Homework Review Key: Chapter 5-- Gases

Homework Review Key: Chapter 5-- Gases

Homework Review Key: Chapter 5-- Gases
o In text book: pp. 205-210 #87, 89, 90, 91*, 93, 103, 104, 127, 135, 137*
*For #91 the reaction produces nitrogen gas and water.
*For #137, do all parts except the very last statement in part (c)(iii)
o Multiple Choice Review
o Optional: In Study Guide: p. 96 # 4, 14, 15, 17, 19, 29, 30, 31 and p. 105 # 4, 13
87.
Discuss the following phenomena in terms of the gas laws:
(a) The pressure increase in an automobile tire on a hot day: Neither the amount of gas in the tire
nor its volume change appreciably. The pressure is proportional to the temperature. Therefore, as
the temperature rises, the pressure increases.
(b) The “popping” of a paper bag: As the paper bag is hit, its volume decreases so that its pressure
increases. The popping sound occurs when the bag is broken.
(c) The expansion of a weather balloon as it rises in the air: As the balloon rises, the pressure
outside decreases steadily, and the balloon expands.
(d) The loud noise heard when a lightbulb shatters: The pressure inside the bulb is greater than 1
atm.
89.
Nitroglycerine decomposes according to the following equation:
4C3H5(NO3)3(s) → 12 CO2(g) + 10 H2O(g) + 6 N2 (g) + O2 (g)
Find the total volume of gases when collected at 1.2 atm and 25°C from 2.6×102 g nitroglycerine.
What are the partial pressures of the gases under these conditions?
? mol products = 2.6 × 10 2 g nitroglycerin ×
1 mol nitroglycerin
29 mol gaseous product
×
227.09 g nitroglycerin
4 mol nitroglycerin
= 8.3 mol gaseous product
Vproducts
L⋅atm
(8.3 mol) (0.0821 mol⋅K
n products RT
)(298 K ) = 1.7 × 102 L
=
=
P
(1.2 atm)
Χ component =
moles component
12 mol CO2
, so Χ CO2 =
= 0.41
total moles all components
29 mol product
Similarly, Χ H2O = 0.34, Χ N2 = 0.21, and Χ O2 = 0.034
PCO2 = Χ CO2 PT
PCO2 = (0.41)(1.2 atm) = 0.49 atm
Similarly, PH2O = 0.41 atm, PN2 = 0.25 atm, and PO2 = 0.041 atm.
Note: if we realize that H2O is a liquid at 25°C, then the solution is actually:
19 moles gaseous products/4 mol nitroglycerine → 5.4 mol gaseous products
V products = 111 L
12 mol CO2
Χ CO2 =
= 0.632; Χ N 2 = 0.316; Χ O2 = 0.0526
19 mol product
PCO2 = (0.632)(1.2 atm) = 0.76 atm; PN2 = 0.38 atm; PO2 = 0.063 atm
2
90.
Chapter 5 Review
0.145 g of a compound with empirical formula CH occupies 97.2 mL at 200.°C and 0.74 atm.
Determine the molecular formula of the gas. Comparing the molar mass to the empirical mass
will allow us to determine the molecular formula.
M=
dRT
=
P
(
0.145 g
97.2 mL
)
L⋅atm
× 10001 LmL (0.0821 mol⋅K
)(200. + 273)K
(0.74 atm)
= 78.3 g/mol
The empirical mass of CH = 13.02 g/mol
Since
91.
78.3 g/mol
= 6.01 ≈ 6 , the molecular formula is (CH)6 or C6H6.
13.02 g/mol
When ammonium nitrite, NH4NO2, is heated, it decomposes to give nitrogen gas. This
property is used to inflate some tennis balls.
(a) Write the balanced equation for the reaction.
→ N2(g) + 2H2O(l)
NH4NO2(s) !!
(b) What mass of NH4NO2 is needed to inflate a tennis ball to a volume of 86.2 mL at 1.20 atm
and 22°C?
1L
T(K) = 22° + 273° = 295 K; V = 86.2 mL ×
= 0.0862 L
1000 mL
n N2 =
PN2V
(1.20 atm)(0.0862 L)
=
= 4.27 × 10−3 mol
L⋅atm
RT
0.0821
(295
K)
(
mol⋅K )
−3
? g NH4NO2 = 4.27 × 10 mol N 2 ×
1 mol NH 4 NO2 64.05 g NH 4 NO2
×
= 0.273 g
1 mol N 2
1 mol NH 4 NO2
93.
The boiling point of liquid nitrogen is –196°C. On the basis of this information alone, do you
think nitrogen is an ideal gas?
No, because an ideal gas cannot be liquefied, since the assumption is that there are no intermolecular
forces in an ideal gas.
