In the previous section,
we saw the dynamic nature of equilibrium in chemical reactions. In this section, we
will see Law of chemical equilibrium and Equilibrium constant
• In the previous section, we have seen a number of graphs related to chemical equilibrium
• The horizontal portion of the graph indicates equilibrium
♦ ‘Horizontal portion’ is a horizontal line
♦ Any horizontal line in a graph will be at a particular height from the x axis
♦ In our present case, this ‘particular height’ is the ‘concentration at equilibrium’
• If we start the experiment with different concentration of reactants, we will get a different horizontal line
♦ That is., a horizontal line at a different height from the x-axis
◼ So it is clear that, equilibrium depends upon concentration of reactants and products
◼ Norwegian scientists Cato Maximillian and Peter Waage discovered the relation between concentration and equilibrium
• Their findings were based on experiments
• We will now discuss about one of those experiments. Some basics about the experiment can be written in 5 steps:
1. Consider the reversible reaction: H2(g) + I2(g) ⇌ 2HI(g)
2. This reaction was carried out 6 times
• Each time, different concentrations of reactants and/or products were used
3. In all six experiments, the reaction was carried out in a closed vessel
• The concentration of reactants and products is specified in the units: mol L-1
♦ That is., number of moles present in one litre
4. So for convenience, we will assume that, the reactions were carried out in a closed vessel of capacity 1 litre
♦ By this assumption, we can drop the ‘L-1’ part from the units
5. The experiment was carried out 6 times. That means, 6 trials were done
♦ In the first four trials, H2 and I2 were taken initially
♦ In the last two trials, HI was taken initially
• We will now write the details of each trial
Trial 1:
Details can be written in 5 steps:
1. Initial quantities:
2.4 × 10-2 mol H2 and 1.38 × 10-2 mol I2 was taken
2. We know that, 1 mol H2 needs 1 mol I2
• Here, the quantity of I2 is lesser. So we would expect all the 1.38 × 10-2 mol of I2 to be used up
• But it was found that, at equilibrium, 0.12 × 10-2 mol I2 remained
• That means, only (1.38 – 0.12) = 1.26 × 10-2 mol I2 was used up
3. If 1.26 × 10-2 mol I2 is used up, the same 1.26 × 10-2 mol H2 must have been used up
• That means, (2.4 – 1.26) = 1.14 × 10-2 mol H2 must be remaining
• Indeed, the quantity of H2 at equilibrium was found to be 1.14 × 10-2 mol
4. Now, every 1 mol H2 or I2 would give 2 mol HI
• So 1.26 × 10-2 mol H2 or I2 would give (2 × 1.26) = 2.52 × 10-2 mol HI
• Indeed, the quantity of HI at equilibrium was found to be 2.52 × 10-2 mol
5. So we can write the final quantities at equilibrium:
♦ Quantity of I2: 0.12 × 10-2 mol (from 2)
♦ Quantity of H2: 1.14 × 10-2 mol (from 3)
♦ Quantity of HI: 2.52 × 10-2 mol (from 4)
• The initial and final quantities in this trial are written in the first row of the table 7.1 given further below
Trial 2:
Details can be written in 5 steps:
1. Initial quantities:
2.4 × 10-2 mol H2 and 1.68 × 10-2 mol I2 was taken
2. We know that, 1 mol H2 needs 1 mol I2
• Here, the quantity of I2 is lesser. So we would expect all the 1.68 × 10-2 mol of I2 to be used up
• But it was found that, at equilibrium, 0.20 × 10-2 mol I2 remained
• That means, only (1.68 – 0.20) = 1.48 × 10-2 mol I2 was used up
3. If 1.48 × 10-2 mol I2 is used up, the same 1.48 × 10-2 mol H2 must have been used up
• That means, (2.4 – 1.48) = 0.92 × 10-2 mol H2 must be remaining
• Indeed, the quantity of H2 at equilibrium was found to be 0.92 × 10-2 mol
4. Now, every 1 mol H2 or I2 would give 2 mol HI
• So 1.48 × 10-2 mol H2 or I2 would give (2 × 1.48) = 2.96 × 10-2 mol HI
• Indeed, the quantity of HI at equilibrium was found to be 2.96 × 10-2 mol
5. So we can write the final quantities at equilibrium:
♦ Quantity of I2: 0.20 × 10-2 mol (from 2)
♦ Quantity of H2: 0.92 × 10-2 mol (from 3)
♦ Quantity of HI: 2.96 × 10-2 mol (from 4)
• The initial and final quantities in this trial are written in the second row of the table 7.1 below:
 |
Table 7.1
|
