Answer: D
Examine the first reaction and realize that SO 2 (g) needs to be on the reactant side. Reverse the
equation and change the sign of ∆H°. When you examine the second reaction, you notice that
SO 3 (g) is on the correct side, so there is no need to reverse this equation. At this point, your two
reactions can be added together.
/
/
gs g
sgg
ggg
32
12
SO S O
SOSO
SO O SO
22
23
223
"
"
"
+
+
+
_ ^ _
^ __
___
i h i
h ii
iii ∆H =-^100 kJ/mole
%
295
395
∆
∆
kJ/mole
kJ/mole
H
H
=
=-
%
%
But before concluding that this is your answer, note that the question asks for ∆H°in terms of 2
moles of SO 2 (g). Doubling the ∆H°gives the answer, −200. kJ/mole.
- In expanding from 3.00 to 6.00 liters at a constant pressure of 2.00 atmospheres, a gas
absorbs 100.0 calories (24.14 calories = 1 liter ⋅atm). The change in energy, ∆E, for the
gas is
A. −600. calories
B. −100. calories
C. −44.8 calories
D. 44.8 calories
E. 100. calories
Answer: C
The first law of thermodynamics states that ∆E = q + w. Since w = –Pext⋅∆V, the equation can
be stated as
∆E =∆H −Pext⋅∆V
∆E = 100.0 calories −`j2 00../atmospheres:: 3 liters24 14cal L atm = −44.8 calories
10. A gas which initially occupies a volume of 6.00 liters at 4.00 atm is allowed to expand to
a volume of 14.00 liters at a pressure of 1.00 atm. Calculate the value of work, w, done
by the gas on the surroundings
A. −8.00 L ⋅atm
B. −7.00 L ⋅atm
C. 6.00 L ⋅atm
D. 7.00 L ⋅atm
E. 8.00 L ⋅atm
Thermochemistry