HW 3D
1. A certain gas effuses 4 times as fast as oxygen gas. What is the molar mass of the unknown gas?
Oxygen is diatomic (O2) and its molar mass is 32.0 g/mol.
“Certain Gas” = fast/light and Oxygen = slow/heavy
Set rate expression equal to molar mass expression
2. A compound effuses through a porous cylinder 3.20 times faster than helium. What is its molar mass?
Helium = fast/light and “Compound” = slow/heavy
Set rate expression equal to molar mass expression
**Note: To my knowledge, there is NO gas with a molar mass less than that of hydrogen (H2 = 2.0
g/mol), but this is the correct answer given the problem! We took some creative license and chose a
gas you would not expect!
3. If 0.00251 mol of NH3 effuse through a hole in 2.47 min, how much HCl would effuse in the same
time?
A rate can be expressed as distance over time but in the case of reactions, it can be measured as
amount over time.
NH3 = fast/light and HCl = slow/heavy
Set distance expression equal to the molar mass expression as we know the formulas (and therefore
the masses also)
4. A sample of N2 effuses through a hole in 38 seconds. What must be the molecular weight of gas that
effuses in 55 seconds under identical conditions?
Nitrogen = fast/light and “Gas” = slow/heavy
Set time expression equal to the molar mass expression
5. At a certain temperature, molecules of methane gas, CH4, have an average velocity of 0.098 meters
per second (m/s). What is the average velocity of carbon dioxide, CO2 molecules at this same
temperature?
CH4 = fast/light and CO2 = slow/heavy
Set rate expression equal to the molar mass expression
6. Carbon dioxide gas (1.00 mole) at 373 K occupies 536 mL at 50.0 atmosphere pressure. What is the
calculated value of the pressure using each equation listed below and then calculate the % deviation
of each value from that observed
(a) Ideal gas equation
(b) Van der Waals equation
[Data - Van der Waals constants for carbon dioxide: a = 3.61 L2 atm mol-2; b = 0.0428 L mol-1]
At high pressures, it is convenient to quote pressure in atmospheres rather than kPa. The
value of the gas constant R depends on the units used for P and V.
If P is expressed as atmospheres and V as litres,
R = 0.0821 L atm K-1 mol-1.
(a) Using the Ideal Gas Equation
V = 0.536 L
n = 1.00 mol
T = 373 K
PV = nRT
P = nRT/V = 1.00 x 0.0821 x 373/0.536
= 57.1 atm
Actual pressure = 50.0 atm
% deviation = (7.1/50) x 100 = 14.2 %
(b) Using Van der Waals equation.
(P + an2/V2)(V - nb) = nRT.
(P + 3.61 x (1.00/0.536)2)(0.536 - 1.00 x 0.0428) = 1.00 x 0.0821 x 373)
(P + 12.57)(0.493) = 30.62
P + 12.57 = 62.12
P = 49.6 atm
% deviation = (0.4/50) x 100 = 0.8%
7. Predict which of the substances, NH3, N2, CH2Cl2, Cl2, CCl4 has
(a) the smallest van der Waals "a" constant
(b) the largest "b" constant.
(a) N2
The value of the Van der Waals constant, a, of a given gaseous substance depends on the
strength of attractions between its component molecules. Molecules experiencing the
weakest attractive forces will have the smallest a constant while those with the strongest
attractive forces will have the largest a values.
Of the molecules NH3, N2, CH2Cl2, Cl2, CCl4, the two elements nitrogen and chlorine alone
have non-polar bonds between their component atoms. This results is weaker attractions
between molecules, so N2 and Cl2 will have the smallest a values. The remaining molecules
all have polar bonds between their component atoms and in the case of NH3, hydrogen
bonds between molecules. These are factors that lead to increased a constants. As the
N2 molecule is smaller than the Cl2 molecule and therefore has less electrons present, the
attractive forces between N2 molecules will be weaker than those between Cl2 molecules and
N2 will have the smallest a constant.
(b) CCl4
The value of the Van der Waals b constant is merely the actual volume of a mole of the
molecules and this can be deduced by comparing the volumes of the molecules in the list.