• Using the values given in the table, the reader may write the five steps each for trial 3 and trial 4 and become convinced about the correctness of the values
Trial 5:
Details can be written in 5 steps:
1. Initial quantity: 3.04 × 10-2 mol HI
2. We know that, 2 mol HI decomposes to give 1 mol H2 and 1 mol I2
•So 1 mol HI decomposes to give 1⁄2 mol H2 and 1⁄2 mol I2
3. So we would expect 3.04 mol HI to give:
♦ (3.04⁄2) = 1.52 mol H2
♦ (3.04⁄2) = 1.52 mol I2
• But it was found that, at equilibrium, the quantities were:
♦ 0.345 mol H2
♦ 0.345 mol I2
4. If 0.345 mol H2 was formed, (0.345 × 2) = 0.690 mol HI would have decomposed
• That means, (3.04 – 0.690) = 2.35 × 10-2 mol HI must have remained
• Indeed, the quantity of HI at equilibrium was found to be 2.35 × 10-2 mol
5. So we can write the final quantities at equilibrium:
♦ Quantity of I2: 0.345 × 10-2 mol (from3)
♦ Quantity of H2: 0.345 × 10-2 mol (from 3)
♦ Quantity of HI: 2.35 × 10-2 mol (from 4)
• The initial and final quantities in this trial are written in the fifth row of the table 7.1 above
Trial 6:
Details can be written in 5 steps:
1. Initial quantity: 7.58 × 10-2 mol HI
2. We know that, 2 mol HI decomposes to give 1 mol H2 and 1 mol I2
•So 1 mol HI decomposes to give 1⁄2 mol H2 and 1⁄2 mol I2
3. So we would expect 7.58 mol HI to give:
♦ (7.58⁄2) = 3.79 mol H2
♦ (7.58⁄2) = 3.79 mol I2
• But it was found that, at equilibrium, the quantities were:
♦ 0.86 mol H2
♦ 0.86 mol I2
4. If 0.86 mol H2 was formed, (0.86 × 2) = 1.72 mol HI would have decomposed
• That means, (7.58 – 1.72) = 5.86 × 10-2 mol HI must have remained
• Indeed, the quantity of HI at equilibrium was found to be 5.86 × 10-2 mol
5. So we can write the final quantities at equilibrium:
♦ Quantity of I2: 3.79 × 10-2 mol (from 3)
♦ Quantity of H2: 3.79 × 10-2 mol (from 3)
♦ Quantity of HI: 5.86 × 10-2 mol (from 4)
• The initial and final quantities in this trial are written in the sixth row of the table 7.1 above
• Table 7.1 gives accurate values of the quantities (concentrations) at equilibrium
• Now we can do some calculations and find if there is any relation between those concentrations
• The first calculation can be written in 8 steps:
1. Write the concentrations of the reactants at equilibrium:
• Concentration is denoted by writing the chemical formula of the substance within square brackets. So we get:
♦ Concentration of H2(g) at equilibrium = [H2(g)]eq
♦ Concentration of I2(g) at equilibrium = [I2(g)]eq
2. Find the product of the above concentrations
• We get: [H2(g)]eq × [I2(g)]eq
3. Write the concentrations of the products at equilibrium:
• Concentration of HI(g) at equilibrium = [HI(g)]eq
4. Find the product of the above concentrations
• Since there is only one product, we get: [HI(g)]eq
5. Divide the result (4) by result (2). We get: $\mathbf\small{\rm{\frac{[HI(g)]_{eq}}{[H_2(g)]_{eq}[I_2(g)]_{eq}}}}$
6. The expression in (5) is applied to each of the 6 trials in table 7.1
• The results are tabulated in the second column of table 7.2 below:
 |
Table 7.2
|
• Let us see a sample calculation. We will consider the second trial
• Substituting the values from the second row of table 7.1 into the expression in (5), we get:
$\mathbf\small{\rm{\frac{2.96 \times 10^{-2}}{0.92 \times 10^{-2} \;\; \times \;\;2.96 \times 10^{-2}}}}$ = 1608.7
7. We see that, the values in the second column of table 7.2 are different from each other
• So we can conclude that, the expression in (5) will not give a constant value
8. The above 7 steps do not help us because, we do not get a constant value. We must look for another expression
• The second calculation can be written in 4 steps:
1. We consider another expression: $\mathbf\small{\rm{\frac{[HI(g)]^2_{eq}}{[H_2(g)]^1_{eq}[I_2(g)]^1_{eq}}}}$
• This expression is similar to the expression in (5)
♦ In the numerator, we multiply the product concentrations
♦ In the denominator, we multiply the reactant concentrations
• But the difference is that:
♦ The concentrations are raised to the respective stoichiometric coefficients
2. The expression in (1) is applied to each of the 6 trials in table 7.1
• The results are tabulated in the third column of table 7.2 above
• Let us see a sample calculation. We will consider the second trial
