GENERAL CHEMISTRY I
CHEM 1311.001
EXAM 4
Thursday July 2, 1998


Problems Involving Thermochemical Terminology and Basic Principles


1. Reactions which release heat to the surroundings are said to be
a) endothermic b) exothermic
2. Reactions which absorb heat from the surroundings are said to be
a) endothermic b) exothermic
3. For exothermic reactions qreaction is
a) positive (qreaction > 0) b) negative (qreaction < 0) c) zero (qreaction = 0)
4. For endothermic reactions qreaction is
a) positive (qreaction > 0) b) negative (qreaction < 0) c) zero (qreaction = 0)
5.Endothermic reactions

a) cause their immediate surroundings to become warmer

b) cause their immediate surroundings to become cooler

c) have no effect on their immediate surroundings

6. Exothermic reactions

a) cause their immediate surroundings to become warmer

b) cause their immediate surroundings to become cooler

c) have no effect on their immediate surroundings

7. Four (4) systems A, B, C, and D were all in thermal contact with each other, and all were initially at different temperatures. After a redistribution of heat among these 4 sytstems, they reached thermal equilibrium. The following is a description of the heat redistribution among systems A, B and C:

System A lost 5 kJ of heat
System B gained 3 kJ of heat
System C lost 6 kJ of heat

What happened to System D?

a) System D lost 8 kJ of heat

b) System D lost 4 kJ of heat

c) System D gained 2 kJ of heat

d) System D gained 4 kJ of heat

e) System D gained 8 kJ of heat

Problems Using q = s . m . Dt


8. How much heat is required to raise the temperature of 68.21 g of water from 14.3 oC to 75.9 oC? Note that the specific heat of water is 4.184 J / g oC.
a) 4.08 kJ b) 6.11 kJ c) 12.8 kJ d) 17.6 kJ e) 21.7 kJ
9. If 206.92 g of water at 14.19 oC absorbs 15.81 kJ of heat, what is the final temperature of the water? The specific heat of water is 4.184 J / g oC.
a) 18.26 oC b) 32.45 oC c) 54.21 oC d) 72.18 oC e) 93.41 oC
10. A 75.12 g sample of water initially at 91.5 oC lost 6.15 kJ of heat. What was the final temperature of the water? The specific heat of water is 4.184 J / g oC.
a) 19.6 oC b) 38.7 oC c) 59.7 oC d) 71.9 oC e) 111.1 oC
11. A 6.196 g sample of zinc at 99.7 oC is dropped into 28.950 g of water at 25.2 oC. What will the final temperature be after the system has reached equilibrium? Assume that all of the heat lost by the zinc is gained by the water. Note that the specific heat of zinc is 0.390 J / g oC and the specific heat of water is 4.18 J / g oC.

Tip to avoid errors: Remember that the specific heats of zinc and water are different, so you can't ignore them. Use the "full formula" for calculating the final temperature.
a) 26.7 oC b) 38.3 oC c) 47.8 oC d) 77.1 oC e) 98.2 oC
12. A 53.11 g sample of water at 61.3 oC was mixed with a 38.52 g sample of water at 17.9 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 4.18 J / g oC.

Hint: Since the specific heats are the same, you can ignore them by using the "condensed formula" if you like.
a) 36.1 oC b) 39.6 oC c) 43.1 oC d) 49.4 oC e) 57.8 oC
Problems Using Hess' Law


13. Given the reaction

CaC2(s)   +   2H2O(l)   ----->   C2H2(g)   +   Ca(OH)2(aq)      DHrxn = -128 kJ

What is DH for the following reaction:

2C2H2(g)   +   2Ca(OH)2(aq) ----->   2CaC2(s)   +   4H2O(l)      DHrxn = ?
a) -256 kJ b) -128 kJ c) +64 kJ d) +128 kJ e) +256 kJ
14. Given the following thermochemical equation:

CH3COOH(l)   +   2O2(g)   ---------->   2CO2(g)   +   2H2O(g)      DHrxn = -783.6 kJ

What is DH for the following reaction?

