DO
Grader
Points Q1:
/16
NOT WRITE IN THIS SECTION
Q2:
/38
Q3:
Name
/22
Date graded
TOTAL:
Section number
Chemistry 130 (1,2) Dr. J. F. C. Turner
Quiz 1
31st January 2007
30 minutes
Please answer all parts of all questions and show all your work
You may use:
Non­programmable calculators
You may not use:
Any written materials (cards or notes etc) or any other electronic device
1(a)
Write down the enthalpy and volume of mixing for a ideal solution.
 H mix = 0
 V mix = 0
2 points
(b)
Define an ideal solution.
An ideal solution is one in which there is no net difference between the solute­solute and solvent­solvent forces in the pure components and the solute­solute, solvent­solvent and solvent­solute forces in the solution. Partial credit:  H mix = 0 and  V mix = 0 do not define an ideal solution but are consequences of the definition
4 points
(c)
There are four criteria that must be satisfied for a gas to be ideal. Write down two of them.
Criterion 1:
Criterion 2:
Any two from No particle volume, no intermolecular or interatomic forces, perfectly elastic collisions and straight­
line trajectories
4 points
(d)
The boiling points and melting points of the noble gases are given below, together with atomic number. He
Ne
Ar
Kr
Xe
Rn
m.p. / °C [­272.2]
­248.59
­189.3
157.36
­111.7
71 b.p. / °C
­268.93
­246.08
­185.8
­153.22
­108
­61.7
Z
2
10
18
36
54
86
What is the reason for the trend in melting point and boiling point?
The polarizability rises with atomic number as the outer electrons become more weakly bound and are therefore more easily perturbed. The increased polarizability increases the van der Waals (dispersion + induced dipole forces etc) and therefore increases the m.p. and b.p.
2 points
(e)
Do intermolecular forces exist in
(i)
An ideal or perfect gas?
(ii)
An ideal solution?
Yes
Yes
No
No
(underline your answer)
(underline your answer)
2 points
(f)
Write down the equation of state for an ideal gas that relates temperature, pressure and volume to the quantity of gas.
P V =n R T or rearrangements
2 points
16 points
2(a)
A mixture of three components, A, B and C, is prepared. Write down the equation for the pressure of the vapor in equilibrium with this mixture, using Raoult's Law.
P= x A P°A  x B P°B x C P°C
4 points
(b)
When may Raoult's law be applied to a real or non­ideal solution?
When a solution is very dilute.
2 points
Methanol, CH 3 OHl  , and water are fully miscible and solutions may be considered to be ideal. The vapor °
°
pressure of pure methanol and pure water are P  CH 3 OH  =14.4 kPa and P  H2 O =2.3 kPa respectively. The relative (c)
molecular masses of methanol and water are RMM  CH 3 OH =32 g mol and RMM  H2 O  =18 g mol respectively. A a mixture of 4 g of methanol and 36 g of water is prepared. Calculate the pressure of the vapor that is in equilibrium with this mixture.
−1
−1
1. Mole calculation
RMM  CH 3 OH =32 g mol so 4 g of MeOH≡
−1
RMM  H2 O =18 g mol−1 so 36 g of H2 ≡
4
moles=0.125 moles
32
36
moles=2 moles
18
2. Mole fraction calculation in the solution
n MeOH =0.125 moles and nH O =2 moles so ntotal =2.125 moles
2
x
L
H2 O
2
0.125
L
=
=0.94 and xMeOH =
=0.06
2.125
2.125
3. Vapor pressure calculation
Ptotal = x LH O P°H Ox LMeOH P°MeOH
2
2
Ptotal =  0.94×2.3 kPa    0.06×14.4 kPa  =2.1620.864=3.026 kPa
8 points
(d)
Calculate the composition of the vapor that is in equilibrium with this mixture.
1. Mole fraction calculation in the vapor
In order to calculate the composition of the vapor, we need to calculate the mole fractions
in the gas or vapor phase, which we can do from a knowledge of the partial pressures, calculated in (c).
The partial pressure of A is given by p A = x a P°A and we have calculated this above
Ptotal =3.026 kPa and p H O=2.162 kPa and p MeOH =0.864 kPa
2
x
V
H2 O
2.162
0.125
V
=
=0.71 and x MeOH
=
=0.29
3.026
2.125
8 points
(e)
5 g of sodium chloride, NaCl s  , is dissolved in this mixture, to form a solution that obeys Raoult's Law. The relative formula mass of sodium chloride is RMM  NaCl  =58.5 g mol −1 . Calculate the new pressure of the vapor that is in equilibrium with this solution, given that sodium chloride is involatile.
