Hf = -1273.1 kJ/mol
∆Hf = -2225.5 kJ/mol
Both glucose and sucrose are very stable as evidenced by their large negative enthalpies of
formation. Sucrose is more stable than glucose based on the large negative enthalpy of formation.
L. Farrell - Enthalpy of Reaction (∆Hrx) and Hess’ Law – Answers – Page 3 of 9
6) What mass of hydrogen gas should be burned (under standard conditions) to heat 1.00 L of water
from 10.0°C to 80.0°C?
q = mc∆t
q = (1.00 kg)(4.18 J/gºC)(70.0ºC)
q = 296.6 kJ
∴the water requires 292.6 kJ
∴the combustion of hydrogen must supply 292.6 kJ
2H2(g) + O2(g)  2H2O(l)
∆Hrxn = (2 mol)( H2O(l)) – (2 mol)( 2H2(g)) – (1 mol)( O2(g))
∆Hrxn = (2 mol)(-285.8 kJ/mol) – (1 mol)(0 kJ/mol) – (1 mol)(0 kJ/mol)
∆Hrxn = -571.8 kJ
2 mol −571.8 kJ
=
x
− 292.6 kJ
x = 1.02 mol
mass = (mol)(mol mass ) = (1.02 mol)(20.1588 g mol) = 2.06 g
hydrogen
7) What would be the final temperature of 885 g water, initially at 25.0°C, if it were heated by the
burning of 20.0 g of sulfur in excess oxygen to produce sulfur dioxide?
S(s) + O2(g)  SO2(g)
∆Hrxn = (1 mol)(-296.8 kJ/mol) – (1 mol)(0 kJ/mol) – (1 mol)(0 kJ/mol)
∆Hrxn = -296.8 kJ
n=
20.0g
mass
=
= 0.624molSulfur
molmass 32.066 g mol
1 mol
− 296.8 kJ
=
0.624 mol
x
x = −185 kJ
∴ when 20.0 g of sulfur burns, 185 kJ is released
q = mcΔc
185 kJ = (0.885 kg )(4.18 J g°C )(T − 25.0°C )
185 kJ
(0.885 kg)(4.18 J g°C )
= T − 25.0°C
50.0°C = T − 25.0°C
T = 75.0°C
L. Farrell - Enthalpy of Reaction (∆Hrx) and Hess’ Law – Answers – Page 4 of 9
8) When 4.00 g of ammonium chloride is formed from its elements, 23.5 kJ of energy is released.
a) Write the thermochemical equation showing the formation of ammonium chloride.
½N2(g) + 2H2(g) + ½Cl2(g)  NH4Cl(s)
b) Calculate the standard heat of formation for ammonium chloride.
n=
4.00 g
mass
=
= 0.0748 mol
mol mass 53.49146 g mol
0.0748 mol − 23.5 kJ
=
1 mol
x
x = −314 kJ
∴ ΔHf NH4 Cl(s) = - 314 kJ/mol
9) What volume of water would be needed to absorb the energy released in burning 1.00 g of
magnesium and show a temperature increase (of the water) of 10.0°C?
2Mg(s) + O2(g)  2MgO(s)
∆Hrxn = (2 mol)(601.7 kJ/mol) – (2 mol)(0 kJ/mol) – (1 mol)(0 kJ/mol)
∆Hrxn = -1203.4 kJ
n=
1.00 g
mass
=
= 0.0411 mol
mol mass 24.305 g mol
2 mol
− 1203.4 kJ
=
0.0411 mol
x
x = −24.5 kJ
∴ 24.8 kJ is released when 1.00 g of magnesium burns
∴ 24.8 kJ is absorved by the water
q = mc∆t
24.8 kJ = m(4.18 kJ kg°C )(10.0°C )
m=
24.8kJ
= 0.592kg
(4.18 kJ kg°C )(10.0°C )
∴ 0.592 L or 592 mL of water is needed
L. Farrell - Enthalpy of Reaction (∆Hrx) and Hess’ Law – Answers – Page 5 of 9
Hess’ Law
A.
B.
C.
