Chemistry 12
Resource Exam B
Scoring Guide
1. (4 marks)
Consider the following values for a catalyzed reaction that goes to completion:
PE ( products ) = 250 kJ
E a = 175 kJ
ΔH = +50 kJ
Sketch a PE diagram for this reaction on the grid provided, then use a dotted line to show how
removing the catalyst would change the PE diagram.
400
300
PE
(kJ) 200
100
Progress of reaction
Solution:
For Example:
400
300
PE
(kJ) 200
1 mark not catalyzed
1 mark PE activated complex = 375 kJ
1 mark ∆ H = +50 kJ
100
1 mark PE reactants = 200 kJ
Progress of reaction
Chemistry 12 – Resource Exam B Scoring Guide
Page 1
2. (4 marks)
Consider the following equilibrium:
← C ( s ) + 2Cl ( g )
CCl 4 ( g ) →
2
A student added 2.40 mol CCl 4 to a 2.00 L flask and monitored the [ Cl 2 ] . The following
graph was produced.
0.20
[Cl2] mol/L
0.12
0.10
0
mol
L
Time
Calculate the value of K eq .
Solution:
For Example:
CCl 4 ( g )
I
← C ( s ) + 2Cl ( g )
→
2
1.20 mol L
0
0
C
− 0.06
+ 0.12
E
1.14
0.12
2
Cl 2 ]
[
K eq =
[CCl 4 ]
2
0.12 )
(
=
⎫
⎪
⎪
⎪
⎬ ← 2 marks
⎪
⎪
⎪⎭
⎫
⎪
⎬ ← 1 mark
⎪
⎭
1.14
= 0.013
Chemistry 12 – Resource Exam B Scoring Guide
← 1 mark
Page 2
3. (4 marks)
A 0.15 g sample of solid PbF2 is recovered from 300.0 mL of its saturated solution.
What is the K sp of PbF2 ?
Solution:
For Example:
moles PbF2 = 0.15 g ×
1 mol
= 6.12 × 10 −4 mol
245.2 g
1
= 2.04 × 10 −3 M
0.300 L
[ PbF2 ] = 6.12 × 10 −4 mol ×
← 1 mark
← Pb2+ ( aq ) + 2F − ( aq )
PbF2 ( s ) →
[ Pb2 + ]
= 2.04 × 10 −3 M ×
1
= 2.04 × 10 −3 M
1
[F− ]
= 2.04 × 10 −3 M ×
2
= 4.08 × 10 −3 M
1
← 2 marks
K sp = [ Pb2 + ] [ F − ]
2
= ( 2.04 × 10 −3 ) ( 4.08 × 10 −3 )
= 3.4 × 10 −8
Chemistry 12 – Resource Exam B Scoring Guide
2
← 1 mark
Page 3
4. (3 marks)
Complete the Brønsted–Lowry acid base equation below and predict whether reactants or products
will be favoured at equilibrium, and justify your answer.
←
HCO3− + H 3C6 H 5O 7 →
_________________________________________________
Solution:
For Example:
← H CO + H C H O −
HCO3− + H 3C6 H 5O 7 →
2
3
2 6 5 7
Since K a ( H 3C6 H 5O 7 ) > K a ( H 2 CO3 ) , products are favoured
Chemistry 12 – Resource Exam B Scoring Guide
← 1 mark
} ← 2 marks
Page 4
5. (5 marks)
Calculate the pH of 0.45 M H 2 CO3 . Start by writing the predominant equilibrium equation.
Solution:
For Example:
H 2 CO3
+
+ H 2O ( ) →
← H 3O
+ HCO3−
I
0.45 M
0
0
C
−x
+x
+x
x
x
E 0.45 − x
← 1 mark
← 1 mark
(assume x is negligible)
⎡⎣ H 3O + ⎤⎦ ⎡⎣ HCO3− ⎤⎦
Ka =
H 2 CO3
4.3 × 10 −7 =
( x )( x )
0.45
x = ⎡⎣ H 3O + ⎤⎦ = 4.4 × 10 −4 M
pH = 3.36
Chemistry 12 – Resource Exam B Scoring Guide
} ← 1 mark
← 1 mark
← 1 mark
Page 5
6. (3 marks)
The following two experiments were conducted:
•
Titration A: a strong base was titrated with a strong acid.
•
Titration B: a weak base was titrated with a strong acid.
How does the pH at the equivalence point of Titration B compare with the pH at the equivalence
point of Titration A? Explain.
Solution:
For Example:
The pH at the equivalence point of Titration B is lower
than that of Titration A.
← 1 mark
In Titration A, a neutral salt is formed ( pH = 7 ).
← 1 mark
In Titration B, an acidic salt is formed ( pH < 7 ).
← 1 mark
Chemistry 12 – Resource Exam B Scoring Guide
Page 6
7. (4 marks)
Balance the following in basic solution:
I 2 + NO3− → NH 3 + H 3 IO6 2−
( basic )
Solution:
For Example:
14 ( NO3− + 9H + + 8e − → NH 3 + 3H 2 O )
14NO3−
14NO3−
← 1 mark
← 1 mark
8 ( I 2 + 12H 2 O → 2H 3 IO6 2 − + 18H + + 14e − )
2−
+
+ ← 1 mark
+ 126H + 96H 2 O + 8I 2 → 14NH 3 + 42H 2 O + 16H 3 IO6 + 144H
for
balancing
2−
−
+
−
+ 54H 2 O + 8I 2 + 18 OH → 14NH 3 + 16H 3 IO6 + 18H + 18 OH
14NO3− + 36H 2 O + 8I 2 + 18 OH − → 14NH 3 + 16H 3 IO6 2 −
7NO3− + 18H 2 O + 4 I 2 + 9 OH − → 7NH 3 + 8H 3 I O6 2 −
Chemistry 12 – Resource Exam B Scoring Guide
← 1 mark
Page 7
8. (3 marks)
Draw and label the parts of an electrolytic cell capable of copper plating an inert carbon electrode.
Solution:
For Example:
DC
Power
–
+
inert carbon
1 mark
Cu(s)
1 mark
Cu(NO3)2(aq)
1 mark
Chemistry 12 – Resource Exam B Scoring Guide
Page 8