Key to 1st term Examination (2008-09)
1.
S.6B
ALE&T08.605a
(a) C
(b) C
(c) C
(d) A
(e) D
(f)
D
(g) C
(h) B
(i)
A
(j)
B
(k) D
(l)
2.
A
(a) Equilibrium shift to the right.
(b) Equilibrium shift to the left.
(c) Equilibrium shift to the right.
(d) No shifting will be found.
3.
Let v be the volume of the container
At start
At equil.
CH3COOH + C2H5OH
2
0 .5
v
v
2x
0 .5  x
v
v
CH3COOC2H5 + H2O
0
0
x
v
x
v
xx
vv
=4
2  x 0.5  x
v
v
x2
=4
(2  x)(0.5  x)
x2 = 4(2-x)(0.5-x)
3x2 - 10x + 4 = 0
x = 0.4648 or x = 2.869 (rejected)
Ans. 0.4648 mole of ester will be formed at equilibrium
P.1
4.
(a)
PCl5(g)
At start
PCl3(g)
x
At equil. x(1-0.86)
+
0
0.86x
Cl2(g)
0
moles
0.86x
moles
= 0.14x
Total No. of moles = 0.14x + 0.86x + 0.86x = 1.86x
χ PCl5 =
χ PCl3 =
0.14 x
= 0.075
1.86 x
0.86 x
= 0.462
1.86 x
χ Cl 2 = 0.462
PCl5 = 0.075 x 101 = 7.575 (kPa)
PCl3 = 0.462 x 101 = 46.66 (kPa)
PCl2 = 46.66 (kPa)
46.66  46.66
Kp =
7.575
= 287.4 (kPa)
(b) (i)
As Kp is dependent on temperature only, an increase in pressure has no effect on
the value of Kp.
(ii) As the total number of moles of the gaseous products is greater than that of the
gaseous reactants, by Le Chatilier's Principle, an increase in pressure shifts the
equilibrium position to the side with smaller gaseous volumes, in this case, to the
left. So the percentage dissociation of PCl5(g) decreases.
5.
(a) (i)
Mass of CH3COOH in 100 cm3 0.50 M solution
100
= 0.5 ×
× 60
1000
= 3.0 (g)
3m
3.05 = 100
m
200
2(3  m)
3.05 =
m
3.05m = 6 - 2m
5.05 m = 6
6
m=
= 1.188 (g)
5.05
P.2
5.
(a) (i)
Efficiency =
1.188
= 39.6 %
3
(ii) First Extraction:
3  m1
3.05 = 100
m1
100
3  m1
3.05 =
m1
3.05 m1 = 3 - m1
4.05m1 = 3
3
m1 =
4.05
= 0.741 (g)
% of CH3COOH extracted =
0.741
= 24.7 (%)
3
Mass of CH3COOH remains = 3 - 0.741 = 2.259 (g)
Second extraction = 2.259 × 24.7% = 0.558 (g)
Total extract = 0.741 + 0.558 = 1.299 (g)
1.299
Efficiency =
× 100% = 43.3%
3
(b) (a)(ii) is the better extraction method because the extraction efficiency for (a)(ii) is
larger than (a)(i).
In fact, it is better to extract an organic compound by small lots of 2-methylpropan-1-ol
instead of a single amount of 2-methylpropan-1-ol.
6.
H+ + A-
HA + heat
Ka
When the temperature is lowered from 35oC to 10oC, the equilibrium constant (Ka) for citric
acid is also decreased because dissociation of acid proceeds with absorption of energy. At
lower temperature, the concentration of hydrogen ions is comparatively smaller and this
causes the lower degree of sourness of the citrus fruits.
[Or when temperature decreases, equilibrium shifts to the left hand side.]
7.
CH3CH2COOH
CH3CH2COO- +
H+
At start
0.05
0
0
At equil.
0.05-y
y
y
pH = 3.10
-log y = 3.10
y = 7.943 x 10-4
(7.943  10 4 )(7.943  10 4 )
Ka =
0.050  7.943  10 4
= 1.282 x 10-5 (mol dm-3)
P.3
8.
(a)
CH3CH2COO- +
CH3CH2COOH
At start
0.1
At equil.
