GENERAL CHEMISTRY I
CHEM 1311.001
EXAM 3
Friday June 28, 1996


Name____________________________

SSN_____________________________

1. Reactions which have negative values of DH (the same thing as q) cause their immediate surroundings to

a) become warmerb) become coolerc) neither
2. Reactions which have positive values of DH cause their immediate surroundings to
a) become warmerb) become coolerc) neither
3. Reactions which cause their immediate surroundings to become warmer are said to be
a) endothermicb) exothermic
4. Reactions which cause their immediate surroundings to become cooler are said to be
a) endothermicb) exothermic
5. How much heat is required to raise the temperature of a 6.193 g sample of copper metal from 21.4 oC to 48.9 oC? Note that the specific heat of copper is 0.384 J / g oC.
a) 50.9 Jb) 65.4 Jc) 116 Jd) 321 Je) 713 J
6. A 52.11 g sample of aluminum at 18.8 oC absorbs 3.850 kJ of heat. What is the temperature of the aluminum after absorbing this heat? Note that the specific heat of aluminum is 0.901 J / g oC.
a) 10.7 oC b) 29.5 oC c) 82.0 oC d) 100.8 oC e) 120.6 oC
7. A 91.60 g sample of magnesium at 95.8 oC loses 4.50 kJ of heat. What is the temperature of the magnesium after losing this heat? Note that the specific heat of magnesium is 1.02 J / g oC.
a) -48.2 oC b) 47.6 oC c) 60.3 oC d) 84.0 oC e) 144.0 oC
8. 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
9. 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
10. Given the thermochemical equation

N2(g)   +   O2(g)   -----> 2NO(g)      DH = 181 kJ

what is DH for the following reaction

(1/2)N2(g)   +   (1/2)O2(g)   ----->   NO(g)      DH = ?
a) -362 kJ b) -181 kJ c) -90.5 kJ d) 90.5 kJ e) 181 kJ
11. Given the thermochemical equation

H2O2(l)   ----->   H2O(l)   +   (1/2)O2(g)      DH = -98.0 kJ

what is DH for the following reaction

2H2O(l)   +   O2(g)   ----->   2H2O2(l)      DH = ?
a) -196 kJ b) -49.0 kJ c) 49.0 kJ d) 98.0 kJ e) 196 kJ
12. Given the thermochemical equation

N2H4(l)   +   O2(g)   ----->   N2(g)   +   2H2O(l)      DH = -622 kJ

what is DH for the following reaction:

2N2H4(l)   +   2O2(g)   ----->   2N2(g)   +   4H2O(l)      DH = ?
a) -1244 kJ b) -311kJ c) 311 kJ d) 622 kJ e) 1244 kJ
13. The balanced thermochemical equation (in lowest terms) for the combustion of hydrogen is

2H2(g)   +   O2(g)   ----->   2H2O(g)      DH = -484 kJ

What is DH for the combustion of 3.175 g of H2?
a) -762 kJ b) -381 kJ c) 381 kJ d) 484 kJ e) 762 kJ
14. When 21.45 g of potassium nitrate was dissolved in water in a calorimeter, the temperature fell from 25.00 oC to 14.14 oC. What is DH for the reaction

KNO3(s)   ----->   K+(aq)   +   NO3-(aq)      DH = ?

The heat capacity for the calorimeter and its contents was 682 J / oC. Hint: Use q = C . m . 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) -34.9 kJ b) -7.41 kJ c) 7.41 kJ d) 17.5 kJ e) 34.9 kJ
15. The following are the thermochemical equations for the combustion of liquid hydrazine (N2H4) and hydrogen gas.

N2H4(l)   +   O2(g)   ----->   N2(g)   +   2H2O(l)      DH = -622 kJ

2H2   +   O2(g)   ----->   2H2O(l)      DH = -572 kJ

Using the above thermochemical equations and applying Hess' Law, calculate the DH for the following reaction

N2(g)   +   2H2(g)   ----->   N2H4(l)      DH = ?
a) -1194 kJ b) -597 kJ c) -50 kJ d) 50 kJ e) 1194 kJ
16. Use standard enthalpies of formation listed in Table 6.2 to calculate DH for the following reaction:

2C6H6(l)   +   15O2(g)   ----->   12CO2(g)   +   6H2O(g)      DH = ?
a) -6534.8 kJb) -6270.8 kJc) -733.3 kJ
d) 6270.8 kJe) 6534.8 kJ
ANSWERS:

1 a      2 b      3 b      4 a      5 b     

6 d      7 b      8 a      9 c      10 d     

11 e      12 a      13 b      14 e      15 d     

16 b

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