39
10.78)
u =
3RT
=
⎛ 3RT ⎞
⎜
⎟
⎝
⎠
1/ 2
At constant T, u = (1/ )1/2
a) The larger the molar mass, the slower the avg. speed (at constant T).
SF6 < HBr < Cl2 < H2S < CO
(g/mol)
146.06
80.91
Dec
70.91 34.08
28.01
, Inc speed (u)
b) Calculate the rms speeds of CO and Cl2 at 300 K. Use eqn above and make sure
R, T and are in SI units (8.314 J/molCK or kgCm2/s2CmolCK, Kelvin, kg/mol,
respectively) to give SI units for u (m/s)
3 (8.314 kgCm2/s2CmolCK)(300 K) 1/2
uCO = (-------------------------------------------)
= 516.85 m/s = 516 m/s
28.01 x 10-3 kg/mol
3 (8.314 kgCm2/s2CmolCK)(300 K) 1/2
uCl = (-------------------------------------------)
= 324.84 m/s = 325 m/s
2
70.91 x 10-3 kg/mol
As expected, the lighter CO molecules move as the greater rms speed.
10.81)
40
10.81) (cont.)
10.82)
41
10.82) (cont.)
10.83)
a) Gases behave non-ideally at HIGH Pressure and LOW Temp
(Real gases behave ideally at low P and high T)
b) Real gases behave non-ideally because they have:
Finite Volumes: real gas particles occupy some of the volume of the
container so the volume of free space for the molecules to
move in is less than the container (causes Pmeasured > Pideal)
- Positive deviations from ideality
Intermolecular Attractive Forces:
Real gases have AF and attract each other - causes the
measured pressure to be less than ideal (Pm < Pi)
- Negative deviations from ideality
c)
For an ideal gas, PV/RT = n, the number of moles of gas particles, which
should be a constant for all pressure, volume and temperature conditions (for
1 mole of gas PV/RT = 1 or PV/nRT = 1 is equivalent). If this ratio changes
with increasing pressure the gas is not behaving ideally (i.e. if PV/nRT = 1
the gas is behaving ideally, if not then it is exhibiting non-ideal behavior). If
PV/nRT > 1 the gas is exhibiting a positive deviation from ideality (due to
the volume of the molecules). If PV/nRT < 1 the gas is exhibiting a negative
deviation from ideality (due to the IAF between molecules).
42
10.86) The van der Waals equation is given below:
or
n2a
( P + -----) ( V - nb ) = nRT
V2
P=
nRT
n2a
--------- - -----V - nb
V2
causes (+)
deviation
causes (!)
deviation
Pm > Pi
Pm < Pi
Pm = measured (real) P;
Pi = ideal P
due to vol
due to IAF
of molecules
a: corrects for attractive forces between particles - real gas particles have AF
- ideal gas particles have NO AF (a=0)
b: corrects for finite molecular volume of particles
- real gases occupy some vol of the container (we are interested in free vol)
- ideal gas part. Have NO vol (b=0)
Both tend to inc. with inc. in Mol. Wt. And structural complexity. The “a” constant
depends on AF between particles and these will be discussed further in Chapter 11.
Vcont - nb < Vcont so
nRT/(V-nb) > nRT/Videal
n2a/V2 accounts for fact that AF cause particles to hit walls w. less force than ideal gases,
ˆ smaller P then expected for an ideal gas (negative deviation from ideality)
43
10.88)
Under these conditions this gas exhibits a negative deviation from ideality
(Pmeasured will be closer to PVDW than Pideal & PVDW < Pideal)
The intermolecular attractive forces reduce the effective number of particles and
the real pressure. This is reasonable for 1 mole of gas at relatively low temp.
and pressure. The molecules hit the wall with less force than in ideal case.
c)
10.89)
10.95)
CCl4 has larger a and b values (20.4 and 0.1383, respectively) than Cl2 (6.49
and 0.0562). Thus, one would expect CCl4 to exhibit a larger deviation
from ideality.