0.1-x
x x
=1.3×10-5
0.1  x
x2
=1.3×10-5 (Assume x<<0.1)
0 .1
H+
0
0
x
x
x2 = 1.3×10-6
x = 1.14 × 10-3 (The assumption is valid)
[H+] = 1.14 × 10-3
pH = -log 1.14 × 10-3
= 2.94
CH3CH2COO- + H2O
(b)
At start
CH3CH2COOH + OH-
0.1
At equil.
0.1-y
14
10
Kh =
= 7.69×10-10
5
1.3  10
y y
=7.69×10-10
0 .1  y
0
0
y
y
y2
= 7.69×10-10 (Assume y<<0.1)
0 .1
y2 = 7.69×10-11
y = 8.77 x 10-6 (The assumption is valid)
[OH-] = 8.77 x 10-6
pOH = -log 8.77 x 10-6
= 5.06
pH = 14 - 5.06 = 8.94
9.
Let the solubility of MgF2 in 0.1 mol dm-3 Mg(NO3)2 be x.
MgF2
Mg2+
At start
0.1
At equil.
0.1+x
+
2F2x
(0.1+x) (2x)2 = 8.0×10-8
(0.1) (4x2) = 8.0×10-8 (∵ 0.1>>x)
x2 = 2 × 10-7
x = 4.47 × 10-4
The solubility of MgF2 in 0.1 mol dm-3 Mg(NO3)2 is 4.47 × 10-4 mol dm-3 at 25 oC.
P.4
H+
10. HIn
red
+ In-
Ka = 4.7×10-4
yellow
[ H  ][ In  ]
[ HIn ]
K a [ HIn ]
[H+] =
[ In  ]
Ka =
4.7  10 4  5
= 2.47×10-5
95
95% yellow: [H+] =
pH = - log 2.47×10-5 = 4.61
4.7  10 4  95
+
95% red: [H ] =
= 8.93×10-3
5
pH = - log 8.93×10-3= 2.05
Change in pH = 2.05 - 4.61 = -2.56
11. (a) NH4+(aq)
NH3(aq) + H+(aq)
[ NH 3 ][ H  ]
KX =

[ NH 4 ]
[CH 3 COOH ][ NH 3 ]
(b) KI =

[CH 3COO  ][ NH 4 ]
=
[CH 3 COOH ][ NH 3 ] [ H  ]


[CH 3COO  ][ NH 4 ] [ H ]
=
[ NH 3 ][ H  ]
[CH 3COOH ]

[CH 3COO  ][ H  ]
[ NH 4 ]
[ NH 3 ][ H  ]
1
×

[CH 3COOH ]
[ NH 4 ]
[CH 3COO  ][ H  ]
1
=
× KX
KY
=
=
5.59  10 10
1.76  10 5
= 3.176x10-5
(no unit)
CH3COO- + NH4+
(c)
CH3COOH + NH3
At start
0.10
0.10
0
0
At equil.
0.10-y
0.10-y
y
y
KI =
yy
= 3.176 × 10-5
( 0 .1  y ) ( 0.1  y )
y2
= 3.176 × 10-5 (Assume y << 0.1)
0 .1  0 . 1
y2 = 3.176 × 10-7
y = 5.636 × 10-4 (M)
(The assumption is valid)
Ans. The concentration of NH3 is 5.636×10-4 M
P.5
12. (a) For solution A:
pH = -log 1.8×10-5 + log
5.00
= 4.74
5.00
For solution B:
pH = -log 1.8×10-5 + log
0.05
= 4.74
0.05
Conclusion: Both solutions have the same pH as they have the same
CH 3COO 
CH 3COOH
ratio.
i.e For a particular buffer system, all solutions that have the same ratio
salt
will have the same pH.
acid
(b) For solution A:
No. of moles of HCl added = 0.01
No. of moles of CH3COOH formed = 0.01
No. of moles of CH3COONa reacted = 0.01
Total no. of moles of CH3COOH = 5 + 0.01 = 5.01
No. of moles of CH3COONa remained = 5 – 0.01 = 4.99
4.99
pH = -log 1.8 x 10-5 + log
= 4.74
5.01
For solution B:
No. of moles of HCl added = 0.01
No. of moles of CH3COOH formed = 0.01
No. of moles of CH3COONa reacted = 0.01
Total no. of moles of CH3COOH = 0.05 + 0.01 = 0.06
No. of moles of CH3COONa remained = 0.05 – 0.01 = 0.04
0.04
pH = -log 1.8 x 10-5 + log
= 4.57
0.06
Conclusion: Solution A is the better buffer.