• Substituting the values from the second row of table 7.1 into the expression in (1), we get:
$\mathbf\small{\rm{\frac{(2.96 \times 10^{-2})^2}{(0.92 \times 10^{-2})^1 \;\; \times \;\;(2.96 \times 10^{-2})^1}}}$ = 47.6
3. We see that, the values in the third column of table 7.2 are comparable to each other
• So we can conclude that, the expression in (1) will give a constant value
4. The constant value obtained in this way is called Equilibrium constant
♦ It is denoted by the symbol Kc
✰ The subscript 'c' indicates that, Kc is expressed in terms of 'concentrations'
✰ The units of concentrations should be mol L-1
• Thus Kc for the reversible reaction H2(g) + I2(g) ⇌ 2HI(g) is given by the expression:
$\mathbf\small{\rm{K_c=\frac{[HI(g)]^2_{eq}}{[H_2(g)]^1_{eq}[I_2(g)]^1_{eq}}}}$
• Note that:
♦ The subscript 'eq' can be omitted
♦ This is because, the concentrations in the expression for Kc, are usually the equilibrium concentrations
♦ If the power is '1', it need not be written in the expression
♦ The symbol for phases (s, l, g) are generally ignored |
◼ Thus Kc for the reversible reaction H2(g) + I2(g) ⇌ 2HI(g) can be modified as:
$\mathbf\small{\rm{K_c=\frac{[HI]^2}{[H_2][I_2]}}}$
• Now we can write the Law of Chemical Equilibrium. It can be written in 4 steps:
1. The concentrations of the products are raised to their individual stoichiometric coefficients (obtained from balanced equation) and then multiplied
2. The concentrations of the reactants are raised to their individual
stoichiometric coefficients (obtained from balanced equation) and then
multiplied
3. Result in (1) is divided by result in (2)
4. The result in (3) will be a constant value. This is known as the Law of Chemical Equilibrium
◼ In general, we can write:
Kc for the reaction a A + b B ⇌ c C + d D is given by:
Eq.7.1: $\mathbf\small{\rm{K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}}}$
An example:
Kc for the reaction 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g) is given by:
$\mathbf\small{\rm{K_c=\frac{[NO]^4 [H_2O]^6}{[NH_3]^4[O_2]^5}}}$
Next, we
will see some calculations related to this Kc
Inverse of Kc
• This can be explained in 4 steps:
1. We have seen that, Kc for the reaction H2(g) + I2(g) ⇌ 2HI(g) is given by:
$\mathbf\small{\rm{K_c=\frac{[HI]^2}{[H_2][I_2]}}}$
2. Now consider the reverse reaction: 2HI(g) ⇌ H2(g) + I2(g)
• Here, HI comes on the reactant side. So the equilibrium constant will be given by:
$\mathbf\small{\rm{K'_c=\frac{[H_2][I_2]}{[HI]^2}}}$
3. Note that, to differentiate between forwards and backward reactions, we use Kc and K'c
• We see that, K'c is the inverse of Kc. That is: $\mathbf\small{\rm{K'_c=\frac{1}{K_c}}}$
4. So we can write:
♦ Equilibrium constant of the backward reaction
♦ is the inverse of the
♦ Equilibrium constant of the forward reaction
When stoichiometric coefficients change
• This can be explained in 5 steps:
1. We know that, a balanced equation can be multiplied throughout by a factor 'n'
2. For example, the balanced equation a A + b B ⇌ c C + d D can be multiplied throughout by n
• The modifies balanced equation is: an A + bn B ⇌ cn C + dn D
3. The equilibrium constant for this reaction is denoted as Knc
• We can write: $\mathbf\small{\rm{{K^n}_c=\frac{[C]^{cn}[D]^{dn}}{[A]^{an}[B]^{bn}}}}$
3. Consider the reaction: H2(g) + I2(g) ⇌ 2HI(g)
• Here, a = 1, b = 1 and c = 2
4. Let us multiply throughout by (n = 1⁄2)
• We get: 1⁄2H2(g) + 1⁄2I2(g) ⇌ HI(g)
5. Now, the equilibrium constant will be given by:
$\mathbf\small{\rm{{K^{\frac{1}{2}}}_c=\frac{[HI]^{2
\times \frac{1}{2}}}{[H_2]^{{1 \times \frac{1}{2}}}[B]^{{1 \times
\frac{1}{2}}}}}}$
⇒ $\mathbf\small{\rm{{K^{\frac{1}{2}}}_c=\frac{[HI]}{[H_2]^{{\frac{1}{2}}}[B]^{{\frac{1}{2}}}}}}$
• We see that:
Kc, K'c and Knc all have different values
◼ So it is important to write the correct form of the balanced equation while quoting the value of equilibrium constant |
We will now see some solved examples
Solved example 7.1
The following concentrations were obtained for the formation of NH3 from N2
and H2 at equilibrium at 500K. [N2 ] = 1.5 × 10-2 M. [H2 ] = 3.0 × 10-2 M and
[NH3 ] = 1.2 × 10-2 M. Calculate equilibrium constant.