44
10.97)
We can use Avogadro’s Law to show a relationship between the molar masses (molecular
weights) and macroscopic masses (measured in the lab) of two gases.
Volume is directly proportional to moles (at constant P & T)
V = kCn
or
V/n = k
V2
V1
----= ----n1
n2
or
V2
n2
----= ----V1
n1
(At constant P & T)
We are using the same flask so V2 = V1 so,
n2
----= 1
n1
so,
n 2 = n1
Also, from the definition of molar mass (grams/mol) we have,
m
= ----n
so,
m
n = ------
Substituting into n2 = n1 we get,
m2
m
-----= -----1
2
1
or
m2
2
------ = -----m1
1
The ratio of molar masses is the same as the ratio of the macroscopic masses.
a) The mass of Ar is 3.224 g and the mass of the unknown is 8.102 g.
2
m2
= ------ *
m1
1
8.102 g
= ----------- * 39.948 g/mol = 100.39 g/mol
3.224 g
b) You have to assume that P, V and T of the two gases are the same.
45
10.98) It is simplest to calculate the partial pressure of each gas as it expands into the
total volume, then sum the partial pressures. Use Boyle’s Law to calculate the
new pressures of each gas in the mixture after the expansion. The final vol is:
V2 = 1.0 L + 1.0 L + 0.5 L = 2.5 L
10.99) Use mole fraction to calculate moles of O2 in air (nO = PO nair).
2
2
46
10.102) Calculate the average molar mass of the mixture of O2 and Kr. Then relate the
average molar mass to the molar masses of O2 and Kr using mole fractions .
Remember mole fractions add to 1. Also, the mass of 1 mole of any substance
is its molar mass.
10.106)
To answer this question we need to look at the sizes of the molecules given and
based on this which might have the largest “a” and “b” values in the van der Waals
Eqn. (a and b values given in Table 10.3 shown).
CH4 < Ne < N2 < SF6
(g/mol)
a (L2-atm/mol2)
b (L/mol)
a)
16.04
2.25
0.0428
20.18 28.01
0.211
1.39
0.0171 0.0391
146.06
-------
Assumption 3: Attractive & repulsive forces between gas molecules are
negligible.
Which one will likely have the strongest attractive forces (AF)? The “a”
value is related to the AF. Generally, the larger the molecules the larger the
AF and the bigger the “a” value. As it turns out (ch. 11), all these molecules
are nonpolar and have only London Forces (LF), which depend on the size
(and shape) of the molecules. Since SF6 is the largest molecule one would
expect it to have the strongest AF and the largest “a” value and depart the
most from assumption 3 of KMT.
47
10.106)
b)
Based on the sizes of the atoms and molecules and “a” and “b” values (which
depend on size and shape), one would expect Ne to behave the most ideally.
c)
The rms speed is inversely proportional to the square root of the molar
mass,
u =
3RT
=
⎛ 3RT ⎞
⎜
⎟
⎝
⎠
1/ 2
At constant T, u = (1/ )1/2
Thus, CH4 has the highest rms speed, u, since it has the smallest molar mass.
d)
Since SF6 is the largest molecule (or atom) one would expect it to have the
largest molecular volume (largest “b” value).
e)
ALL molecules at the same T & P have the SAME average KE (KEavg) per
mole. As shown in class, the average KE per mole is 3/2 RT. This means all
gases at the same T will have the same KEavg (per mole).
f)
Rate of effusion is directly proportional to the rms speed and thus is
inversely proportional to the molar mass,
reff ∝ u =
3RT
=
⎛ 3RT ⎞
⎜
⎟
⎝
⎠
1/ 2
Since both CH4 and Ne are lighter than N2 they should effuse more rapidly
than N2.
48
10.108)
49
10.108) (cont.)
50
10.111)
(Enough moles of gas so they occupy a fairly significant volume
compared to volume of the container.)