Although the pH change for solution B is small, a change did occur, in contrast to
solution A, which shows no observable change in pH.
These results show that solution A, which contains much larger quantities of the
buffering components, has a much higher buffering capacity than solution B.
13. (a)
P.6
13. (a) 1.
Hydrogen gas at 1 atmospheric pressure is passed into the standard hydrogen
electrode and an equilibrium is set up between the adsorbed layer of hydrogen gas
and the hydrogen ion in solution.
2.
The platinum electrode is connected by a salt bridge of aqueous potassium
chloride to the other half-cell of copper electrode immersing in solution
containing copper(II) ions so that the overall e.m.f. can be measured from the
voltmeter.
3.
Since the standard electrode potential for hydrogen is arbitrarily taken as zero, the
reading taken in the voltmeter will be the standard potential electrode of copper.
(b) Ecello = Ecathodeo - Eanodeo
= 0.80 - 0.34
= +0.46 V
Overall cell reaction:
2Ag+(aq) + Cu(s)  2Ag(s) + Cu2+(aq)
14. (a)
(b) At anode:
At cathode:
Pb(s) + SO42-(aq)  PbSO4(s) + 2e2H+(aq) + PbO2(s) + H2SO4(aq) + 2e-  PbSO4(s) + 2H2O(l)
(c) Pb(s)│PbSO4(s)│H2SO4(aq)│PbO2(s),PbSO4(s)│Pb(s)
Or Pb(s)│PbSO4(s)│[PbO2(s)+2H+(1M)+H2SO4(1M)],[PbSO4(s)+2H2O(l)]│Pb(s)
(d) Advantage: Rechargeable, it can be reused. (or Very large current, it can supply a high
chemical energy.)
Disadvantage: Large and heavy. (or not leakproof.)
(e) 1.
2.
As car batteries.
As an emergency source of electricity in hospitals. (or cinemas or large building)
P.7
15. (a) Oxygen cathode: O2(g) + 2H2O(l) + 4e-  4OH-(aq)
Hydrogen anode: H2(g)  2H+(aq) + 2eOverall reaction: 2H2(g) + O2(g)  2H2O(l)
(b) Increase in concentration of H2SO4 increases the concentration of H+(aq), this shift the
equilibrium position to the left at the hydrogen anode.
Increase in concentration of H2SO4 decreases the concentration of OH-(aq) at the
oxygen cathode, this shifts the equilibrium position to the right.
(c) The reading of the voltmeter increases.
When weights are added to the pistons of both chambers, the pressures of the gases are
increased, this shifts the equilibria of the oxidation and reduction to the right, the
position of equilibrium of the overall reaction is shifted to the right. More electrical
energy is liberated and thus the voltmeter shows a higher reading.
16. MnO4-(aq) + 8H+(aq) + 5eFe3+(aq) + e-
Mn2+(aq) + 4H2O(l) ------------------ (1) E1o = +1.51 V
Fe2+(aq) ------------------------------------------------- (2) E2o = +0.77 V
From (2): 5Fe2+(aq)
5Fe3+(aq) + 5e- -------------------------------- (3) E3o = -0.77 V
-----------------------------------------------------------------------------------------------------------(1) + (3): MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)
Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Eo = +1.51 V + (-0.77 V) = +0.74 V
Since the above reaction has a positive Eo value, therefore, the following reaction is
energetically favourable and spontaneous.
MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
Conclusion: Fe2+ is oxidized to Fe3+, and MnO4- is reduced to Mn2+.
17. (a) Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)
(b)
o
E cell
= (+0.34) – (-0.76) = +1.1 (V)
(c) Ecell = +1.1 –
0.059
(0.001)
log
2
(2.0)
= +1.1 – (-0.10) = +1.2 0(V)
18. (a) B is the first activated complex of A. (First transition state)
E is the second intermediate product.
