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Summary (CAIE) Cambridge A Level Chemistry (9701)
Summary & 15’s Chemical Equilibria Problems with Solution

Chemical Equilibria

Equilibrium Concept
REVERSIBLE REACTION

IRREVERSIBLE REACTION


An irreversible reaction is a reaction
that takes place in the same
direction or a reaction that cannot
reversed
...
In this case the product
can react again to form reactants
...

It requires a closed system

Equilibrium Type
HOMOGENEOUS EQUILIBRIUM

HETEROGENEOUS EQUILIBRIUM



Homogeneous Equilibrium:
Equilibrium in which all the
components are in one phase
...




N2(g) + 3 H2(g)  2 NH3(g)



CaCO3(s)  CaO(s) + CO2(g)



CH3COOH(aq)  CH3COO-(aq) + H+(aq)



Ag2CrO4(s)  2 Ag+(aq) + CrO42-(aq)

Laws and Provisions of Equilibrium
The product of the equilibrium concentrations of a substance on the right
side divided by the product of the equilibrium concentrations of a substance
on the left side, each raised to the power of its reaction coefficient, has a
certain value at a certain temperature
...

The total pressure of the gases is
the sum of the partial pressures
of all the gases

Shifting Equilibria: Le Chatelier’s Principle
If an action is performed on an equilibrium, the system will carry out a reaction
that tends to reduce the effect of that action
...
How does change in concentration affect the position of equilibrium?

Substance
concentration on the
left side
Substance
concentration on the
right side

Plus

Minus

The equilibrium
shifts to the right

The equilibrium
shifts to the left

The equilibrium
shifts to the left

The equilibrium
shifts to the right

2
...
The effect of temperature on the position of equilibrium
Endothermic Reaction

Exothermic Reaction

Reaction scheme

Reactant  Product H(+)

Reactant  Product H(-)

Energy concept

Need energy/heat

Release energy/heat

Temperature increased

Equilibrium shift to the right

Equilibrium shift to the left

Temperature reduced

Equilibrium shift to the left

Equilibrium shift to the right

Note: Especially for changes in temperature,
besides shifting the equilibrium, it also changes
the value of the equilibrium constant [Kc]

Equilibrium Shift
To the right
To the left

Effect on Kc
Up
Down

4
...

No Catalyst

With Catalyst

Concentration

Concentration

Product

Product

Reactant

Reactant

Time

Time

Equilibrium time

Equilibrium time

Shifting Equilibria
Treatment

Plus

Minus

Kc value

Left side substance
concentration
Right side substance
concentration

shift to the right

shift to the left

Not change

shift to the left

shift to the right

Not change

Volume
Pressure
Temperature
Catalyst

Shift to larger number
Shift to smaller number
coefficients
coefficients
Shift to smaller number
Shift to larger number
coefficients
coefficients
Endothermic: shift to right Endothermic: shift to left
Exothermic: shift to left
Exothermic: shift to right
Not shift

Not change
Not change
Change
Not change

The Equilibrium Constant for the Related Reaction
1
...
If the equilibrium reaction coefficient is multiplied by n, then the value
of the equilibrium constant is raised to the power of n
...
If several equilibrium reaction equations are added together, then the
value of the equilibrium constant is the product of the equilibrium
constant for the reactions that make it up
...
Provides Information on Reaction Completeness
 Kc is very large: the reaction proceeds completely or almost
completely to the right (irreversible reaction)
 Kc is very small: the reaction proceeds completely or almost
completely to the left
2
...

 Qc < Kc: reaction goes to the right
 Qc > Kc: reaction goes to the left
 Qc = Kc: equilibrium state

Degree of Dissociation
The degree of dissociation () is the ratio between the number of moles
of the substance that decomposes (reacts) to the number of moles
initially
...
Making Ammonia By Process of Haber-Bosch
N2(g) + 3 H2(g)  2 NH3(g)
 Performed at moderate temperature (500°C)
...

 Performed at high pressure (200 – 400 atm) so that the
equilibrium shifts to the right
...


