CHEM 1411 PRACTICE PROBLEM OF Chapters 6
Multiple Choices: Select one best answer.
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
Hint: p. 232: Example 6.2. ΔE=q+w=+127-325=-198J
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
(e) +2905.6
Hint: p. 236: Guidelines or p. 249: Example 6.9: concepts to use in multiple or reversing
equation. If you reverse a reaction, the sign of ΔH changes. If you multiply both sides of
the equation by a factor n, then DH must change by the same factor n.
3. 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
Hint: p. 237: Example 6.3. First determine the number of mole of nitrogen dioxide produced.
n=m/M= 12.6*1000g/(46g/mol)=273.91mol,
q=n*∆H=273.91mol*(-114.6 kJ) = -1.57 x 104kJ
2mol NO2
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
Hint: p. 240: Example 6.5. q = m x s x Δt =4.18 J/g.K*250g*(98-22)=79kJ
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
Hint: p. 243: Example 6.7. Or heat absorbed by the colder object gold = - heat released by the
hotter object iron. p. 256: problem 6.36. qgold=qiron, q = m x s x Δt, 10.0g*0.129J/g.oC(T-18.0)=
20.0g*0.444 J/g.oC*(55.6-T), solve T=50.7
6. Calculate the heat of combustion (kJ/mol) for the following reaction:
1
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
Hint: p. 246: equation 6.17 or 6.18, p. 247: in-text example, and p. 250: Example 6.10.
∆H=2∆Hf 0SO2+2∆Hf 0H2o -2∆Hf 0H2S -3∆Hf 0O2
7. From the following data,
C(graphite) + O2(g)  CO2(g)
H2(g) + 1/2O2(g)  H2O(l)
2C2H6(l) + 7O2(g)  4CO2(g) + 6H2O(l)
∆H = -393.5 kJ/mol
∆H = -285.8 kJ/mol
∆H = -3119.6 kJ/mol
Calculate the enthalpy change (kJ/mol) for the reaction
2C(graphite) + 3H2(g)  C2H6(g)
(a) -84.6
(b) -42.3
(c) +84.6
(d) +42.3
(e) -67.2
Hint: p.p. 248-250: in-text example and Example 6.9.
2C(graphite) + 2O2(g)  2CO2(g)
∆H = -393.5*2 kJ/mol
+3H2(g) + 3/2O2(g)  3H2O(l)
∆H = -285.8*3 kJ/mol
+2CO2(g) + 3H2O(l)  C2H6(l) + 7/2O2(g)
∆H = 3119.6 /2kJ/mol
2C(graphite) + 3H2(g)  C2H6(g)
∆H=-393.5*2+(-285.8 kJ/mol)*3+( 3119.6 /2)kJ/mol=-84.6kJ/mol
8. *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)
2Fe(s) + 3/2O2(g)  Fe2O3(s)
∆H = -1669.8 kJ/mol
∆H = -822.2 kJ/mol
(a) -637.1
(b) -847.6
(c) -984.6
(d) -1120.3
(e) +847.6
Hint: p.p. 248-250: in-text example and Example 6.9.
2Al(s) + 3/2O2(g)  Al2O3(s)
∆H = -1669.8 kJ/mol
+Fe2O3(s)2Fe(s) + 3/2O2(g)
∆H = 822.2 kJ/mol
2Al(s) + Fe2O3(s)  2Fe(s) + Al2O3(s) ∆H = -1669.8 +8222.2=-847.6 kJ/mol
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