G is the final product.
(b) (0-100) = -100( kJ mole-1).
(c) (325-100) = 225 kJ (mole-1).
(d) A  C
(e) E  G
(f)
325 kJ mol-1.
P.8
19. Step 1:
Step 2:
NO(g) + F2(g)  FNO(g) + F
NO(g) + F FNO(g)
Step 1 is the rate determining step.
[Reason: Step 2 is fast comparing with the step 1 because it is a free radical reaction. Hence,
step 1 is the rate determining step.]
20. (a) Step 1: I2(g)
[ I ]2
Kc =
[I 2 ]
2I(g)
(b) Step 2 is the rate determining step, because it is slower than step 1.
From step 2: Rate = k1[H2][I]2
[ I ]2
From step 1: Kc =
[I 2 ]
[I]2 = Kc[I2]
Subst. (2) into (1):
(1)
(2)
Rate = k1 [H2] Kc [I2]
= k1 Kc [H2] [I2]
= k[H2][I2]
∴ The reaction is first order with respect to H2, and first order with respect to I2.
i.e. Overall order of reaction = 2 (second order reaction)
21. For a second order reaction.
1
1
= kt +
[ HI ]t
[ HI ]0
1
Plot
against t
[ HI ]t
Time (min.)
1
[ HI ]t
0
120
240
360
480
1.20
1.41
1.59
1.79
2.00
Because a straight line is obtained, therefore the decomposition of gaseous hydrogen iodide
is a second order reaction.
P.9
22. (a) aX + bY  cC
Let the rate = k [X]m [Y]n
4  10 3
k (0.1) m (0.1) n
=
k (0.1) m (0.5) n
1  10 1
1
1
1
( )n = 4×10-2 = 0.04 =
= ( )2
5
25
5
n=2
∴ The reaction is second order with respect to Y.
4  10 3
k (0.1) m (0.1) n
=
k (0.5) m (0.1) n
2  10 2
1
1
( )m = 2×10-1 = 0.2 =
5
5
m=1
∴ The reaction is first order with respect to X.
i.e. Rate = k[X][Y]2
Rate constant k =
4  10 3
= 4 (dm6 mol-2s-1)
2
0.1  0.1
0 .3
= 0.15 (mol dm-3)
2
0 .6
[Y] =
= 0.30 (mol dm-3)
2
(b) [X] =
Initial rate of disappearance of X = 4×(0.15)(0.3)2 = 5.4×10-2 (mol dm-3 s-1)
Amount of X disappearing during the first 0.5 Sec.
= 0.5 × 5.4x10-2 × 2
= 5.4 × 10-2 (mol)
23. (a) k = Ae
 Ea
RT
ln k = ln A -
Ea
RT
When T = 370 + 273 = 643(K), k = 3.91x10-4
When T = 470 + 273 = 743(K), k = 4.05 x 10-2
Ea
ln(3.91x10-4) = ln A ----------------(1)
643R
Ea
ln(4.05x10-2) = ln A ----------------(2)
743R
E
Ea
(2)-(1): ln(4.05x10-2)-ln(3.91x10-4) = - a +
743R
643R
2
E
1
1
4.05  10

ln
= a(
)
4
3.91  10
R 643 743
Ea
4.640 =
×2.093×10-4
8.31
Ea = 184225 (J mol-1)
= 184 kJ mol-1
P.10
184200
23. (b) 3.91 x 10-4 = A e 8.31643
= A x 1.068 x 10-15
A = 3.661 x 1011
 Ea
RT
k = 3.661 x 1011 e
T = 450 + 273 = 723K
184200
k = 3.661 x 1011 e 8.31723
= 3.661 x 1011 x 4.844 x 10-14
= 1.77 x 10-2 (mol-1 dm3 s-1)
24. (a) Plot
ln
a
ax
vs t
t(min.)
a
ln
ax
k = slope =
0
40
100
160
200
270
0
0.094
0.273
0.459
0.524
0.739
a
0.4
=
b 148
= 2.70x10-3 (min-1) (or 4.50x10-5 S-1)
(b) For first order reaction:
ln 2
t1/2 =
k
0.693
=
4.56  10 5
= 1.54x104 (S)
(or 257 min.)
P.11