1. Mole calculation for sodium chloride
RMM  NaCl  =58.5 g mol so 5 g of NaCl ≡
−1
5
moles=0.085 moles
58.5
2. New mole fraction calculation in the solution
n MeOH =0.125 moles, nH O=2 moles and nNaCl=0.085 moles and so ntotal =2.21
2
x
L
H2 O
2
0.125
0.085
=
=0.905 , x LMeOH =
=0.057 and x LNaCl=
=0.038
2.21
2.21
2.21
3. Vapor pressure calculation
L
L
L
Ptotal = x H O PH Ox MeOH P MeOH  xNaCk P NaCl
°
2
°
°
2
As P NaCl=0
Ptotal =  0.905×2.3 kPa   0.057×14.4 kPa  =2.08150.821=2.9025 kPa
°
8 points
(f)
Calculate the composition of the vapor that is in equilibrium with this new solution.
1. New mole fraction calculation in the vapor
Ptotal =2.902 5 kPa and p H O=2.0815 kPa and pMeOH =0.821 kPa
2
x
V
H2 O
2.0815
0.821
V
=
=0.71 and x MeOH =
=0.29
2.902
2.902
Partial credit: The precise outcome of the last calculation will depend on rounding approaches. Variation in the third place of decimal is OK.
8 points
38 points
3(a)
For the general reaction a A  bB  c C  d D , the stoichiometric constants are given by a, b, c and d. Given that n
m
the rate law for this reaction is given by = k [ A ] [ B ] , what is the relationship, if any, between the stoichiometric coefficients and the order of the reaction with respect to A and B, which are given by n and m respectively?
There is no necessary relationship between the stoichiometric constants and the order of the reaction, either individually with respect to the components or in terms of the overall order.
2 points
NO
Cl
(b)
The balanced stoichiometric equation for the reaction of nitric oxide, g  , and chlorine, 2 g , to form nitrosyl chloride in the gas phase is Cl 2 g 2 NO g   2 NOCl g  , for which the experimental rate law is given by =k [Cl2 ][NO]2
Write down the order of reaction with respect to [NO g  ] and [Cl2 g  ] as well as the overall order of the reaction.
Order of reaction with respect to [ NO g  ]=2
Order of reaction with respect to [Cl 2 g  ]=1
Overall order of reaction =3
3 points
(c)
Aqueous ethyl acetate, MeCO 2 Et aq and sodium hydroxide, NaOH aq , react to give ethanol, EtOH aq , and sodium acetate, NaCO 2 Me aq , according to the stoichiometric equation MeCO 2 Et aq NaOHaq  NaCO2 Meaq EtOHaq . The rate law for this reaction is given by =k [MeCO2 Et  aq ][NaOHaq ] .
Write down the order of reaction with respect to [MeCO2 Et aq ] , [NaOHaq ] as well as the overall order of the reaction.
Order of reaction with respect to [ MeCO2 Et aq  ]=1
Order of reaction with respect to [ NaOHaq  ]=1
3 points
Overall order of reaction =2
(d)
Under a certain set of experimental conditions, the rate constant k is given by k=1.8×10−4 mol L −1 s−1 . Calculate the initial rate of reaction for the reaction between aqueous ethyl acetate and sodium hydroxide when the −1
−1
initial concentrations are given by [MeCO2 Et aq ]=0.32 mol L and [NaOHaq ]=0.5 mol L
= k [MeCO2 Et aq  ][NaOH aq ]
k =1.8×10 mol L s
−4
=  1.8×10
−4
(e)
−1
−1
,[MeCO2 Et aq  ]=0.32 mol L
−1
,[NaOH aq ]=0.5 mol L
−1
 × 0.32  × 0.5  =2.88×10−5 mol L−1 s−1
6 points
If the concentration of either reactant is doubled, what is the new initial rate in terms of the old initial rate?
For a first order reaction, the reaction rate will double if the concentration of either of the reactants is doubled
Partial credit: Students may explicitly calculate both new rates explicitly or, the smarter answer, recognize the first order rate relationship as above.
4 points
(f)
If both the concentrations of the reactants are doubled in the same experiment, what is the new initial rate, in terms of the old initial rate New rate will be four times the old rate.
4 points
22 points