CO2(g) + H2(g)  H2O(l) + CO(g)
H2(g) + ½O2(g)  H2O(l)
CO(g) + ½O2(g)  CO2(g)
∆H = –285.9 kJ
∆H = –393.5 kJ
CO2(g)  CO(g) + ½O2(g)
∆H = (–1)(–393.5 kJ)
H2(g) + ½O2(g)  H2O(l)
∆H =
CO2(g) + H2(g)  H2O(l) + CO(g)
∆H = +107.6 kJ
2C2H5OH(l) + O2(g)  2C2H4O(l) + 2H2O(l)
C2H5OH + 3O2(g)  2CO2(g) + 3H2O(l)
C2H4O(l) + 5 2 O2(g)  2CO2(g) + 2H2O(l)
–285.9 kJ
∆H = –1370.7 kJ
∆H = –1167.3 kJ
2C2H5OH + 6O2(g)  4CO2(g) + 6H2O(l)
∆H = (2)(–1370.7 kJ)
4CO2(g) + 4H2O(l)  2C2H4O(l) + 5O2(g)
∆H = (2)(–1167.3 kJ)
2C2H5OH(l) + O2(g)  2C2H4O(l) + 2H2O(l)
∆H = –406.8 kJ
2C(gr) +
5
2
H2(g) + ½Cl2(g)  C2H5Cl(g)
H2(g) + ½O2(g)  H2O(l)
C(gr) + O2(g)  CO2(g)
C2H4(g) + HCl(g)  C2H5Cl(g)
H2(g) + Cl2(g)  2HCl(g)
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l)
∆H
∆H
∆H
∆H
∆H
= –285.9 kJ
= –393.5 kJ
= –72.0 kJ
= –184.6 kJ
= –1410.8 kJ
2C(gr) + 2O2(g)  2CO2(g)
∆H = (2)(–393.5 kJ)
C2H4(g) + HCl(g)  C2H5Cl(g)
∆H =
2CO2(g) + 2H2O(l)  C2H4(g) + 3O2(g)
∆H = (–1)(–1410.8 kJ)
½H2(g) + ½Cl2(g)  HCl(g)
∆H = ( ½)( –184.6 kJ)
2H2(g) + O2(g)  2H2O(l)
∆H = (2)(–285.9 kJ)
2C(gr) +
∆H = -112.3 kJ
5
2
H2(g) + ½Cl2(g)  C2H5Cl(g)
L. Farrell - Enthalpy of Reaction (∆Hrx) and Hess’ Law – Answers – Page 6 of 9
( –72.0 kJ)
D.
E.
Na(s) + ½Cl2(g)  NaCl(s)
2Na(s) + 2HCl(g)  2NaCl(s) + H2(g)
H2(g) + Cl2(g)  2HCl(g)
∆H = –637.4 kJ
∆H = –184.6 kJ
Na(s) + HCl(g)  NaCl(s) + ½H2(g)
∆H = (½ )(–637.4 kJ)
½H2(g) + ½Cl2(g)  HCl(g)
∆H = (½ )(–184.6 kJ)
Na(s) + ½Cl2(g)  NaCl(s)
∆H = –411 kJ
2XO2(s) + CO(g)  X2O3(s) + CO2(g)
XO2(s) + CO(g)  XO(s) + CO2(g)
X3O4(s) + CO(g)  3XO(s) + CO2(g)
3X2O3(s) + CO(g)  2X3O4(s) + CO2(g)
∆H = –83.7 kJ
∆H = +25.1 kJ
∆H = –50.2 kJ
2XO2(s) + 2CO(g) 2 XO(s) + 2CO2(g)
∆H = (2)(–83.7 kJ)
2
3
X3O4(s) +
2XO(s) +
F.
2
1
3
CO2(g)  X2O3(s) +
3 CO2(g)

2
3 X3O4(s)
+
2
1
CO(g)
3
3 CO(g)
∆H = (– 1 3 )(–50.2 kJ)
∆H = (– 2 3 )(+25.1 kJ)
2XO2(s) + CO(g)  X2O3(s) + CO2(g)
∆H = –167.4 kJ
2MnO2(s) + CO(g)  Mn2O3(s) + CO2(g)
MnO2(s) + CO(g)  MnO(s) + CO2(g)
Mn3O4(s) + CO(g)  3MnO(s) +CO2(g)
3Mn2O3 + CO(g)  2Mn3O4(s) + CO2(g)
∆H = –150.6 kJ
∆H = –54.4 kJ
∆H = –142.3 kJ
2MnO2(s) + 2CO(g)  2MnO(s) + 2CO2(g)
∆H = (2)(–150.6 kJ)
2
3
Mn3O4(s) +
2MnO(s) +
1
3
CO2(g)  Mn2O3 +
CO(g)
∆H = (- 1 3 )(–142.3 kJ)
∆H = (- 2 3 )(–54.4 kJ)
2MnO2(s) + CO(g)  Mn2O3(s) + CO2(g)
∆H = -217.5 kJ
3
2
3
Mn3O4(s) +
3
CO(g)
2
CO2(g) 
1
2
3
L. Farrell - Enthalpy of Reaction (∆Hrx) and Hess’ Law – Answers – Page 7 of 9
G.
