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CHM 136 General Chemistry II
Homework 1 – Spring 2009
Name
SOLUTIONS
1. A 295-g aluminum engine part at an initial temperature of 3.00 °C absorbs 85.0 kJ of
heat. What is the final temperature of the part (c of Al = 0.900 J/g·K)?
q = m × c × ΔT
85000 J = (295 g)(0.9000
J
)(Tf − 3.00 o C)
g⋅ o C
320.2 o C = Tf − 3.00 o C
Tf = 323.2 o C = 323 o C
2. What amount of heat is absorbed when 2.50-L of ethylene glycol (d = 1.132 g/mL
and c = 2.42 J/g·K) in an engine, has its temperature raised from 22.0 °C to 174.3 °C?
q = m × c × ΔT
= (2.50 L ×
J
1000 mL 1.132 g
×
)(2.42 o )(174.3 − 22.0 o C)
1L
1 mL
g⋅ C
= 1.043 × 10 6 J
3. Calculate the ΔHorxn at 298 K for the reaction between butane and oxygen (the
equation is UNBALANCED). Then, using your value for ΔHorxn, determine the heat
change when 25.0 g of butane are burned in excess oxygen.
C4H10(g) + O2(g)
CO2(g) + H2O(g)
These values may help:
Substance ΔHof (kJ/mol)
C(g)
715.0
CO2(g)
-393.5
C4H10(g)
-126
H2(g)
0
H2O(g)
-241.8
O2(g)
0
The balanced equation is:
2 C4H10(g) + 13 O2(g)
8 CO2(g) + 10 H2O(g)
o
ΔH rxn
= ∑ ΔH fo(products) − ∑ ΔH fo(reactants)
= [8(−393.5) + 10( −241.8)] − [2(−126) + 13(0)]
= −5314 kJ/mol
·∆
25.0
1143
1
58.12
5314
2
4. A sample of nitrogen gas is confined to a 5.00-L container at 228 torr and 27.0 °C.
The chamber is then compressed to a new volume of 3.86 L at a constant temperature.
What is the new pressure of the nitrogen?
P1 V1 = P2 V2
P2 =
=
P1 V1
V2
(228 torr )(5.00 L )
3.86L
= 295 torr
5. Calculate the molar mass of a gas at 976 torr and 192 °C if 376 mg of the gas
occupies 46.3 mL.
PV = nRT =
m
M
RT
mRT
PV
atm⋅L
(0.376 g )(0.0821 mol
)(
)
⋅K 465 K
=
1 atm
(976 torr × 760 torr )(0.0463 L )
M =
g
= 241 mol
6. A 1.00-L mixture of helium, neon, and argon has total pressure of 662 mmHg at
298 K. If the partial pressures of helium and neon are 341 mmHg and 112 mmHg,
respectively, what mass of argon is present in the mixture?
PAr = Ptotal – PHe – PNe = 662 – 341 – 113 = 209 mmHg
PV = nRT =
m=
=
m
M
RT
PVM
PRT
atm
(1.00 L ) 39.95 molg
209 mmHg × 7601 mmHg
(
)
(
atm⋅L
(0.0821 mol
)(
)
⋅K 298 K
= 0.449 g Ar
)