1. KEY
GENERAL CHEMISTRY-I (1411)
S.I. # 23
1. Calculate the enthalpy change for the reaction
P4O6 (s) + 2 O2 (g)  P4O10 (s)
Given the following enthalpies of reaction:
P4 (s) + 3 O2 (g)  P4O6 (s) ∆H = - 1640.1 kJ
P4 (s) + 5 O2 (g)  P4O10 (s) ∆H = -2940.1 kJ
P4O6 (s)  P4 (s) + 3 O2 (g) ∆H = 1640.1 kJ
P4 (s) + 5 O2 (g)  P4O10 (s) ∆H = -2940.1 kJ
P4O6 (s) + 2 O2 (g)  P4O10 (s) ∆H = - 1300.0 kJ
2. Calculate the value of ∆H° for each of the following reactions. Use Appendix C of
your chemistry book.
a) N2O4 (g) + 4 H2 (g)  N2 (g) + 4 H2O (g)
∆H°rxn = 4 ∆H°f H2O (g) + ∆H°f N2 (g) - ∆H°f N2O4 (g) - 4∆H°f H2 (g)
= 4(-241.82) + 0 – (9.66) – 4(0) = -976.94 kJ
b) 2 KOH (s) + CO2 (g)  K2CO3 (s) + H2O (g)
∆H°rxn = ∆H°f K2CO3 (s) + ∆H°f H2O (g) - 2∆H°f KOH (s) -∆H°f CO2 (g)
= -1150.18 kJ – 241.82 kJ – 2(-424.7)kJ – (-393.5 kJ) = - 149.1 kJ
c) SO2 (g) + 2 H2S (g)  3/8 S8 (s) + 2 H2O (g)
∆H°rxn = 3/8 ∆H°f S8 (s) + 2∆H°f H2O (g) - ∆H°f SO2 (g) - 2∆H°f H2S (g)
= 3/8 (0) + 2(-241.82) – (-296.9) – 2 (-20.17) = -146.4 kJ
d) Fe2O3 (s) + 6 HCl (g)  2 FeCl3 (s) + 3 H2O (g)
∆H°rxn = 2∆H°f FeCl3 (s) + 3∆H°f H2O (g) - ∆H°f Fe2O (g) - 6∆H°f HCl (g)
= 2(-400 kJ) + 3(-241.82 kJ) – (-822.16 kJ) – 6(-92.30 kJ) = -150 kJ
3. a) Calculate the standard enthalpy of formation of gaseous diborane (B2H6) using
the following thermochemical information:
4B (s) + 3 O2 (g)  2B2O3 (s) ∆H° = -2509.1 kJ
2H2 (g) + O2 (g)  2 H2O (l) ∆H° = -571.7 kJ
B2H6 (g) + 3 O2 (g)  B2O3 (s) + 3 H2O (l) ∆H° = -2147.5 kJ
2B (s) + 3/2 O2 (g)  B2O (s) ∆H° = ½ (-2509.1 kJ)
3H2 (g) + 3/2 O2 (g)  3 H2O (l) ∆H° = 3/2 (-571.7 kJ)
B2O3 (s) + 3 H2O (l)  B2H6 (g) + 3 O2 (g) ∆H° = -(-2147.5 kJ)
2 B (s) + 3 H2 (g)  B2H6 (g) ∆H° f = +35.4 kJ
4. a) what is meant by the term fuel value? b) Which is a greater source of energy as
food, 5 g of fat or 9 g of carbohydrates?
a) Fuel value is the amount of heat produced when 1 gram of a substance (fuel) is combusted.
b) The fuel value of fats is 9 kcal/g and of carbohydrates is 4 kcal/g. Therefore, 5 g of fat
produce 45 kcal, while 9 g of carbohydrates produces 36 kcal; 5 g of fat are a greater energy
source.
5. The heat of combustion of ethanol, C2H5OH (l) is -1367 kJ/mol. A batch of wine

2. KEY
contains 10.6% ethanol by mass. Assuming the density of the wine to be 1.0g/mL,
what caloric content does the alcohol (ethanol) in a 6-oz glass of wine (177 mL)
have?
(177 mL ) (1.0g wine / 1 mL) (0.106 g ethanol / 1 g wine) (1mol ethanol/46.1g ethanol)
(1367 kJ / 1 mol ethanol) (1 Cal / 4.184 kJ) = 133 = 1.3x102 Cal
A typical 6 oz. glass of wine has 150-250 Cal, so this is a reasonable
result. Note that alcohol is responsible for most of the food value of wine.
6. a) Why are fats well suited for energy storage in the human body? b) A
particular chip snack food is composed of 12% protein, 14% fat, and the rest
carbohydrate. What percentage of the calorie content of this food is fat? c) How
many grams of protein provide the same fuel value as 25g of fat?
a) Fats are appropriate for fuel storage because they are insoluble in water (and body
fluids) and have high fuel value.
b) assume 100 g of chips for convenience.
(12g Protein) (17kJ/ 1g protein) (1 Cal / 4.184 kJ) = 48.76 = 49 Cal
(14g Fat) (38kJ / 1g Fat) (1 Cal / 4.184 kJ) = 127.15 = 130 Cal
(74 g Carbs) (17 kJ / 1g Carbs) (1 Cal / 4.184 kJ) = 300.67 = 301 Cal
total Cal = 48.76 + 130 + 300.67 = 476.58 = 480 Cal
% Cal from fat = (127.15 Cal Fat / 476.58 total Cal) x 100 = 26.68 = 27%
c) (25 g fat) (38 kJ / 1g fat) = (X g protein) (17 kJ / g protein) ; X = 56 g protein
7. Write balanced equations that describe the formation of the following
compounds from their elements in their standard states, and use Appendix C to
obtain the values of their standard enthalpies of formation: a) HBr (g), b) AgNO3
(s), c) Hg2Cl2 (s), d) C2H5OH (l).
a) ½ H2 (g) + ½ Br2 (l)  HBr (g) ∆H°f = -36.23 kJ
b) Ag(s) + ½ N2 (g) + 3/2 O2 (g)  AgNO3 (s) ∆H°f = -124.4 kJ
c) 2 Hg (l) + Cl2 (g)  Hg2Cl2 (s) ∆H°f = -264.9 kJ
d) 2 C (s,gr) + ½ O2 (g) + 3H2 (g)  C2H5OH (l) ∆H°f = -277.7 kJ
8. Suppose that the gas-phase reaction 2NO (g) + O2 (g)  2 NO2 (g), were carried
out in a constant –volume container at constant temperature. Would the measured
heat change represent ∆H or ∆E? If there is a difference, which quantity is larger
for this reaction? Explain.
At constant volume (∆V = 0), ∆E = qv. According to the definition of enthalpy, H = E + PV, so
∆H = ∆E + ∆(PV). For an ideal gas at constant temperature and volume, ∆PV = V∆P = RT∆n.
For this reaction, there are 2 mol of gaseous product and 3 mol of gaseous reactants, so ∆n = -1.
Thus V∆P or ∆(PV) is negative. Since ∆H = ∆E + ∆(PV), the negative ∆(PV) term means that
∆H will be smaller or more negative than ∆E. ∆H<∆E