3V2O3(s)  V2O5(s) + 4VO(s)
2V(s) + 3 2 O2(g)  V2O3(s)
2V(s) +
5
2
O2(g)  V2O5(s)
V(s) + ½O2(g)  VO(s)
V(s) + O2((g)  VO2(s)
3V2O3(s)  6V(s) +
2V(s) +
5
2
9
2
O2(g)
O2(g)  V2O5(s)
4V(s) + 2O2(g)  4VO(s)
3V2O3(s)  V2O5(s) + 4VO(s)
H.
5V(s) +
J.
∆H = –1561 kJ
∆H = –418 kJ
∆H = –720 kJ
∆H = (-3)(–1213 kJ)
∆H = –1561 kJ
∆H = (4)(–418 kJ)
∆H = 406 kJ
V2O5(s) + 3V(s)  5VO(s)
Data as in G.
V2O5(s)  2V(s) +
I.
∆H = –1213 kJ
5
2
5
2
O2(g)
O2(g)  5VO(s)
∆H = (-1)(–1561 kJ)
∆H = (5)(–418 kJ)
V2O5(s) + 3V(s)  5VO(s)
∆H = -529 kJ
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
Fe2O3(s) + CO(g)  2FeO(s) + CO2(g)
Fe(s) + CO2(g)  FeO(s) + CO(g)
∆H = –2.9 kJ
∆H = +11.3 kJ
Fe2O3(s) + CO(g)  2FeO(s) + CO2(g)
2FeO(s) + 2CO(g)  2Fe(s) + 2CO2(g)
∆H = –2.9 kJ
∆H = (-2)(+11.3 kJ)
2C(gr) + H2(g)  C2H2(g)
CaO(s) + H2O(l)  Ca(OH)2(s)
CaO(s) + 3C(gr)  CaC2(s) + CO(g)
CaC2(s) + 2H2O(l)  Ca(OH)2 + C2H2(g)
2C(gr) + O2(g)  2CO(g)
2H2O(l)  2H2(g) + O2(g)
∆H
∆H
∆H
∆H
∆H
CaC2(s) + 2H2O(l)  Ca(OH)2 + C2H2(g)
Ca(OH)2(s)  CaO(s) + H2O(l)
CaO(s) + 3C(gr)  CaC2(s) + CO(g)
H2(g) + ½O2(g)  H2O(l)
CO(g)  C(gr) + ½O2(g)
∆H
∆H
∆H
∆H
∆H
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
2C(gr) + H2(g)  C2H2(g)
∆H = -25.5 kJ
= –65.3 kJ
= +462.3 kJ
= –125.5 kJ
= –220.9 kJ
= +571.5 kJ
= –125.5 kJ
= (-1)(–65.3 kJ)
= +462.3 kJ
= (-½)(+571.5 kJ)
= (-½)(–220.9 kJ)
∆H = 226.8 kJ
L. Farrell - Enthalpy of Reaction (∆Hrx) and Hess’ Law – Answers – Page 8 of 9
K.
Cu(s) + ½O2(g)  CuO(s)
Cu2O(s) + ½O2(g)  2CuO(s)
CuO(s) + Cu(s)  Cu2O(s)
∆H = –143.9 kJ
∆H = –11.3 kJ
Cu2O(s) + ½O2(g)  2CuO(s)
CuO(s) + Cu(s)  Cu2O(s)
∆H = –143.9 kJ
∆H = –11.3 kJ
Cu(s) + ½O2(g)  CuO(s)
∆H = -155.2 kJ
L. Farrell - Enthalpy of Reaction (∆Hrx) and Hess’ Law – Answers – Page 9 of 9