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Exam 22 March 2012, questions and answers - midterm quiz
3
Chemical Dynamics (York University)
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YORK UNIVERSITY
NAME:____________________________________
SC/CHEM 1001M
STUDENT NO.:_____________________________
Marks
No
Quiz #3
March 22, 2012
50 min
30 marks
THE USE OF ANY REFERENCE MATERIAL OR ELECTRONIC GADGET IS PROHIBITED
Use blue pen. The final answers must be in ink. Write “No answer” if you do not know the answer.
(6)
1.
Problem
MgF2 is a poorly soluble salt with Ksp = 4  108. What is its solubility (S) in a buffered
solution with pH = 1 if Ka of HF is equal to 103?
You can solve either for a number (complete all calculations) or for a formula (express S
as a function of Ksp, Ka, and pH).
Solution:
MgF2 (s)
 Mg 2 (aq)  2F (aq)
2F (aq)  2H 3 0 (aq)  2HF(aq)  2H 2 0(l )
K sp
1/K a 2

MgF2 (s)  2H 3 0 (aq)  Mg 2 (aq)+ 2HF(aq)  2H 2 0(l )
K c  K sp / K a 2
1. Solution for a number:
Kc 
K sp
Ka 2
4 108

 4 102 M
6
10
[H3O  ]  10 pH  101 M
[Mg 2 ][HF]2
Kc 
[H 3 0 ]2
4 S 3 (M 3 )
 4 10 (M)  2 2 
10 (M )
2
S  3 104  3 103  3 101 =101  (~ 0.5)  5 102 M
2. Solution for a formula:
MgF2 (s)  2H 3 0 (aq)  Mg 2 (aq)+2HF(aq)  2H 2 0(l )
K c  K sp / K a 2
K sp
[Mg 2 ][HF]2 S (2S ) 2
4S 3



Kc 
[H 3 0 ]2
102pH 102pH K a 2
S
3
102pH K sp
4Ka 2
1
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(2)
2.
Define the 2nd Law of Thermodynamics for spontaneous and reversible processes
Answer:
In a spontaneous processes, the Suniv increases or Suniv = Ssys + Ssur > 0
In reversible processes, Suniv does not change or Suniv = Ssys + Ssur = 0
(1)
3.
Define the 1st Law of Thermodynamics for change of enthalpy:
Answer:
Enthalpy change in the system is equal to minus enthalpy change in the surroundings or
Hsys =  Hsur
(1)
4.
Relate standard molar entropies for the same amount of the same substance in different phases
by properly placing the < or > signs:
S°(s) < S°(l) < S°(g)
(4)
5.
Characterize the following processes by a single word as spontaneous, non-spontaneous, or
reversible
a) Gas condensation at melting temperature: spontaneous
b) Liquid freezing at melting temperature: reversible
c) Solid melting at melting temperature: reversible
d) Solid melting at boiling temperature: spontaneous
6.
Reaction 2N 2 O  g   2N 2  g   O 2  g  is characterized by H°  150 kJ/mol and
S° 0.150 kJ/(mol K) at T  300 K. Answer the following 2 questions (do calculations if
required).
(2)
a) Is this reaction spontaneous at 300 K under standard conditions?
G° = H°  TS°   150 kJ/(mol) – 300 K  0.150 kJ/(mol K)   195 kJ/mol < 0
The reaction is spontaneous under standard conditions
(2)
b) Is this reaction spontaneous at 300 K under non-standard conditions characterized
by Q  ( PN 2 PO2 ) / PN2O  2.73 ?
2
2
G = G° + RTlnQ°   195 kJ/(mol) + 8.31  103 kJ/(mol K) 300 K  ln(2.73) 
  195 + 3   192 kJ/mol < 0
The reaction is spontaneous under stated conditions
2
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7.
A figure below illustrates a graphical definition of Gº. Define in a similar way G for
conditions illustrated by the solid arrow
G
8.
A graph below relates the thermodynamic equilibrium constant of a reaction with temperature.
Do rough estimates of values of Hº and Sº from the graph.
1/T, K1
3
2.00E-03
2.010
0.00E+00
0
2.5103
2.50E-03
3.5103
3
3.00E-03
4.00E-03
3.0103 3.50E-03 4.010
2.0
 103
-2.00E-03
6.0
 103
-6.00E-03
~102
-8.00E-03
8.0
 103
Slope =  H°/R  102/(1.7  103) 
5K
Intercept = S°/R = 24  103 (unitless)
S° = Intercept  R =
24  103 (unitless)  8.31 J/(mol K) 
100  103 = 0.2 J/(mol K)
-1.00E-02
10.0
 103
-1.20E-02
12.0
 103
-1.40E-02
14.0
 103
lnKeq =  H°/RT + S°/R
H = slope  R
 5 K  8.31 J/(mol K)  50 J/mol
-4.00E-03
4.0
 103
ln Keq
(6)
~1.7103
3
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(4)
9.
Cu(s) dissolves in HNO3 producing Cu2+and NO2(g). Write a balanced equation for this redox
process.
Cu ½-reaction is trivial:
Cu(s)  Cu2+(aq) + 2e
NO3 ½-reaction:
NO3(aq)  NO2(g) O.S. of N changes from +5 to +4
Balancing electrons
NO3 + e  NO2(g)
Balancing charge
NO3 + e +2H+(aq)  NO2(g)
Balancing H and O atoms NO3 + e +2H+(aq)  NO2(g) + H2O
Balancing electrons with the Cu ½ reaction
2NO3 + 2e +4H+(aq)  2NO2(g) + 2H2O
Adding two ½ reactions Cu(s) + 2NO3 + 2e +4H+(aq)  Cu2+(aq) + 2e + 2NO2(g) + 2H2O
Electrons cancel Cu(s) + 2NO3 + 4H+(aq)  Cu2+(aq) + 2NO2(g) + 2H2O
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