2CH3COOH(l)   +   4O2(g)   ---------->   4CO2(g)   +   4H2O(g)      DHrxn = ?
a) -1567.2 kJ b) -391.8 kJ c) +391.8 kJ
d) +783.6 kJ e) +1567.2 kJ
15. Given the following thermochemical equation:

2C2H6(g)   +   7O2(g)   ---------->   4CO2(g)   +   6H2O(l)      DHrxn = -3119.4 kJ

decide on the value of DH for the following reaction:

4CO2(g)   +   6H2O(l)   ---------->   2C2H6(g)   +   7O2(g)      DHrxn = ?
a) -6238.8 kJ b) +1559.7 kJ c) -1559.7 kJ
d) +3119.4 kJ e) -3119.4 kJ
16. Given the following thermochemical equation:

2CaCO3(s)      ---------->   2CaO(s)   +   2CO2(g)      DHrxn = +356.6 kJ

Decide on the value of DH for the reaction

CaCO3(s)   ---------->   CaO(s)   +   CO2(g)      DHrxn = ?
a) +178.3 kJ b) -178.3 kJ c) +356.6 kJ d) -356.6 kJ e) -89.2 kJ
17. What is DH for the following reaction:

N2(g)   +   2H2(g)   ----->   N2H4(l)      DHrxn = ?

Make use of the following known thermochemical equations:

N2H4(l)   +   O2(g)   ----->   N2(g)   +   2H2O(l)      DHrxn = -622.2 kJ

2H2(g)   +   O2(g)   ----->   2H2O(l)      DHrxn = -571.6 kJ
a) -1193.8 kJ b) -50.6 kJ c) +50.6 kJ d) +672.8 kJ e) +1193.8 kJ
18. Manganese metal (Mn) can be prepared by reacting aluminum metal (Al) with manganese dioxide (MnO2).

4Al(s)   +   3MnO2(s)   ---------->   2Al2O3(s)   +   3Mn(s)      DHrxn = ?

Calculate DH for the above reaction using the following known thermochemical equations:

4Al(s)   +   3O2(g)   ---------->   2Al2O3(s)      DHrxn = -3352 kJ

Mn(s)   +   O2(g)      ---------->      MnO2(s)      DHrxn = -521 kJ
a) -3873 kJ b) -2831 kJ c) -1789 kJ d) +2831 kJ e) +3873 kJ
19. Use standard enthalpies of formation (DHf values) to calculate DH for the following reaction:

2C2H6(g)   +   7O2(g)   ---------->   4CO2(g)   +   6H2O(g)      DHrxn = ?
a) -3194.2 kJ b) -2855.4 kJ c) -720.0 kJ d) -550.6 kJ e) -67.0 kJ
Note: The multiple choice answers to problem 19 were modified on Monday November 19, 2001 to correct an error in the original version. If you printed this test before that date, you need to update your multiple choice answers. Your printed version of this test is outdated if it does not include this correction notice.

Problems Involving Calorimetry


20. The chemical equation for the dissolving of potassium nitrate, KNO3 in water is

KNO3(s)   ---------->   K+(aq)   +   NO3-(aq)

To measure DH for this reaction, 25.184 g of KNO3 was dissolved in 238.439 g of water in a calorimeter. The temperature of both the water and the KNO3 was initially 25.00 oC, but after the KNO3 was added to the calorimeter and allowed to dissolve, the temperature was 16.96 oC. What is DH for the above chemical reaction describing the dissolving process? For simplicity, assume that both the H2O and the KNO3 have the same specific heat -- that of water, which is 4.184 J / ( g oC).

Hints: Since the H2O and KNO3 are assumed to have the same specific heat, you can add their masses and use the total mass in the equation q = s . m . Dt. Recall that this is what I did in class when solving the exercise in which H2SO4 was dissolving in water. You solve this problem using the same technique. Also be aware that what you are initially calculating is qcalorimeter but what you really want is qreaction. Note that qreaction = -qcalorimeter. You want qreaction because it's the same thing as DH. Finally, note that the qreaction you initially calculate is for the actual amount of KNO3 used (25.184 g), not one mole. You need to adjust the qreaction from what it is for 25.184 g of KNO3 to what it would be for one mole of KNO3 because the chemical equation shows one mole.
a) -35.60 kJ b) -8.87 kJ c) +8.87 kJ d) +26.73 kJ e) +35.60 kJ
***   END   OF   TEST   ***


ANSWERS:

1 b      2 a      3 b      4 a      5 b     

6 a      7 e      8 d      9 b      10 d     

11 a      12 c      13 e      14 a      15 d     

16 a      17 c      18 c      19 b      20 e

SELECTED ATOMIC WEIGHTS

N = 14.00674     O = 15.9994     K = 39.0983


SELECTED ENTHALPIES OF FORMATION
CH4(g):DHf = -74.9 kJ / mol
CO2(g):DHf = -393.5 kJ / mol
C2H6(g):DHf = -84.7 kJ / mol
HCl(g):DHf = -92.3 kJ / mol
Cl2(g):DHf = 0.0 kJ / mol
H2O(g):DHf = -241.8 kJ / mol
O2(g):DHf = 0.0 kJ / mol
This page was last modified Monday November 19, 2001