PRACTICE PROBLEM OF Chapters 6
1.
*What is the change in energy (J) of the gas when a gas expends and does P-V work on
the surroundings equal to 325 J and at the same time absorbs 127 J of heat from the surroundings?
(a) -198
(b) + 198
(c) + 157
(d) -157
(e) + 452
2.
*Consider the reaction below:
2CH3OH(l) + 3O2(g)  4H2O(l) + 2CO2(g)
∆H = -1452.8 kJ/mol
What is the value of ∆H for the reaction of
8H2O(l) + 4CO2(g)  4CH3OH (l) + 6O2 (g)?
(a) +1458.2
(b) -1458.2
(c) 3.46
(d) -2905.6
3.
(e) +2905.6
Determine the amount of heat (in kJ) given off when 12.6 kilograms of nitrogen dioxide
are produced according to the equation:
2NO(g) + O2(g)  2NO2(g)
∆H = -114.6 kJ/mol
(a) -1.57 x 104 (b) -1.14 x 103 (c) -3.76 x 102 (d) -1.57 x 104 (e) -1.05 x 103
4. What is the heat absorbed (in kJ) when 250 grams of water is heated from 22oC to 98oC? The
specific heat of water is 4.18 J/g.K (or 4.18 J/g. oC)
(a) 79
(b) 88
(c) 97
(d) 102
(e) 137
5.
*A sheet of gold weighing 10.0 g and at a temperature of 18.0oC is placed flat on a sheet
of iron weighing 20.0 g and at a temperature of 55.6oC. What is the final temperature (oC) of the
combined metals? Assume that no heat is lost to the surroundings. The specific heat of iron and
gold are o.444 and 0.129 J/g.oC , respectively. (Hint: the heat gained by the gold must be equal to
the heat lost by the iron.)
(a) 50.7
(b) 63.1
(c) 72.4
(d) 47.2
(e) 36.8
6.
Calculate the heat of combustion (kJ/mol) for the following reaction:
2H2S(g) + 3O2(g)  2H2O(l) + 2SO2(g)
The standard enthalpies of formation for H2S(g), H2O(l), SO2(g) and O2(g) are
-20.15, -285.8, -296.4 and 0.0, respectively.
(a) +1124
(b) -1124
(c) +562
(d) -562
(e) +281
7.
From the following data,
C(graphite) + O2(g)  CO2(g)
∆H = -393.5 kJ/mol
H2(g) + 1/2O2(g)  H2O(l)
∆H = -285.8 kJ/mol
2C2H6(l) + 7O2(g)  4CO2(g) + 6H2O(l)
∆H = -3119.6 kJ/mol
Calculate the enthalpy change (kJ/mol) for the reaction
2C(graphite) + 3H2(g)  C2H6(g)
(a) -84.6
8.
(b) -42.3
(c) +84.6
(d) +42.3
(e) -67.2
*Calculate the standard enthalpy change (kJ/mol) for the reaction
2Al(s) + Fe2O3(s)  2Fe(s) + Al2O3(s)
Given that
2Al(s) + 3/2O2(g)  Al2O3(s)
∆H = -1669.8 kJ/mol
2Fe(s) + 3/2O2(g)  Fe2O3(s)
∆H = -822.2 kJ/mol
(a) -637.1
(b) -847.6
(c) -984.6
1
(d) -1120.3
(e) +847.6