(b) Restatement: Moles of O 2 (g) at equilibrium.

PV = nRT

.

..

/

n PVRT

liter atm mole K K

atm liters

0 0821 523

550 40

gas

::

==

``

``

jj

jj

= 0.51 mole gas

.

.

mole gas

mole total gas

mole oxygen gas

1 moleO

051

5

1

= (^0102)

(c) Restatement: Grams of solid Cu(NO 3 ) 2 in flask at equilibrium.
moles of Cu(NO 3 ) 2 that decomposed:
.
.
mole O
mole O
moles Cu NO
1 mole Cu NO
010
1
2
(^20202)
2
2
3

= 3

^
^
h
h
mass of Cu(NO 3 ) 2 that decomposed:

. ().
()

mole Cu NO

mole Cu NO

gCu NO

1 gCu NO

020

1

187 57

(^2382)
32
32
3

= 3

^
^
h
h
mass of Cu(NO 3 ) 2 that remains in flask:
250.0 g Cu(NO 3 ) 2 originally −38 g Cu(NO 3 ) 2 decomposed
= 212 g Cu(NO 3 ) 2 remain
(d) Restatement: Equilibrium expression for Kpand value.
Kp= (pressure NO 2 )^4 ×pressure O 2
Dalton’s law of partial pressures:
().
total moles gas.
moles NO g atm
5 atm NO
4
1
(^2550440)

= 2

5.50 atmtot−4.40 atmNO 2 = 1.10 atm O 2
Kp= (4.40 atm)^4 (1.10 atm) = 412 atm^5
(e) Given: 420.0 grams of Cu(NO 3 ) 2 placed in flask.
Restatement: What total pressure at equilibrium?
Because the temperature was kept constant, as was the size of the flask, and because
some of the original 250.0 grams of Cu(NO 3 ) 2 was left as solid in the flask at equilib-
rium, any extra Cu(NO 3 ) 2 introduced into the flask would remain as solid — there
would be no change in the pressure.
Part II: Specific Topics