DSE CHEM MASTERY
THE FINAL
PROBLEM
10. Chemical Equilibrium
Marking Scheme
ATTENTION
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1
DYNAMIC EQUILIBRIUM
3.1 Multiple Choice Question
Question 1
B
Question 2
D
Question 3
A.
X for a reaction system in equilibrium, the forward and backward reactions still continue all the time.
B.
X
C.
√
D.
X The composition of a system equilibrium remains constant. However, the amount of products may NOT
be equal to the amounts of reactants.
C
Question 4
The composition of a system at equilibrium remains constant.
Although the rates of the forward and backward reactions are the same, it does NOT mean that the concentrations
of the reactants and the products are the same at equilibrium.
D
2
Question 5
(1)
√
(2)
X The composition of a system at equilibrium remains constant. Although the rates of the forward and
backward reactions are the same, it does NOT mean that the concentrations of the reactants and the
products are the same at equilibrium.
(3)
√
B
Question 6
The composition of a system at equilibrium remains constant.
Although the rate of the forward and backward reactions are the same, it does NOT mean that the concentrations
of the reactants and products are the same.
C
Question 7
14 DSE MC 35-01
1st statement: incorrect
At equilibrium, forward reaction rate equals to backward reaction rate; but NOT forward reaction rate equals to 0.
2nd statement: incorrect
At equilibrium state, both forward and backward reactions are still undergoing. The concentrations of products and
reactants remain unchanged but they are still reacting.
D
Question 8
PP MCQ 31-01
Both forward and backward reactions will keep happening when the system is in equilibrium.
B
Question 9
A
Question 10
C
3
EQUILIBRIUM CONSTANT
4.1 Multiple Choice Question
Question 11
𝑛 "
2 #
'10* × '10*
𝐾! = 0.60 =
1 $
'10*
𝑛 = 6.12
A
Question 12
8.5 =
𝑥×𝑥
(2 − 𝑥)"
𝑥 = 1.49 𝑜𝑟 3.04 (𝑟𝑒𝑗𝑒𝑐𝑡𝑒𝑑)
C
Question 13 14 MCQ 26-01
K1 =
[&'! &(("(*+)]['# (*+)]
[&'! &(('(*+)]
K2 = [H+(aq)][OH-(aq)]
Required Kc =
[&'! &(('(*+)][('" (*+)]
[&'! &((" (*+)]
=
[&'! &(('(*+)][('"(*+)]['# (*+)]
[&'! &(("(*+)['# (*+)]
=
.$
.%
C
4
Question 14 14 MCQ 26-02
[/(0)]
Kc1 = [1(0)] = 2.00
[&(0)]
Kc2 = [/(0)] = 0.0100
[1(0)]$
Kc = [&(0)]$ = (.
2
2
$
$
&% ) (.&$ )
= (".44)$ (4.4244)$ = 2500
A
Question 15 14 MCQ 26-03
A
Question 16 18 MCQ 29-01
Qc =
[5((0)]$ [&6$ (0)]
[5(&6(0)]$
=
(4.244)$ (4.244)
(4."44)$
= 0.0250 mol dm-3
Qc > Kc means a net backward reaction will occur until equilibrium is attained.
C
Question 17 18 MCQ 29-02
The number of moles of gas increases as the system attains equilibrium.
It can be deduced that a net reaction that increases the number of moles of gas occurs, meaning that a net forward
reaction occurs.
A net forward reaction occurs when Qc < Kc
A
Question 18 18 MCQ 29-03
B
5
Question 19 19 MCQ 29-01
CO
H2
CH3OH
Initial concentration (mol dm-3)
0.1
0.1
0
Change in concentration
-0.01
-0.02
+0.01
Equilibrium concentration (mol dm-3)
0.09
0.08
0.01
[&' ('(0)]
$
$ (0)]
!
Kc = [&((0)]['
4.42
= 4.47×4.49$ = 17.4 mol-2 dm6
D
Question 20
[( (0)]!
Kc = [($ (0)]$ = 1
!
!
[O3(g)] = [O" (g)]$
B
Question 21
Kc = [:(
[:(! (0)]$
$
$ (0)] [($ (0)]
13.0 = [:(
(4.9#4)$
$
$ (0)] (4.#2;)
[SO2(g)] = 0.361 mol dm-3
B
Question 22
"
Initial concentration of N2O4(g) = 4.< = 4 𝑚𝑜𝑙 𝑑𝑚=$
Equilibrium constant 𝐾! =
(">)$
#=>
=
#> $
#=>
𝑚𝑜𝑙 𝑑𝑚=$
D
6
Question 23
Fe3-(aq)
SCN-(aq)
[Fe(SCN)]2+(aq)
Initial concentration (mol dm-3)
0.125
0.09375
0
Equilibrium concentration (mol dm-3)
0.125 – 0.080
= 0.045
0.09375 – 0.080
= 0.01375
0.080
[?@(:&5)]$#
Kc = [?@!#(*+)][:&5" (*+)] =
4.494
4.4#<×4.42$A<
= 129 dm3 mol-1
D
Question 24
D
Question 25
[&((0)]['$ (0)]!
