Lecture April 17
Sec 11.7, Chap. 5 Sec. 5.4-5.6
heating or cooling curve Fig. 11.36
Gases Chap. 5 Sec. 5.4 - 5.6
No in-class quiz that counts until
Thursday, April 24
Mastering Chem due April 24
CHEM131 - Spring 14 - April 17
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Summary of ΔH for
phase changes
Heats of vaporization ΔHvap kJ/mol
liquid→gas
Table 11.7 (note the boiling points so
not ΔHo which assumes 25oC)
Heats of fusion ΔHfus kJ/mol
solid→liquid
Table 11.9 (at the freezing point)
Sublimation solid→gas (CO2)
CHEM131 - Spring 14 - April 17
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Example of Hess’s Law
#112 from Chap 6
Use following ΔH info to calculate the ΔHsub for
I2(s) → I2(g)
1) ΔHof TiI3(s) = -328 kJ/mol
2) 2 Ti(s) + 3I2(g) → 2TiI3(s) ΔHo = -839 kJ/mol
Write the balanced equations and combine
Ans: 61 kJ/mol
From Appendix II B get 62.42 kJ/mol
CHEM131 - Spring 14 - April 17
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Example of Heating Curve
What energy is required to heat 50.0 g
of water from 25oC to 100oC (gas)?
1) q = C m ΔT
C = 4.18 (J/goC)
Table 6.4
= (4.18) (50.0) (75) = 15.7 kJ
2) ΔHvap = 40.7 kJ/mol at 100oC
Table 11.7
Now add an extra (50.0/18.0) x 40.7 kJ/mol
= 113 kJ to vaporize the water
total energy required = 15.7 + 113 =129 kJ
CHEM131 - Spring 14 - April 17
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Heating Curve Example Fig. 11.361
Calculate for 1.00 mol water (or 18.0 g)
Heat required to go from -25.oC to 125.oC
1) -25.0oC to 0.0oC:
q = CmΔT = (2.09J/(g-oC))(18.0.g)(25.0oC) = 0.941 kJ
2) Melt the ice: (6.02 kJ/mol)(1.00 mol) = 6.02 kJ
3) Heat from 0.0oC to 100.0oC:
q = CmΔT = (4.18J/(g-oC))(18.0g)(100.oC) = 7.52 kJ
4) Vaporize the water: (40.7kJ/mol)(1.00mol) = 40.7 kJ
5) Heat the vapor from 100. to 125.oC:
q = CmΔT = (2.01J/(g-oC))(18.0g)(25.oC) = 0.904 kJ
TOTAL heat required = 56.1 kJ
CHEM131 - Spring 14 - April 17
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Heating Curve
See Sec. 11.7
T oC
Note these are at
constant T
5
100
4
3
0
-10
1
2
Heat (kJ)
CHEM131 - Spring 14 - April 17
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Another example
Supposed you have a liter of water and want to cool it from
32.0 oC down to 20.0 oC. Assuming a constant volume and
no heat transfer with the surroundings, estimate the
volume of water you need to evaporate?
Need q = cmΔT = (4.184 J/g-K) x (1.00 x 103g)
x (-12 K)
q = -50.21 kJ
Evaporation requires heat
ΔHovap (heat of vaporization) = 44.0 kJ/mol
q= ΔHovap x mass
mol water = 50.21 kJ / (44.0 kJ/mol) = 1.14 mol
or mass ~ 20 g or 20 mL of water
CHEM131 - Spring 14 - April 17
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Another Example
How much heat needs to be removed to
take 35.0 g of water from 25.0oC to ice at
-5.0oC? Express your answer in kJ.
Heat capacities: J/(g-oC)
Heat of fusion of water = 6.02 kJ/mol
liquid water 4.18
ice 2.09
moles of water = 35.0/18.0 = 1.94 moles
Ans:
1) Cool from 25oC to 0oC: q = (4.18)(35.0)(25.0) = 3.658 kJ
2) Freeze the water: q = (6.02)(1.94) = 11.68 kJ
3) Cool ice from 0oC to -5oC: q =(2.09)(35.0)(5.0)=0.366 kJ
TOTAL = 15.7 kJ
CHEM131 - Spring 14 - April 17
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Ideal Gas Law Sec. 5.4
PV=nRT
R gas constant = 0.08206 L-atm/(mol-K)
P = pressure (atm)
V = volume (L)
n = moles (mol)
T = temperature (K)
Other units of Pressure:
1 atm = 760 mm Hg
1 mm Hg = 1 torr
1 atm = 760 torr
1 atm = 101.325 kPa
CHEM131 - Spring 14 - April 17
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Manipulating the Ideal Gas
Equation
density proportional to n/V
from PV = nRT ☞ n/V = P/RT
density = mass/volume ☞ mass = n x molar mass
Density of N2 at STP
(standard temperature of 0oC and 1 atm)
Density = (n/V)(molar mass) = P/RT (molar mass)
molar mass of N2 28.014 g/mol
density = {1 atm/[0.08206(L-atm/mol-K)x273.15K]}
x 28.014 g/mol
Ans. = 1.250 g/L at STP Note: 22.4 L/mol at STP
CHEM131 - Spring 14 - April 17
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Another Simple Example
How many moles of N2 gas are in a 10.0 L
container that is at a pressure of 10.0 atm at
25oC?
PV = nRT
☞
n = PV/RT
T = 298 K
n=
(10.0 atm)(10.0 L)/(0.08206 L-atm/mol-K)(298 K)
n = 4.09 moles of N2 gas
Note: n independent of the identity of the gas!
CHEM131 - Spring 14 - April 17
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6
Composition of Air
Composition of Earth’s Atmosphere
Compound % by Volume Mole Fraction*
Nitrogen
78.08
0.7808
Oxygen
20.95
0.2095
Argon
0.934
0.00934
CO2
0.033
0.00033
CH4
2 x 10-4
2 x 10-6
Hydrogen
5 x 10-5
5 x 10-7
*mole fraction A = #moles A/total # moles
Note: n (#moles) is proportional to V (volume)
CHEM131 - Spring 14 - April 17
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3
Partial Pressure
Dalton’s Law of Partial Pressure
Total pressure = sum of the individual pressures
Ptotal = Patm = PN2 + PO2 + P Ar + PCO2 ....
Partial pressure of (a) = Pa/Ptotal
Since PV = nRT ☞ P proportional to n (# of moles)
Mole fraction χa = na / ntotal
(at constant T and V)
= partial pressure of (a)
For air had mole fraction of O2 = 0.2095 ☞
partial pressure of O2 in air = .2095 atm
CHEM131 - Spring 14 - April 17
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Example Using Partial Pressures
Carbon monoxide at a pressure of 680 mm Hg reacts
completely with O2 at a pressure of 340 mm Hg in a sealed
vessel to produce CO2. What is the final pressure in the flask?
2CO(g) + O2(g) → 2CO2(g)
2 moles
1 mole
2 moles
Answer: 680 mm Hg
Now what if had 500 mm Hg (2 sig fig) of O2
instead?
What is the limiting reagent?
CO
Final pressure = 680 + 160 = 840 mm Hg
CO2 made + O2 left
CHEM131 - Spring 14 - April 17
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