GENERAL CHEMISTRY I
CHEM 1311.001
EXAM 4
Thursday July 3, 1997
1. Endothermic reactions cause their immediate surroundings to
| a) become cooler | b) become warmer | c) neither |
2. Exothermic reactions cause their immediately surroundings to
| a) become cooler | b) become warmer | c) neither |
3. The value of q, when defined with respect to an endothermic reaction is
| a) a positive number | b) a negative number | c) zero |
4. The value of q, when defined with respect to an exothermic reaction is
| a) a positive number | b) a negative number | c) zero |
5. How much heat is required to raise the temperaure of a
9.179 g sample of silver metal from 22.4 oC to 88.1 oC?
Note that the specific heat of silver is 0.235 J / (g oC).
| a) 48.9 J | b) 83.6 J | c) 102.4 J | d) 141.7 J |
| e) 173.5 J |
6. A 533.18 g sample of lead at 12.6 oC absorbs 2.176 kJ of heat.
What is the temperature of the lead after absorbing this heat?
Note that the specific heat of lead is 0.128 J / (g oC).
| a) 15.2 oC | b) 31.9 oC | c) 44.5 oC | d) 60.0 oC |
| e) 72.3 oC |
7. A 95.50 g sample of aluminum at 98.7 oC loses 5.132 kJ of heat.
What is the temperature of the aluminum after losing this heat? Note that
the specific heat of aluminum is 0.901 J / g oC.
| a) 21.4 oC | b) 39.1 oC | c) 59.6 oC | d) 74.2 oC |
| e) 88.2 oC |
8. A 5.193 g sample of copper at 100.0 oC is dropped into 24.016 g of water at 24.3 oC.
What will the final temperature be after the system has reached equilibrium?
Assume that all of the heat lost by the copper is gained by the water.
Note that the specific heat of copper is
0.387 J / (g oC) and the specific heat of
water is 4.1796 J / (g oC). Tip to avoid
errors: Remember that the specific heats of copper and water are
different, so you can't ignore them. Use the "full formula" for
calculating the final temperature.
| a) 25.8 oC | b) 26.5 oC | c) 27.0 oC | d) 28.3 oC |
| e) 29.4 oC |
9. A 48.66 g sample of water at 85.2 oC was mixed with a 21.15 g sample of water at 21.1 oC.
What was the final temperature of the combined masses of water?
Assume that all of the heat lost by the hot water is gained by the
cold water. Note that the specific heat of water (whether hot or cold) is
4.1796 J / (g oC). Hint: Since the
specific heats are the same, you can ignore them by using the
"condensed fromula" if you like.
| a) 24.9 oC | b) 36.7 oC | c) 50.0 oC | d) 65.8 oC |
| e) 74.4 oC |
10. Given the thermochemical equation
2H2(g)
+
O2(g)
----->
2H2O(l)
DH = -571.8 kJ
what is DH for the following reaction?
H2O(l)----->
H2(g)
+
(1/2)O2(g)
DH = ?
| a) -1146.3 kJ | b) -571.8 kJ | c) -285.9 kJ |
| d) 285.9 kJ | e) 571.8 kJ |
11. Given the thermochemical equation
4Fe(s)
+
3O2(g)
----->
2Fe2O3(s)
DH = -1644.4 kJ
what is DH for the following reaction?
8Fe(s)
+
6O2(g)
----->
4Fe2O3(s)
DH = ?
| a) -3288.8 kJ | b) -411.1 kJ | c) 822.2 kJ |
| d) 1644.4 kJ | e) 3288.8 kJ |
12. Given the thermochemical equation
N2(g)
+
3H2(g)
----->
2NH3(g)
DH = -92.38 kJ
what is DH for the following equation
2NH3(g)
----->
N2(g)
+
3H2(g)
DH = ?
| a) -184.76 kJ | b) -92.38 kJ | c) -46.19 kJ |
| d) 46.19 kJ | e) 92.38 kJ |
13. The balanced equation (in lowest terms) for the combustion of methane is
CH4(g)
+
2O2(g)
----->
CO2(g)
+
2H2O(g)
DH = -802.3 kJ
What is DH for the combustion of 8.195 g of CH4?
| a) -621.0 kJ | b) -409.8 kJ | c) -312 kJ |
| d) 312 kJ | e) 409 kJ |
14. When 18.75 g NaOH was dissolved in water in a calorimeter,
the temperature rose from 24.89 oC to 29.55 oC.
What is DH for the reaction
NaOH(s)
----->
Na+(aq)
+
OH-(aq)
The heat capacity for the calorimeter and its contents was 4330 J / oC.
Hint: Use q = C . Dt rather than
q = s . m . Dt for this problem.
Remember that what you are calculating initially is qcalorimeter but what you want is qreaction.
Note that qreaction = -qcalorimeter.
| a) +88.2 kJ | b) -88.2 kJ | c) +43.0 kJ |
| d) -43.0 kJ | e) +11.1 kJ |
15. The following are the thermochemical equations for the formation of
lead(II) sulfate from lead, lead(IV) oxide and sulfuric acid, and the
formation of sulfuric acid from sulfur trioxide and water:
Pb(s)
+
PbO2(s)
+
2H2SO4(l)
----->
2PbSO4(s)
+
2H2O(l)
DH = -509.2 kJ
SO3(g)
+
H2O(l)
----->
H2SO4(l)
DH = -130 kJ
Using the above thermochemical equations, and applying Hess's Law,
calculate the DH for the following reaction:
Pb(s)
+
PbO2(s)
+
2SO3(g)
----->
2PbSO4(s)
DH = ?
| a) -3.77 x 103 kJ | b) 3.77 x 103 kJ | c) -639 kJ |
| d) -521 kJ | e) -769 kJ |
Note: The target equation (the one you are to calculate
DH for) has been modified to include a coefficeint
of 2 in front of SO3. This coefficient was missing in the
original version. The correction was made Wednesday June 9, 1999. Your
printed version of this exam is outdated if it does not include this
correction notice.
16. Use standard enthalpies of formation listed in Table 5.2 to calculate
DH for the following reaction:
Fe2O3(s)
+
3CO(g)
----->
2Fe(s)
+
3CO2(g)
DH = ?
| a) 26.8 kJ | b) -26.8 kJ | c) 145.7 kJ |
| d) -145.7 kJ | e) 218.3 kJ |
Note: Table 5.2 refers to the table on page 209 of Chemistry the Study
of Matter and its Changes by Brady and Holum (John Wiley, 1996), the
textbook in use at the time this test was written. If you use a different
text to solve this problem, you may get a slightly different answer, due to
minor differences in various published sources of thermochemical information.
ANSWERS:
1 a
2 b
3 a
4 b
5 d
6 c
7 b
8 a
9 d
10 d
11 a
12 e
13 b
14 d
15 e
16 b
This page was last modified Wednesday June 9, 1999