Kc = [&'
' (0)]['$ ((0)]
The value of Kc for the reaction is small. Thus, the ratio of products to reactants at equilibrium is small.
A.
X
B.
X according to the equation for the reaction between CH4(g) and H2O(g), 3 moles of H2(g) will form when 1
mole of CO(g) forms. Therefore, 3[CO(g)] = [H2(g)]
C.
X [CO(g)] should be smaller than [CH4(g)]
D.
√ [CH4(g)] should be greater than [H2(g)]
D
Question 26
A
7
Question 27
Qc =
[&! '( ((0)]['$ (0)]
[&! ') ((0)]
=
4."<4×4."44
#
= 0.0125 mol dm-3
As Qc < Kc, concentrations of the products must increase and that of the reactant must decrease until Qc = Kc.
A
8
4.2 Long Question
Question 28 18 LQ 13-01
Mark
(a)
None of the final concentration of X(g), Y(g) and Z(g) is equal to zero./The concentration of the
reactant Y(g) is still not equal to zero after a long period of time.
1
(b)
2Y(g) ⇌ 3X(g) + Z(g)
3
Kc =
(c)
[B(0)]! [C(0)]
[D(0)]$
=
(4.;)! (4."4)
(4.$4)$
= 0.48 mol2 dm-6
The statement is incorrect. At the 25th minute after the reaction has started, the reaction attained
dynamic equilibrium. The rate of forward reaction equals to the rate of backward reaction and
not equal to zero.
1
9
THE EFFECT OF CHANGES IN CONCENTRATION
& TEMPERATURE ON CHEMICAL EQUILIBRIA
5.1 Multiple Choice Question
Question 29 2014 Q31
D
Question 30 2015 Q31
(1)
✘ According to the equation, the mole ratio of 𝐶𝑂 to 𝐶𝑙" is 1 ∶ 1. Hence, the rate of change in the
concentration of 𝐶𝑂(𝑔) is equal to that of 𝐶𝑙" (𝑔) as the reaction proceeds. Therefore, 𝐶𝑂(𝑔) and 𝐶𝑙" (𝑔) are
of the same concentration at equilibrium only if they are of the same initial concentration.
(2)
✔ According to the equation, the mole ratio of 𝐶𝑂𝐶𝑙" to 𝐶𝑂 is 1 ∶ 1. Hence, the rate of decomposition of
𝐶𝑂𝐶𝑙" (𝑔) 𝑖𝑠 equal to the rate of formation of 𝐶𝑂(𝑔).
(3)
✘ The value of 𝐾! remains unchanged since the temperature remains unchanged.
B
Question 31 2016 Q26
C
Question 32
B
10
Question 33 16 MCQ 27-01
If the concentration of one of the species in an equilibrium mixture is changed, the position of equilibrium will shift
so as to reduce the change in concentration.
Option
Adding
System responds by
Direction of net
reaction
Yellow-brown colour
of system
A
H2SO4
Consuming more
H+(aq) ions
Backward
More intense
B
NaBr
Consuming more Br(aq) ions
Backward
More intense
C
AgNO3, will form
AgBr ppt
Producing more Br(aq) ions
Forward
Fade
D
KOBr
Consuming more OBr(aq) ions
Backward
More intense
C
Question 34 16 MCQ 27-02
C
Question 35 16 MCQ 26-01
More N2O4(g) forms when the position of the equilibrium shifts to the left.
(1)
The system contains more gas molecules, thus its volume increases
(2)
A net forward reaction occurs. Heat is absorbed by the system as the forward reaction is endothermic.
(3)
The average relative molecular mass of the gas molecules in the system decreases as more N2O4(g) is
consumed.
D
11
Question 36 14 MCQ 31-01
Forward reaction is endothermic.
A
Question 37 PP MCQ 26-01
A
Question 38
A
Question 39
Shift of position of equilibrium:
When some C2H4(g) is injected into the system, the system responds by reducing the change. A net backward
reaction occurs to consume some C2H4(g), meaning that the equilibrium position shifts to the left.
