Part 1: Thermochemistry Case Study: Why do citrus growers spray their trees with water to protect them
during a freeze? 1 (An application of .dHtus)
During freeze warnings in certain agricultural areas, we sometimes hear that farmers spray their crops with
water to provide protection from the cold temperatures. But if the water freezes on the crops, won't that do
more harm than good?
Citrus crops become threatened when temperatures fall below 28°F for four hours or more. Using heaters in an
orchard to increase the surrounding air temperature is the most effective way to avoid crop damage. However,
if trees are sprayed with water under conditions where freezing occurs, this produces ice on the leaves and
releases enough heat in the process to protect the crop by keeping the fruit temperature at 32°F. The physical
(phase) change taking place is:
H20 (I)~ H20 (s), LlHtreezing (= -LlHtus) = -6.01 kJ/mol
Imagine that a tree is sprayed with 4 liters of water (and the density of water is 1 g/mL). How much heat would
be released upon freezing this water? Where does the released heat go and how does it help to protect the
tree?
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Adapted from: K.K. Karukstis and G.R. Van Heeke, "Chemistry Connections, 2"d Ed.," Academic Press, Amsterdam, 2003.
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Part II: Thermochemistry Case Study: How Do Instant Ice Packs Work?" 1
Many first-aid kits contain "instant ice packs," which can be activated by bending or squeezing the pack. The
pack becomes immediately cold to the touch, and can be applied to injuries. The chemical ice pack consists of an
inner compartment containing water and an outer compartment containing an ionic salt such as ammonium
nitrate. By bending or squeezing the chemical ice pack, you break the inner compartment, thus allowing the
water to come into contact with the salt, which subsequently dissolves. The process of dissolving the salt
requires heat (is this an endothermic or exothermic process?), which is withdrawn from the surroundings (the
water in the pack). Thus, the temperature of the ice pack is lowered, and it feels cold to the touch.
If an instant ice pack contains 75.0 ml of water and 25.0 g of ammonium nitrate, which are mixed when the pack
is activated, and if the thermochemical equation for the dissolution reaction can be written as:
then how much heat is (evolved? consumed?) by the specific reaction taking place in this particular ice pack?
Assuming the density of the solution in the activated cold pack is 1.0 g/ml, and assuming the specific heat
capacity of the solution is the same as that of water (4.184 J/g/ 0 C), then calculate the temperature change that
can be expected to be generated by this ice pack?
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Part Ill: Review questions, Ch. 5 "Thermodynamics"
1) For most chemical reactions
A) L'!H is equal to L'!U.
B) the difference between L'!H and !!U is very small.
C) L'!U is much larger than !!H.
D) L'!H is much larger than L'!U.
2) What is expected when the reaction shown below takes place in a thermally-insulated container outfitted with
a movable piston at a constant atmospheric pressure of 1 atm?
2 C2H6(g) + 7 02(g) .... 4 C02(g) + 6 H20(g)
A) Volume will decrease and work will be done by the system.
B) Volume will decrease and work will be done on the system.
C) Volume will decrease and work will be done on the system.
D) Volume will increase and work will be done by the system.
3) For the reaction shown below, what can be said about PL:::. V and L:::.U?
N2(g) + 3 H2(g) .... 2 NH3(g) L:::.Ho =- 92.2 kJ
A)PL:::.V>
B) P1:::. V >
C) P1:::. V <
D) PL:::. V <
OandL:::.U< -92.2kJ
0 and 1:::. U >- 92.2 kJ
0 and 1:::. U > - 92.2 kJ
0 and L:::.U < -92.2 kJ
4) How much heat is absorbed/released when 20.00 g of NH3(g) reacts in the presence of excess 02(g) to produce
NO(g) and H20(l) according to the following chemical equation?
4 NH3(g) + 5 02(g) .... 4 NO(g) + 6 H20(l)
L'!Ho = +1168 kJ
A) 342.9 kJ of heat are absorbed.
B) 1372 kJ of heat are absorbed.
C) 1372 kJ of heat are released.
D) 342.9 kJ of heat are released.
5) Sodium metal reacts with water to produce hydrogen gas and sodium hydroxide according to the chemical
equation shown below. When 0.0300 mol of Na is added to 100.00 g of water, the temperature of the resulting
solution rises from 25.00°C to 37.90°C. If the specific heat of the solution is 4.18 J/(g · oq, calculate L'!H for the
reaction, as written.
2 Na(s) + 2 H20(l) .... 2 NaOH(aq) + H2(g)
!!H=?
A) -90.0 kJ
B) -5.41 kJ
C) -362 kJ
D) -180 kJ
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6) Calculate the total quantity of heat required to convert 25.0 g of liquid CC4(1) from 25.0°C to gaseous CCI4 at
76.8°C (the normal boiling point for CCl4)? The specific heat of CCI4(l) is 0.857 JfgJoC, its heat of fusion is 3.27
kJ/mol, and its heat of vaporization is 29.82 kJ/mol.
A) 1.64 kJ
B) 5.96 kJ
C) 6.49 kJ
D) 1.11 kJ
7) The specific heat of copper is 0.385 J/(g · 0 C). If 34.2 g of copper, initially at 25°C, absorbs 4.689 kJ, what will be
the final temperature of the copper?
A) 356°C
B) 381 oc
C) 27.8°C
D) 25.4°C
8) It takes 11.2 kJ of energy to raise the temperature of 145 g of benzene from 25.0°C to 70.0°C. What is the specific
heat of benzene?
A) 1.10 J/(g · 0 C)
B) 5.41 J/(g · 0 C)
C) 1.72 J/(g · 0 C)
D) 3.48 J/(g · 0 C)
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