Energy And Enthalpy Worksheet With Answers Page 3
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1
2
3
4
5(a) H
C=CH
(g) + Br
(g) ➝ CH
BrCH
Br(l)
2
2
2
2
2
(g) ➝ 4 CO
(b) 2 C
H
(g) + 7 O
(g) + 6 H
O(g)
2
6
2
2
2
Look at each reaction and try to decide whether molecular randomness increases or
decreases. Reactions that increase the number of gaseous molecules generally have a
positive ΔS, while reactions that decrease the number of gaseous molecules have a
negative ΔS. For (a) the amount of molecular randomness in the system decreases
when 2 mol of gaseous reactants combine to give 1 mol of liquid product, so the
reaction has a negative ΔS°. In (b) the amount of molecular randomness in the
system increases when 9 mol of gaseous reactants give 10 mol of gaseous products,
so the reaction has a positive ΔS°.
9. What is the equation for Gibbs free-energy change (ΔG) and what does each quantity
define? Remember that the value of the free-energy change ΔG is a general criterion for
the spontaneity of a chemical or physical process.
Gibbs free-energy change (ΔG), ΔG = ΔH - T ΔS, determines in a chemical reaction
or other process is spontaneous. ΔG stands for free-energy change, while ΔH,
enthalpy, is the heat of the reaction. T stands temperature and is measured in
Kelvins. Lastly ΔS is the entropy change.
10. The two gases BF
(g) and BCl
(g) are mixed in equal molar amounts. All B⎯F
3
3
bonds have about the same bond enthalpy, as do all B⎯Cl bonds. Explain why the
mixture tends to react to form BF
Cl(g) and BCl
F(g).
2
2
A system containing the “chemically mixed” B halides has greater entropy than a
system of BCl
and BF
. It has the same number of gas phase molecules, but more
3
3
distinguishable kinds of molecules, hence more microstates and higher entropy.
11. Tetraphenylgermane, (C
H
)
Ge, has a melting point of 232.5°C, and its enthalpy
6
5
4
-1
increases by 106.7 J g
during fusion. Calculate the molar enthalpy of fusion and molar
entropy of fusion of tetraphenylgermane.
m
= 232.5°C = 505.5 K
p
ΔH = 106.7 J / g
MW(C
H
)
Ge = 380.59 g / mol
6
5
4
4
molar ΔH = 106.7 J / g × 380.59 g / mol = 4.061 × 10
J / mol
4
-1
-1
molar ΔS = ΔH / T = 4.06 × 10
J / mol ÷ 505.5 K = 80.34 J K
mol
12. Suppose 1.00 mol of water at 25°C is flash-evaporated by allowing it to fall into an
-1
-1
iron crucible maintained at 150°C. Calculate ΔS
, if c
(H
O(l)) = 75.4 J K
mol
and
tot
p
2
-1
-1
-1
c
(H
O(g)) = 36.0 J K
mol
. Take ΔH
= 40.68 kJ mol
for water at its boiling point
p
2
vap
of 100°C.
O(l) ➝ H
O(l) ➝ H
O(g) ➝ H
H
O(g)
2
2
2
2
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