2
...

 The reaction is carried out at normal pressure because at high
pressure it does not give adequate results
...


Problems & Solutions
01
...
N2(g) + H2(g)  NH3(g)
b
...
CaCO3(s)  CaO(s) + CO2(g)
d
...
N2(g) + 3H2(g)  2NH3(g)

b
...
CaCO3(s)  CaO(s) + CO2(g)

d
...
A total of 0
...
When equilibrium is reached, 0
...
Determine the Kc and Kp values for the following reactions:
2 HCl(g)  H2(g) + Cl2(g)

Ans:
2 HCl(g)  H2(g) + Cl2(g)
I

0,8

C

0,4

E

0,4



-

I = initial reaction

0,2  0,2

C = change

0,2

E = equilibrium

0,2

Kp = Kc (RT)n
= 0,25(0,082
...
In a room reacted gases A2 and B2 with a mole ratio of A2 and B2 = 2:3,
according to the reaction equation:
A2(g) + 2 B2(g)  2 AB2(g)
If at a temperature of 127°C and a pressure of 9 atm an equilibrium is
reached when the mole ratio of AB2 and B2 = 1:2, determine the values
of Kc and Kp under these circumstances
...
Put in initial reaction
...


4y = 3x – 2y
x = 2y
A2(g) + 2 B2(g)  2 AB2(g)
I

4y

E

6y

y  2y

C

We substitute x with 2y

3y

4y



2y
2y

pA2 : pB2 : pAB2 = 3y : 4y : 2y = 3 : 4 : 2
p total = 9 atm
pA2 =

3
9

4

2

9

9


...
9 = 4 atm; pAB2 =
...
400)2-3

Kc = 2,73

04
...

Ans:
N2O4(g)  2 NO2(g)

Kc =

1
𝑏

2NO2(g)  2 NO(g) + O2(g)
N2O4(g)  2 NO(g) + O2(g)

05
...

Ans:
1

XY + Z2  XZ2 + Y

Kc =

2 AB + 2 Z2  2 AZ2 + 2 B

Kc = (4√2 )2 = 32

XZ2 + 2 AZ2  XA2 + 3 Z2

Kc = 22 = 4

2B + Y  B2Y

Kc =

XY + 2 AB  XA2 + B2Y

Kc =

16

1
8
1

1


...
4
...
The Kc value for the reaction:
H2(g) + I2(g)  2 HI(g)
at a certain temperature is 49
...

a
...

b
...
Determine the number of moles of each gas at equilibrium
...
the mixture is not in equilibrium
because equilibrium is reached when
Qc = Kc
b
...

H2(g) + I2(g)  2 HI(g)
I

2

2

C

x  x 

2x

E

2-x

4+2x

2-x

4

This is an equilibrium concentration whose relative value = Kc, so
14 – 7x = 4 + 2x
x=

𝟏𝟎
𝟗

mol,

c
...
In a 1 liter chamber and a temperature of 27 °C, the following
equilibrium occurs:
2 SO3(g)  2 SO2(g) + O2(g)
In the initial conditions there were 0
...
Determine the Kc
and Kp values for the above reactions
...
300)1 = 10,93

Kp = 10,93

𝟒
𝟗𝟎

08
...
05 mol PCl5 and 0
...
Determine
the degree of PCl5 dissociation and the Kc value for the above reaction
...
02 mol PCl3, so there is 0,02 mol PCl3 react,
0,02 mol PCl5 and 0,02 mol Cl2 react
...
At a certain temperature and pressure, the AB decomposition reaction
2 AB(g)  A2(g) + B2(g)
is carried out in 2 tubes of the same size
...

After the two tubes reach equilibrium, it is known that the degree of
AB dissociation in tube 1 is 40% and the degree of dissociation in tube
2 is half
...

