Energy And Enthalpy Worksheet With Answers Page 4
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1
2
3
4
525°C
100°C
100°C
150°C
-1
q
= ΔH = m CΔT q
= ΔH = 40.68 kJ mol
q
= ΔH = m CΔT
1
2
3
m = 1.00 mol H
O
2
T
= 25°C = 298 K
i
T
= 150°C = 423 K
f
-1
-1
c
(H
O(l)) = 75.4 J K
mol
p
2
-1
-1
c
(H
O(g)) = 36.0 J K
mol
p
2
-1
ΔH
= 40.68 kJ mol
for H
O at T = 100°C.
vap
2
-1
-1
-1
-1
ΔH = q = (1.00 mol)(75.4 J K
mol
)(100 – 25) + (40680 J K
mol
)(1.00 mol) +
-1
-1
(1.00 mol)(36.0 J K
mol
)(150-100) = 48135 J
ΔS
= n c
(H
O(l)) ln T
/ T
+ n ΔH
/T + n c
(H
O(g) ln T
/ T
= (1.00 mol)(75.4
H2O
p
2
2
1
vap
p
2
2
1
-1
-1
-1
J K
mol
)(ln 373 K/298 K) + (1.00 mol)(40680 J mol
/ 373 K) + (1.00 mol)(36.0 J
-1
-1
mol
)(ln 423 K / 373 K) = 130. 5 J K
-1
ΔS
= q/T = -48135 J ÷ 423 K = -113.8 J K
iron
-1
-1
-1
ΔS
= ΔS
+ ΔS
= 130.5 J K
– 113.8 J K
= 16.7 J K
total
H2O
iron
-1
-1
13. Iron has a heat capacity of 25.1 J K
mol
, approximately independent of
temperature between 0 and 100°C. Calculate the enthalpy and entropy change of 1.00
mol of iron as it is cooled at atmospheric pressure from 100°C to 0°C.
-1
-1
C = 25.1 J K
mol
m = 1.00 mol
T
= 100 °C = 373 K
i
T
= 0°C = 273 K
f
-1
-1
ΔH = m C ΔT = (1.00 mol)(25.1 J K
mol
)(-100 K) = -2510 J
-1
-1
-1
ΔS = n C ln T
/ T
= (1.00 mol)(25.1 J K
mol
)(ln 273 K / 373 K) = 7.83 J K
f
i
14. A water bath is filled with lots of ice and water to maintain a constant temperature of
0°C. You dump a 10 g block of 200°C aluminum into the bath. The heat capacity of
-1
-1
-1
aluminum is 24.35 J K
mol
and the enthalpy of fusion of water is 6 kJ mol
.
(a) How much heat is transferred from the block to the water bath?
n = 10 g × 1mol / 27 g = 0.37 mol Al
-1
-1
ΔH
= q = n C ΔT = (0.37 mol)(25.35 J K
mol
)(-200 K) = -1801.9 J
Al
(b) How many grams of ice melt?
q
= -q
= n ΔH
Al
H2O
fus
-1
1801.9 J = n(6000 J mol
)
n= 0.30 mol × 18 g / 1mol = 5.4 g H
O
2
(c) What is the entropy change of the aluminum block?
-1
-1
-1
ΔS
= n C ln T
/T
= (0.37 mol)(24.37 J K
mol
)(ln 273 K / 473 K) = -4.95 J K
Al
f
i
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