Small molecular volume results in small b values and a large molecular volume corresponds
to a large b constant. From the compound in the list, CCl4 is the largest and so it will have the
greatest b constant.
8.
[Data - Van der Waals constants for helium: a = 0.0341 L2 at mol-2; b = 0.0237 L mol-1]
a. Using Van der Waals equation, calculate the temperature of 20.0 mole of helium in a 10.0 liter
cylinder at 120 atmosphere pressure.
b. Compare this value with the temperature calculated from the ideal gas equation.
(i) Using the Van der Waals equation:
Substituting in
(P + an2/V2)(V - nb) = nRT,
P = 120 atm
n = 20.0 mol
V = 10.0 L
(120 + 0.0341 x (20.0/10.0)2)(10.0 - 20.0 x 0.0237) = 20.0 x 0.0821 x T
[Note the value of R = 0.0821 because P is in atm and V in L]
(120 + 0.1364)(10.0 - 0.5) = 1.64 x T
T = 696 K
[Note that the correction to P is not significant as T is well above the temperature at which
helium gas will liquefy while the volume correction is significant due to the high pressure.]
(ii) Using the Ideal Gas Equation:
Substituting in
PV = nRT,
120 x 10.0 = 20.0 x 0.0821 x T
T = 1200/1.642
= 731 K
9. A particular reaction in the gas phase has an activation energy of 8.000 kJ mol-1. For 1.00 mole of gas,
calculate the number of molecules which exceed this activation energy at
(a) 300 K
(b) 400 K
For any collection of N molecules, the number NE having an energy E or greater is given by
the expression
NE = Ne-E/kT
where k is the Boltzmann constant = 1.381 x 10-23 J K-1
and T is the temperature in Kelvin.
Alternatively, by multiplying both sides of the expression by the Avogadro number, NA, on a
molar basis this then becomes nE = ne-E/RT
where R = 8.314 J K-1 mol-1
Note that in order for the units of E to be consistent with those of R (J K-1 mol-1), E must be
expressed in J rather than kJ.
(a) At 300 K:
n = 1.00 mol
E = 8.000 kJ mol-1 = 8000 J mol-1
Therefore n = 1.00 x e-8000/ (8.314 x 300)
= 4.05 x 10-2 mol
The number of molecules, N = 4.05 x 10-2 x NA
= 4.05 x 10-2 x 6.02 x 1023
= 2.4 x 1022 molecules.
(b) At 400 K:
n = 1.00 mol
E = 8.000 kJ mol-1 = 8000 J mol-1
Therefore n = 1.00 x e-8000/ (8.314 x 400)
= 9.02 x 10-2 mol
The number of molecules, N = 9.02 x 10-2 x NA
= 9.02 x 10-2 x 6.02 x 1023
= 5.4 x 1022 molecules.
10. At 250 K, the activation energy for a gas phase reaction was determined to be 6.500 kJ mol-1. What
percentage of gaseous molecules would be expected to have less than this energy at 250 K?
The expression in terms of moles for the distribution of molecular energies,
nE = ne-E/RT,
can be rewritten to give the fraction of the total moles, (n), that have energy E or greater, (nE),
as nE/n = e-E/RT
At T = 250 K and for E = 6.500 kJ mol-1 = 6500 J mol-1,
this becomes nE/n = e-6500/(8.314 x 250)
[Note that E must be expressed as J to match the units of R] = 0.044 or 4.4%.
Therefore the % of molecules that have less than 6.500 kJ mol-1 energy = 100.0 - 4.4
=95.6%.
11. For 1.00 mole of a particular gas, the average molecular energy is found to be 1.300 kJ mol-1 at 298
K. Approximately how many molecules have at least five times the average molecular kinetic energy.
The expression in terms of moles for the distribution of molecular energies is
nE = ne-E/RT
The average value of energy = 1.300 kJ mol-1
Therefore five times this energy = 5 x 1.300 = 6.500 kJ mol-1
E = 6.500 kJ mol-1 = 6500 J mol-1
T = 298 K
Then nE = 1.00 x e-6500/(8.314 x 298)
= 0.0725 mol
Number of molecules with this energy or greater
= 0.0725 x NA
= 0.0725 x 6.022 x 1023
= 4.4 x 1022