51
10.112) Cyclopropane is composed of 85.7% C and 14.3% H, by mass.
a) What is the molecular formula of cyclopropane if 1.56 g occupies a volume of
1.00 L at 0.984 atm and 50°C?
Remember the molecular formula (MF) is an integer multiple of its empirical
formula (EF) and the molecular weight (MW) is the same multiple of the empirical
formula weight (EFW). Also, the molecular weight (amu) is the same value as the
molar mass (g/mol). So if you know the EF you know the EFW and if you can then
determine the MW you can find the MF.
MF = (EF)n
MW
n = --------EFW
MW = n (EFW)
MW /
m
mass
= ----- = ----------n
moles
Have mass ==> need moles
Given mass % comp ==> empirical form
Given P, V, T ==> moles
Give mass & moles ==> (molar mass) & MW
1) Get Empirical Formula & Empirical Formula Weight (EFW)
a) Convert mass % to moles (assume 100 g cyclopropane)
1 mol C
? mol C = 85.7 g C × -------------- = 7.1357mol C
12.01 g C
1 mol H
? mol H = 14.3 g H × -------------- = 14.186 mol H
1.008 g H
52
10.112) (cont.)
b) Get mole ratio - divide by smallest number of moles
7.1357mol C
C : ------------------ = 1.00 = 1 (mol C/mol C)
7.1357mol C
14.186 mol H
H : ------------------ = 1.988 = 2 (mol H/mol C)
7.1357mol H
The empirical formula of cyclopropane is CH2.
EFW for CH2 = 12.01 + 2 (1.008) = 14.026 amu
2) Get Molecular Weight (MW) and Molecular Formula
Have mass (1.56 g). Need moles. Use the IGL to get moles.
PV
(0.984 atm) (1.00 L)
n = ------ = ---------------------------------------------- = 3.710 x 10-2 mol
RT
(0.08206 L Catm/mol CK) (323.15 K)
m
1.56 g
= ----- = ---------------------= 42.04 g/mol = 42.0 g/mol
(amu)
n
3.710 x 10-2 mol
MW
42.0 amu
n = --------- = --------------- = 3.00 = 3
EFW
14.0 amu
MF = (EF)n = (CH2)3 = C3H6
53
10.112) (cont.)
b) Which gas, Ar or C3H6, deviates more from ideal-gas behavior.
AWAr = 39.95
amu
MWC2H6 = 42.0 amu
The following are the van der Waals constants for Ar and C3H6:
Ar:
a = 1.34 L2atm2/mol2
b = 0.0322 L/mol
CH6:
a = 2.25 L2atm2/mol2
b = 0.0428 L/mol
Would expect C3H6, a larger and more complex molecule than CH4, to have
larger a and b values than CH4.
54
10.116) Ammonia, NH3(g), and hydrogen chloride, HCl(g), react to form solid
ammonium chloride, NH4Cl(s), according to the following equation,
NH3 (g) + HCl (g)
v
NH4Cl (s)
Two 2.00-L flasks at 25°C are connected by a stopcock, as below:
The stopcock is opened and the gases react until one is completely consumed.
a) Which gas remains in the system (i.e. which is the limiting reactant)?
This is a stoichiometry problem so we need to convert to moles.
1 mol NH3
nNH = 5.00 g NH3 × ----------------- = 0.29359 mol NH3
3
17.03 g NH3
1 mol HCl
nHCl = 5.00 g HCl × ----------------- = 013713 mol HCl
36.46 g HCl
Since the reactants react in a 1:1 mole ratio, the HCl is the limiting reactant and all
013713 moles of it reacts with 013713 moles of NH3 leaving 0.15646 moles of NH3
as the only gas present.
b) What is the final pressure of the system after the reaction is complete (assuming
the volume of the NH4Cl(s) can be ignored)? Total volume is 4.00 L.
n = 0.15646 mol; V = 4.00 L; T = 25°C + 273.15 = 298.15 K
nRT
(0.15646 mol) (0.08206 L Catm/mol CK) (298.15 K)
P = ------- = ------------------------------------------------------------------V
4.00 L
P = 0.95701 atm = 0.957 atm
55
Have 10.7 x 109 ft3 of methane, CH4, at methanol, 25°C and atmospheric
pressure (1.00 atm, assuming at least 3 s.f.).
