Chem 1000 Midterm Five
November 23rdth, 2004
You have 30 minutes.
R = 0.08205 L•atm/mol•K
Prof. Peter W. Dibble
Name:
m
n
PM
mRT
3RT
M =
d =
M =
Pn = X n PT urms 
V
V
RT
PV
M
T in K = C + 273.15 R = 0.08206 L • atm/mol \ • K = 62.4 L • torr/mol • K
PV  nRT
d=
1.00 atm = 760. torr
1 J = 1 kg•m2 sec-2
At STP, 1 mole of an ideal gas has V = 22.4 L.
No = 6.022 x 1023 mol-1
R = 8.314 J/mol•K
Question One (three marks)
In a sealed 1.00 L flask, 8.2 g of NaHCO3 is mixed with excess HCl. The following reaction
occurs:
HCO3 (aq) + H3O+ (aq) —> CO2 (g) + 2 H2O (l)
If the temperature is 23°C, what is the pressure in the flask when the reaction is complete?
Rate1
Rate2

M2
M1
We first need to calculate the no. moles of CO2 gas produced – the only gas present according to the reaction.
The stoichiometry of this reaction is such that this will equal the no. moles of NaHCO3 we started with.
No. mol CO2 = m/M = 8.2 g/(23.0 + 1.008 + 12.01 + 48.0) g/mol = 9.76 x 10-2 mol
Now we can solve for the pressure.
P = nRT/V = 9.76 x 10-2 mol x 0.08206 L•atm/mol•K x 300. K/1.00 L = 2.4 atm
Question Two (two marks)
Natural gas is a mixture of several hydrocarbons: CH4, C2H6, C3H8 and C4H10 in the ratio 188 : 8 : 3 : 1
respectively. In a natural gas line where the total pressure is 9.8 atm, what is the partial pressure of methane,
CH4?
Pn = XnPt
Xn = 188/(188+8+3+1) = 0.94
Pn = 0.94 x 9.8 atm = 9.2 atm
Question Three (five marks)
How does the kinetic molecular theory rationalize:
i)
Boyle’s Law (PV)
When volume is decreased, there will be more gas particle collisions and more collisions with the wall of
the vessel thereby increasing the pressure.
ii)
Charles’ Law (VT)
At higher T, gas particles are moving faster and with greater KE. There will more collisions and more
energetic collisions with the vessel walls, hence greater pressure.

Question Four (three marks)
Rate1

Rate2
1.59 
M2
M1
MX
2
MN
2
X 2  (1.59) 2 • 28g /mol  71g /mol
A diatomic halogen gas effuses at a rate 1.59 times more slowly than
N2 gas. What colour is the gas? Show your work.
The diatomic halogen with this mass is chlorine, which is yellow-green.
In the other exam it works out to be bromine which is orange.
Question Five (three marks)
On the grid below, sketch a Boltzmann distribution for Ne gas at temperatures of 500K and 1000K. Don’t
forget to label your diagram.
5 0 0 K or Xe
N
o
.
o
f
m
o
le
c
u
le
s
1 0 0 0 K o r Ne
Sp e e d , u (m / s )
Question Six (five marks)
Of the following gases, H2, N2, Br2 and Xe:
i) which has the highest urms at 300K?
He or H2
ii) which has the highest urms at 600K?
He or H2
iii) which has the highest density?
Br2
iv) Which two can be separated by effusion most efficiently?
v) Which will have the greatest
average kinetic energy at a given T?
He/H2 and Br2
All will be the same
Question Seven (two marks)
At what temperature is the urms of Xenon gas 45 m/s?
Recall that the molar mass must be converted to kg so that the units cancel properly.
3RT
M
urms 
urms • M  3RT
2
2
3
u • M (45m / s) 131.29g/ mol •10 kg /g
T  rms

 11K
3R
3 • 8.31J /molK
2