Ans:
V1 = V2 =V; 1 = 40%; 2 = 20%
Tube 1

Tube 2
2 AB(g)  A2(g) + B2(g)

2 AB(g)  A2(g) + B2(g)
I

x

-

-

C

0,4x 

0,2x  0,2x

E

0,6x

0,2x

Kc =

𝟏
𝟗

0,2x

I

x

-

y

C

0,2x 

0,1x  0,1x

E

0,8x

0,1x

y+0,1x

and x : y = 18 : 11

In tubes 1 and 2 the reactions are the same and at the same temperature
and pressure so the Kc values are the same in both tubes
...
The equilibrium equation is known:

Ag+(aq) + Fe2+(aq)  Ag(s) + Fe3+(aq)
In which direction will the equilibrium shift if:
a
...

b
...

c
...

d
...

Ans:
a
...

b
...

c
...
Left side 2 and right side 1 (because Ag(s) is not
counted)
...

d
...

11
...


a
...

b
...

c
...

Ans:
a
...
In an exothermic reaction that shifts to the right,
the temperature decreases, so T1 > T2
...


4 NH3(g) + 5 O2(g)  4NO(g) + 6 H2O(g)
I

4

C

1

E

3 mol

7


1,25

5


5,75 mol

1

6


6 mol

1,5
7,5 mol

equilibrium (1)
reaction
equilibrium (2)

c
...

Decrease [NO] and [H2O]
...

Decrease temperature
...
In a tank with a volume of 20 L, gases A and B are reacted according to
the reaction equation:

A(g) + B(g)  2 C(g)
At a certain temperature, an equilibrium is reached when gas A is 2
mol, gas B is 8 mol, and gas C is 4 mol
...
In which direction will the equilibrium shift?
b
...

Ans:
a
...

b
...


nA = 8 nB = 8 nC = 4 +

𝟒
𝟑
𝟒
𝟑
𝟖
𝟑

=
=
=

𝟐𝟎
𝟑
𝟐𝟎
𝟑
𝟐𝟎
𝟑

13
...


H2(g) + I2(g)  2 HI(g)
I

0,15

C

-x

E

0,15-x

0,2


-x
0,2-x



2x
2x

Now we use the quadratic equation to determine the value of x:

The first root cannot be used because it would afford a negative
amount of H2 (namely, 0,15 – 0,25 = -0,1)
...

14
...
The first two reactions
of the glycolysis cycle are:
C6H12O6(aq) + ATP(aq)  G6P(aq) + ADP(aq)

H° = -19,74 kJmol-1

G6P(aq)  F6P(aq)

H° =

2,84 kJmol-1

Calculate the equilibrium concentration of F6P(aq) generated in the
glycolysis cycle at normal body temperature, 37°C, starting with
[C6H12O6(aq)] = 1,20 x 10-6 M; [ATP(aq)] = 10-4 M; and [ADP(aq)] = 10-2 M
...
During a fever body temperature
increases
...
K2 = 1278
The equilibrium concentrations of the reactants and products is
determined as follows:
C6H12O6(aq) +
I

1,2 x 10-6

C

-x

E

1,2 x 10-6 - x

ATP(aq)



1 x 10-4


-x
1 x 10-4- x

ADP(aq) + F6P(aq)
1 x 10-2





+x

1 x 10-2 + x

x

+x

Expanding and rearranging the above equation yields the following
second order polynomial: 1277x2 – 0,1393x + 1,534 x 10-7 = 0
Using the quadratic equation to solve for x, we obtain two roots: x =
1,113 x 10-6 and 1,080 x 10-4
...
Therefore, [F6P]eq =
1,113 x 10-6
...
Since the net reaction above
is exothermic, Le Chatelier’s principle would force the equilibrium to
the left, reducing the amount of F6P generated
...
A mixture of 1,00 g H2 and 1,06 g H2S in a 0,500 L flask comes to
equilibrium at 1670 K:

2 H2(g) + S2(g)  2 H2S(g)
...
Determine the
value of Kp at 1670 K
...


2 H2(g)
I

136

C

+0,00438

E

136

+

S2(g)





0,00219
0,00219

2 H2S(g)
8,52



-0,00438
8,52



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