103.
Nitric oxide (NO) reacts with molecular oxygen as follows:
2NO(g) + O2(g) → 2NO2(g)
Initially NO and O2 are separated as shown below. When the valve is opened, the reaction
quickly goes to completion. Determine what gases remain at the end and calculate their partial
pressures. Assume that the temperature remains constant at 25°C.
4.00 L NO at
0.500 atm
2.00 L O2 at
1.00 atm
Calculate the initial number of moles of NO and O2 using the ideal gas equation.
Chapter 5 Review
3
n NO =
PNOV
(0.500 atm)(4.00 L)
=
= 0.0817 mol NO
L⋅atm
RT
(0.0821 mol⋅K
)(298 K)
n O2 =
PO2V
(1.00 atm)(2.00 L)
=
= 0.0817 mol O2
L⋅atm
RT
(0.0821 mol⋅K
)(298 K)
2 NO(g) + O2(g) → 2 NO2(g)
Determine which reactant is the limiting reagent. The number of moles of NO and O2 calculated above
are equal; however, the balanced equation shows that twice as many moles of NO are needed compared
to O2. Thus, NO is the limiting reagent.
Determine the molar amounts of NO, O2, and NO2 after complete reaction.
mol NO = 0 mol (All NO is consumed during reaction)
mol NO2 = 0.0817 mol NO ×
2 mol NO2
= 0.0817 mol NO2
2 mol NO
mol O2 remaining = 0.0817 mol O2 initial − 0.0409 mol O2 consumed = 0.0408 mol O2
Calculate the partial pressures of O2 and NO2 using the ideal gas equation.
VTotal = 2.00 L + 4.00 L = 6.00 L;
T(K) = 25° + 273° = 298 K
L⋅atm
(0.0408 mol) (0.0821 mol⋅K
)(298 K) = 0.166 atm
n O2 RT
PO2 =
=
V
(6.00 L)
PNO2
L⋅atm
(0.0817 mol) (0.0821 mol⋅K
)(298 K) = 0.333 atm
n NO2 RT
=
=
V
(6.00 L)
104.
Consider the apparatus shown here. When a small amount of water is
introduced into the flack by squeezing the bulb of the medicine dropper,
water is squirted upward out of the long glass tubing. Explain this
observation (Hint: Hydrogen choride gas is soluble in water.)
When the water enters the flask from the dropper, some hydrogen chloride
dissolves, creating a partial vacuum. Pressure from the atmosphere forces more
water up the vertical tube.
127.
Under the same conditions of temperature and pressure, why does one liter
of moist air weigh less than one liter of dry air? In weather forecasts, an
oncoming low-pressure front usually means imminent rainfall. Explain.
At the same temperature and pressure, the same volume contains the same number of moles of gases.
Since water has a lower molar mass (18.02 g/mol) than air (about 29 g/mol), moisture-laden air weighs
less than dry air. Under conditions of constant temperature and volume, moist air exerts a lower
pressure than dry air. Hence, a low-pressure front indicates that the air is moist.
4
135.
Chapter 5 Review
Venus’s atmosphere is composed of 96.5% CO2, 3.5 percent N2, and 0.015% SO2 by volume. Its
standard atmospheric pressure is 9.0×106 Pa. Calculate the partial pressures of the gases in
pascals.
PCO2 = (0.965) × (9.0 × 106 Pa) = 8.7 × 106 Pa
PN2 = (0.035) × (9.0 × 106 Pa) = 3.2 × 105 Pa
PSO2 = (1.5 × 10 4) × (9.0 × 106 Pa) = 1.4 × 103 Pa
−
137.
Apply your knowledge of the kinetic theory of gases to the following situations:
(a) Two flasks of volumes V 1 and V 2 (V 2 > V 1) contain the same number of helium atoms at
the same temperature
(i) Since the two He samples are at the same temperature, their rms speeds and the average
kinetic energies are the same.
(ii) The He atoms in V1 (smaller volume) collide with the walls more frequently. Since the
average kinetic energies are the same, the force exerted in the collisions is the same in both
flasks.
(b) Equal numbers of He atoms are placed in two flasks of the same volume at temperatures T 1
and T 2 (T 2 > T 1).
(i) The rms speed is greater at the higher temperature, T2.
(ii) The He atoms at the higher temperature, T2, collide with the walls with greater frequency and
with greater force.
(c) Equal numbers of He and Ne atoms are placed in two flasks of the same volume and a
temperature of 74°C.
(i) The rms speed of He is equal to that of Ne? False. The rms speed is greater for the lighter
gas, He.
(ii) The average kinetic energies of the two gases are equal? True. The gases are at the same
temperature.