Chemistry SL/HL Equilibrium Worked Examples and Practice
Writing equilibrium constant expressions
Example 1: Derive an equilibrium constant (K c
) expression for the Haber Process
Step 1: Write a balanced equilibrium equation
N
2 (g)
+ 3H
2 (g)
⇌ 2NH
3 (g)
Step 2: Set the equation to determine a ratio by placing the concentration of products as the numerator and the concentration of products as the denominator
__[NH
3
]__
[N
2
][H
2
]
Step 3: Raise each product and reactant to the power of their molar coefficients (found in balanced equation)
__[NH
3
] 2 __
[N
2
][H
2
] 3
Deducing the value of equilibrium constant expressions
Example 2: Determine the value of the equilibrium constant (K c
) for the equilibrium equation below:
H
2 (g)
+ I
2 (g)
⇌ 2HI
(g)
[H
2 (g)
]= 0.300, [I
2 (g)
]= 0.300, [HI
(g)
]= 3.00
Step 1: Set up an equilibrium constant expression, as seen in Example 1 above
__[HI] 2 __
[H
2
][I
2
]
Step 2: Substitute in concentration values
__[3.00] 2 __
[0.300][0.300]
Step 3: Calculate K c
K c
= 1.00 x 10 2
Deducing the value of the equilibrium constant expression for the reverse reaction
Example 3: Determine K c
for the reverse reaction for Example 2 above
Step 1: Set up an inverse expression (K c (reverse)
= 1/K c (forward)
__[H
2
][I
2
[HI] 2
]__
Step 2: Substitute in concentration values
__[0.300][0.300]__
[3.00] 2
Step 3: Calculate K c
K c
= 1.00x 10 -2 ( notice the negative. The sign is the inverse of the sign for the forward reaction )

Chemistry SL/HL Equilibrium Worked Examples and Practice
Deducing the value of the equilibrium constant expressions with differing coefficients
Example 4: Given that coefficients are halved, determine K c
for the following reaction (two options):
½ H
2 (g)
+ ½ I
2 (g)
⇌ HI
(g)
[H
2 (g)
]= 0.300, [I
2 (g)
]= 0.300, [HI
(g)
]= 3.00
Option 1: Take the square root of K c
as determined in Example 2
√1.00 x 10 2 = 1.0 x 10 1
Option 2: Replace powers from (Example 2, step 1) with adjusted coefficient values
__[3.00]__
[0.300] 1/2 [0.300] 1/2
Step 2: Calculate K c
K c
= 10.0 or 1.0 x 10 1
Example 5: Given that coefficients are doubled, determine K c
for the following reaction (two options):
2H
2 (g)
+ 2I
2 (g)
⇌ 4HI
(g)
[H
2 (g)
]= 0.300, [I
2 (g)
]= 0.300, [HI
(g)
]= 3.00
Option 1: Take K c
from Example 2 and find the square of that value
(1.0 x 10 2 ) 2 = 10,000 or 1.00x 10 4
Option 2: Replace powers from (Example 2, step 1) with adjusted coefficient values
__[3.00] 4 __
[0.300] 2 [0.300] 2
Step 2: Calculate K c
K c
= 10,000.00 or 1.0x 10 4
………………………………………………………….AHL Material………………………………………………………………
Calculating equilibrium constants using concentration data
Example 6: The Haber process is utilized in the production of ammonia, as seen in the equation below: N
2 (g)
+ 3H
2 (g)
⇌ 2NH
3 (g)
0.20 mol of N
2 (g)
and 0.2 mol of H
2 (g)
were allowed to reach equilibrium in a 1 dm 3 closed container. At equilibrium the concentration of NH
3 (g)
was 0.060 mol dm -3 . Determine the equilibrium concentrations of N
2 (g)
and H
2 (g)
and calculate the value of K c
.
Step 1: Write out a balanced equation for the reaction
N
2 (g)
+ 3H
2 (g)
⇌ 2NH
3 (g)

Chemistry SL/HL Equilibrium Worked Examples and Practice
Step 2: Write out an ICE table
N
2 (g)
+ 3H
2 (g)
⇌ 2NH
3 (g)
I (initial)
C (change)
E (equilibrium)
Step 3: Insert all known values. Initially, [products]=0
N
2 (g)
+ 3H
2 (g)
⇌
I (initial)
C (change)
E (equilibrium)
0.20 0.20 0.0
0.060
Step 4: Change values (represented by a variable) must be consistent with molar ratio coefficients
2NH
3 (g)
N
2 (g)
I (initial)
C (change)
E (equilibrium)
+
0.20
-α
0.20-α
3H
2 (g)
⇌
0.20
-3α
0.20-3α
2NH
3 (g)
0.0
+2α
0.060
Step 5: Using known values, solve for α
2α= 0.060 α=0.030
Step 6: Using the value for α, solve for equilibrium concentrations for the remaining unknowns
N
2 (g)
= 0.20-0.030= 0.17 mol dm -3 H
2 (g)
= 0.20-(3 x 0.030)= 0.11 mol dm -3
Step 7: Deduce an expression for the equilibrium constant:
K c
= __[NH
3
] 2 __
[N
2
][H
2
] 3
Step 8: Replace chemical symbols with equilibrium concentrations from step 6:
K c
= __[0.060] 2 __
[0.17][0.11] 3 K c
= 15.9
Calculating an equilibrium constant using Gibb’s Free Energy (ΔG θ )
Example 7: Calculate the equilibrium constant at 300K for the oxidation of iron:
2Fe
(s)
+ 3/2 O
2 (g)
⇌ Fe
2
O
3 (s)
Step 1: If not provided, calculate ΔH θ and ΔS θ (in this case, they are provided)
ΔH θ = -824.2 kJ mol
Step 2: Find ΔG θ
-1 ΔS θ = -270.5 J K -1 mol -1
ΔG
ΔG
θ
θ
= ΔH θ - T ΔS θ
= (-824.2 kJ mol -1 )- 300K (-270.5 J K -1 mol -1 )
ΔG θ = -743.1 kJ mol -1

Chemistry SL/HL Equilibrium Worked Examples and Practice
Step 3: Use the equation ΔG θ = -RTlnK to solve for K lnK= (-ΔG θ )/ RT lnK= -((743.1x10
K= e 298 = 2.6x10
3
129
J mol -1 )/(8.31 J K -1 mol -1 x 300K))= 298