Change of [C2H4(g)]:
The final concentration of C2H4(g) is below the maximum value when C2H4(g) is added and above the value at the
previous equilibrium.
A
12
Question 40
Colour intensity of mixture:
When the pressure of the system is increased, the concentration of I2(g) increases. Thus, the colour intensity of the
mixture increases.
Value of Kc:
As long as he temperature remains constant, the value of Kc does NOT change.
C
Question 41
Rates of forward and backward reaction:
When the temperature is increased, the particles have more energy and collide more often. A larger portion of the
particles have an energy equal to or greater than the activation energy. Thus, the rates of the forward and backward
reactions increase.
Kc:
When the temperature is increased, the system responds by reducing the temperature. As the forward reaction is
endothermic, the system undergoes a net forward reaction. More CO(g) and H2(g) are produced. So, the value of
Kc becomes higher.
C
Question 42
Sign of Δ𝐻:
When the temperature is increased, the value of Kc increases. It can be deduced that when the temperature is
increased, the system will undergo a net forward reaction so as to lower the temperature. Hence the forward
reaction should be an endothermic reaction. The sign of Δ𝐻 for the forward reaction should be positive.
Product yield as temperature increases:
When the temperature is increased, the value of Kc increases. Thus, the product yield also increases.
A
13
Question 43
Temperature of Experiment 2:
The tangent to the curve for Experiment 2 at the start is steeper than that for Experiment 1. Thus, the initial rate
of Experiment 2 is higher than that of Experiment 1. It can be deduced that Experiment 2 was carried out at a higher
temperature than Experiment 1.
Forward reaction:
When the temperature is increased, the yield of CH3OH(g) decreases. It can be deduced that when the temperature
is increased, the system will undergo a net backward reaction so as to lower the temperature. Hence the backward
reaction is endothermic, and the forward reaction is exothermic.
D
Question 44
When some O2(g) is added to the equilibrium system, the system responds by reducing the change. A net forward
reaction occurs to consume some O2(g).
(1)
False
The concentration of N2H4(g) decreases.
(2)
True
The final concentration of O2(g) is higher relative to the previous equilibrium.
(3)
True
(4)
True
D
Question 45
(1)
True
HA(aq), H+(aq) and A-(aq) are present in the system.
(2)
True
OH-(aq) ions react with H+(aq) ions to form H2O(l). The system responds by producing more H+(aq) ions.
(3)
False
The system does NOT involve any gas. Thus, changing the pressure has NO effect on the position of
equilibrium of the system.
A
14
Question 46
(1)
True
When the temperature is increased, the system responds by reducing the temperature. As the forward
reaction is endothermic, the system undergoes a net forward reaction. Thus, [N2O4(g)] decreases and [NO2(g)]
increases, as shown by the curves.
(2)
False
When the pressure is changed, both [N2O4(g)] and [NO2(g)] would shown a sudden increase or decrease. This
is NOT consistent with what the curves show.
(3)
False
When [NO2(g)] is changed, [NO2(g)] would show a sudden increase or decrease. This is NOT consistent with
what the curves show.
A
Question 47
An increase in pressure brings about a net reaction that decreases the number of moles of gas. The position of
equilibrium shifts to the side of the equation with fewer number of moles of gas.
C
Question 48
When the temperature is increased, the system will respond by lowering the temperature.
If the forward reaction is exothermic, the system will undergo a net backward reaction, the equilibrium position will
shift to the left.
More reactants will form. So, the equilibrium constant Kc decreases.
D
15
Question 49
A.
False
B.
False
The molar concentrations of CaO(s) and CaCO3(s) are constant. Adding or removing them will NOT cause
the position of equilibrium to shift.
C.
True
The pressure increases when the volume is decreased.
An increase in pressure brings about a net reaction that decreases the number of moles of gas. This helps to
decrease the pressure.
A net backward reaction occurs, the equilibrium position shifts to the left.
D.
False
C
Question 50
No. of moles of CO2(g) present at eqm: 0.250 mol dm-3 x 4.00 dm3 = 1.0 mol
Mass of CO2(g) present at eqm: 1.0 x (12 + 16 x 2) = 44.0 g
D
Question 51
Pressure:
A high pressure favours the formation of NH3(g). An increase in pressure brings about a net reaction that decreases
the number of moles of gas. This helps to reduce the pressure. A net forward reaction occurs and increases the
yield of NH3(g).
Temperature:
A low temperature favours the formation of NH3(g). When a low temperature is used, the system responds by
increasing the temperature. As the production of NH3(g) is exothermic, the system undergoes a net forward
reaction. Thus, the yield of NH3(g) increases.