10.120)
a)
What volume of liquid methanol, CH3OH, is formed from the oxidation of the CH4,
given the density of 0.791 g/mL for CH3OH?
2 CH4 (g) + O2 (g) -----> 2 CH3OH (R)
1) Find moles of CH4 first using IGL (need to convert to L of CH4)
(12)3 in3
(2.54)3 cm3
1L
x
---------------x
-----------= 3.0299 x 1011 L CH4
? L CH4 = 10.7 x 10 ft x -----------1 ft3
1 in3
103 cm3
9
3
PV
(1.00 atm) (3.0299 x 1011 L)
n = ------ = ---------------------------------------------- = 1.2384 x 1010 mol CH4
RT
(0.08206 L Catm/mol CK) (298.15 K)
32.04 g CH4O
1 mL
1L
1 mol CH4O
? L CH3OH = 1.2384 x 1010 mol CH4 x ------------------ x ------------------ x ----------- x ---------1 mol CH4
1 mol CH4O
0.791 g
103 mL
= 5.0165 x 108 L CH3OH = 5.02 x 108 L CH3OH
b)
Write balanced eqn. for the oxidations of methane, CH4 (g), and liquid methanol,
CH3OH (R), to form CO2(g) and H2O(R). Calculate the total enthalpy change for the
complete combustion of the 10.7 x 109 ft3 of CH4 and equivalent amount of
CH3OH.
)Horxn = 3n )Hof (products) ! 3m )Hof (reactants)
1) enthalpy for the combustion of CH4(g)
)H f(kJ/mol)
o
CH4 (g) + 2 O2 (g) -----> CO2 (g) + 2 H2O (R)
-74.8
0
-393.5
-285.83
)Horxn = [{(-393.5) + 2 (-285.83)} ! {(-74.8) + 2 (0)}
)Horxn = !890.36 kJ = !890.4 kJ
56
10.120) (cont.)
What is the enthalpy for the reaction of 10.7 x 109 ft3 of CH4 (1.2384 x 1010 mol)?
!890.36 kJ
)Horxn = 1.2384 x 1010 mol CH4 x ---------------- = ! 1.1026 x 1013 kJ
1 mol CH4
2) enthalpy for the combustion of CH3OH(R)
)H f(kJ/mol)
o
CH3OH (R) + 3/2 O2 (g) -----> CO2 (g) + 2 H2O (R)
-238.6
0
-393.5
-285.83
)Horxn = [{(-393.5) + 2 (-285.83)} ! {(-238.6) + 2 (0)}
)Horxn = !726.56 kJ = !726.6 kJ
What is the enthalpy for the reaction of 1.2384 x 1010 mol of CH3OH?
!726.56 kJ
)Horxn = 1.2384 x 1010 mol CH3OH x ------------------- = ! 8.9977 x 1012 kJ
1 mol CH3OH
c)
Compare the enthalpy change upon combustion of a unit volume of liquid CH4
(density = 0.466 g/mL) and liquid CH3OH (density = 0.791 g/mL). Which
substance has the higher enthalpy of combustion per unit volume?
103 mL
0.466 g
1 mol
!890.36 kJ
? kJ = 1 L CH4 x ------------ x ----------- x ----------- x ---------------- = ! 2.5867 x 104 kJ
1L
1 mL
16.04 g
1 mol
! 2.5867 x 104 kJ/L CH4
103 mL
0.791 g
1 mol
!726.56 kJ
? kJ = 1 L CH3OH x ------------ x ----------- x ----------- x ---------------- = ! 1.7937 x 104 kJ
1L
1 mL
32.04 g
1 mol
! 1.7937 x 104 kJ/L CH3OH
Based on these numbers more energy is given off per liter for CH4. However, the
above analysis uses the enthalpy for gaseous CH4 while that for CH3OH is for it’s
liquid state. The enthalpy of combustion for CH3OH in it’s gaseous state is
!763.96 (larger than above). The enthalpy of combustion per unit volume for
CH3OH would then be !1.89 x 104 kJ/L (a little larger than above).