Solution:
1. We have the general formula:
Kc for the reaction a A + b B ⇌ c C + d D is given by:
Eq.7.1: $\mathbf\small{\rm{K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}}}$
2. In our present case, the balanced equation is: N2(g) + 3H2(g) ⇌ 2NH3(g)
So we can write:
[A] = [N2] = 1.5 × 10-2 M; a = 1
[B] = [H2] = 3.0 × 10-2 M; b = 3
[C] = [NH3] = 1.2 × 10-2 M; c = 2
3. Substituting the above values in (2), we get:
$\mathbf\small{\rm{K_c=\frac{(1.2 \times 10^{-2})^2}{(1.5 \times 10^{-2})^1 \times (3.0 \times 10^{-2})^3}}}$ = 355.5
Solved example 7.2
At equilibrium, the concentrations of N2 = 3.0 × 10-3 M, O2 = 4.2 × 10-3 M and NO= 2.8 × 10-3 M in a sealed vessel at 800K. What will be Kc for the reaction
N2(g) + O2(g) ⇌ 2NO(g)
Solution:
1. We have the general formula:
Kc for the reaction a A + b B ⇌ c C + d D is given by:
Eq.7.1: $\mathbf\small{\rm{K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}}}$
2. In our present case, the balanced equation is: N2(g) + O2(g) ⇌ 2NO(g)
So we can write:
[A] = [N2] = 3.0 × 10-3 M; a = 1
[B] = [O2] = 4.2 × 10-3 M; b = 1
[C] = [NO] = 2.8 × 10-3 M; c = 2
3. Substituting the above values in (2), we get:
$\mathbf\small{\rm{K_c=\frac{(2.8 \times 10^{-3})^2}{(3.0 \times 10^{-3})^1 \times (4.2 \times 10^{-3})^1}}}$ = 0.622
Solved example 7.3
What is Kc for the following equilibrium when the equilibrium concentration of
each substance is: [SO2 ]= 0.60M, [O2 ] = 0.82M and [SO3 ] = 1.90M ?
2SO2(g) + O2(g) ⇌ 2SO3(g)
Solution:
1. We have the general formula:
Kc for the reaction a A + b B ⇌ c C + d D is given by:
Eq.7.1: $\mathbf\small{\rm{K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}}}$
2. In our present case, the balanced equation is: 2SO2(g) + O2(g) ⇌ 2SO3(g)
So we can write:
[A] = [SO2] = 0.60 M; a = 2
[B] = [O2] = 0.82 M; b = 1
[C] = [SO3] = 1.90 M; c = 2
3. Substituting the above values in (2), we get:
$\mathbf\small{\rm{K_c=\frac{(1.90)^2}{(0.60)^2 \times (0.82)^1}}}$ = 12.229
Solved example 7.4For the following equilibrium, K
c = 6.3 × 10
14 at 1000 K
NO
2(g) + O
2(g) ⇌ NO(g) + O
3(g)
Both the forward and reverse reactions in the equilibrium are elementary
bimolecular reactions. What is K
c , for the reverse reaction?
Solution:
1. We have:
♦ Equilibrium constant of the backward reaction
♦ is equal to
♦ Equilibrium constant of the forward reaction
That is: $\mathbf\small{\rm{K'_c=\frac{1}{K_c}}}$
2. Substituting the value of K
c, we get:
$\mathbf\small{\rm{K'_c=\frac{1}{6.3 \times 10^{14}}}}$ = 1.59 × 10
-15
•
In the next section, we will see homogeneous equilibrium
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