B
16
Question 52
Direction of net reaction:
Qc = [5
[5'! (0)]$
!
$ (0)]['$ (0)]
*.) $
E
F
= *.!'* %.,*.,,* ! = 25.5 𝑑𝑚; 𝑚𝑜𝑙="
%.,
×E
%.,
F
Qc < Kc, a net forward reaction must occur until equilibrium is reached.
Pressure of system:
The number of moles of gas decreases and thus the pressure decreases.
B
17
5.2 Long Question
Question 53 14 LQ 13-01
Mark
(a)
(i)
NO(g)
O2(g)
NO2(g)
Initial
concentration
1.02
= 0.0204
50
1.20
= 0.024
50
0
Equilibrium
concentration
0.0204 × 39%
= 0.007956
0.024
− (0.0204 × 61.0%) ×
1
2
1
0.0204 × 61.0%
= 0.012444
= 0.017778
(𝟎.𝟎𝟏𝟐𝟒𝟒𝟒)𝟐
Kc = (𝟎.𝟎𝟎𝟕𝟗𝟓𝟔)𝟐(𝟎.𝟎𝟏𝟕𝟕𝟕𝟖)
1
1
= 137.6 dm3 mol-1
(ii)
(b)
No change, because Kc is independent of concentration/ only depends on temperature.
1
As revealed from the data, when temperature increases, Kc decreases. Therefore the
forward reaction is exothermic./
1
As higher temperature favours endothermic side of the reaction, so the forward reaction is
exothermic.
18
Question 54
Mark
(a)
H2(g)
Equilibrium
2.50 − 0.95 × 2
concentration (mol
1.50
= 0.4
dm-3)
CO(g)
1.2 − 0.95
= 0.167
1.50
CH3OH(g)
0.950
= 0.633
1.50
𝟎.𝟔𝟑𝟑
Kc = (𝟎.𝟒)𝟐(𝟎.𝟏𝟔𝟕) = 23.7 dm6 mol-2
(b)
(c)
(d)
1
1
2
Since the number of mole of gas on the product side is smaller than that on the reactant
side, decreasing pressures shift the equilibrium position to the left.
2
The value of Kc remains unchanged as temperature remains unchanged.
1
(i)
Increasing temperature shift the equilibrium positions to the left shows that the forward
reaction is exothermic.
2
(ii)
The value of Kc decreases.
1
Decreasing the volume of the reaction vessel will cause the concentrations of the reactants
to increase.
1
Hence the rate of the reaction will increase.
1
Question 55
When the reaction is conducted in acidic environment, the excess H+(aq) causes an increase in [H+(aq)]. So, the
equilibrium position shifts to the left until all Br-(aq) and HOBr(aq) are consumed, and the brown colour of reaction
mixture becomes darker.
When the reaction is conducted in alkaline environment, the excess OH-(aq) reacts with H+(aq) causing a decreases
in [H+(aq)]. So, the equilibrium position sifts to the right until all brown Br2(l) are consumed and the reaction
becomes colourless.
19
Question 56 2016 Q10
Mark
(a)
At dynamic equilibrium, the rate of forward reaction is equal to the rate of backward
reaction, and not equals zero./
1
At dynamic equilibrium, reactants are convened to products and products are converted
to reactants at equal rate. No net change is observed.
(b)
652 =
2.2 "
O Q
𝑉
3
2.2
2.50 − 2.2 " 2.5 − 2
O
Q R
S
𝑉
𝑉
V=17.0 𝑑𝑚$
(c)
(i)
increase. The reaction is exothermic. decrease in temperature will cause the equilibrium
position to shift to the right
1
(ii)
No change. A catalyst will increase the rate of forward reaction and that of backward
reaction to the same extent. / A catalyst has no effect on the equilibrium position.
1
Increase. There are larger number of mole of gas in the reactant side, increasing pressure
shift the equilibrium position to the right.
20
Question 57 2021 Q9
Mark
[H" (g)]# [CS" (g)]
[CH# (g)][H" S(g)]"
1
(i)
0.04, 0.14, 0.01
1
(ii)
0.01 # 0.03
Q O
Q
2.5
K Q = 2.5
0.04 0.14 "
O
QO
Q
2.5 2.5
1*
= 6.12 × 10=9 mol" dm=; (Correct unit is required)
1
(a)
(b)
O
(Not accept: M2 or (mol dm-3)2)
(iii)
K & remains unchanged as it only depends on temperature/ is independent of concentration
of reactants and products.
1
(Accept: independent of pressure)
(c)
ANY TWO:
2
Increase the volume
Decrease the pressure
Decrease the tempeature
21
22
23