Cliffs AP Chemistry, 3rd Edition

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.

9. 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