In the previous section, we saw the heterogeneous equilibrium. In this section, we will see applications of equilibrium constant
• Before seeing the applications, we will first summarize the important features of the equilibrium constant. It can be written in 5 steps:
1. We have seen that, during a chemical reaction, the stage at which, the rate of forward and backward reactions become equal is called equilibrium
• At this stage, the concentrations of all reactants and products remain constant
• Only these constant values of concentrations must be used for calculating Kc
• Other values of concentrations will not give the correct value of Kc
2. The value of Kc depends only on the concentrations at equilibrium
• It does not depend on the concentrations at the beginning of the reaction
3. The value of Kc depends on temperature
• If the reaction is carried out at a new temperature, a new equilibrium will be reached
• At that new equilibrium, the concentrations will be different
♦ So the value of Kc will also be different
• So when we report the value of Kc, we must quote the temperature also, along with the balanced equation
4. The value of equilibrium constant (K'c) for the backward reaction is equal to the inverse of the equilibrium constant (Kc)of the forward reaction. That is., K'c = 1/Kc
5. The value of the equilibrium constant will change if we multiply the balanced equation through out by an integer. This is because of the change in exponents of the concentrations
• Now we will see the applications of the equilibrium constant. There are mainly three applications:
♦ Predict the extent of a reaction on the basis of its magnitude
♦ Predict the direction of the reaction
♦ Calculate equilibrium concentrations
Predicting the extent of a reaction
• Consider the expression for Kc: $\mathbf\small{\rm{K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}}}$
• We see that:
♦ the concentrations of the products are in the numerator
♦ the concentrations of the reactants are in the denominator
• So we can write:
♦ If the concentrations of the products are high, Kc will have a high value
♦ If the concentrations of the reactants are high, Kc will have a low value
• Based on experiments, scientists have given the following 3 rules:
1. If value of Kc is greater than 103, it can be considered to be very large
• In such a situation, the system will contain mostly the products
• There will be only some traces of the reactants
• We can say: The reaction has almost reached completion
2. If value of Kc is less than 10-3, it can be considered to be very small
• In such a situation, the system will contain mostly the reactants
• There will be only some traces of the products
• We can say: The reaction has started but is very difficult to proceed
3. If the value of Kc is in the range of 10-3 to 103, there will be appreciable concentrations of both reactants and products
Predicting the direction of the reaction
• This can be explained in steps:
1. We know the expression for Kc: $\mathbf\small{\rm{K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}}}$
• In this expression,
♦ all the concentration values are: the values at equilibrium
2. Suppose that at any time ‘t’, the concentrations are: [A]t, [B]t, [C]t, and [D]t
• We cannot use these concentrations to find Kc because, we do not know whether they are the 'equilibrium concentrations' or not
• However, we can do a trial calculation
♦ That is., we proceed to find the ratio $\mathbf\small{\rm{\frac{[C]_t^c[D]_t^d}{[A]_t^a[B]_t^b}}}$
3. But since we are not sure whether they are equilibrium concentrations or not, we give this ratio a new name: Qc
• So we can write: $\mathbf\small{\rm{Q_c=\frac{[C]_t^c[D]_t^d}{[A]_t^a[B]_t^b}}}$
♦ Qc is called the reaction quotient
◼ Once we obtain Qc, we can compare it with Kc
4. If Qc = Kc, we can say:
Equilibrium is already reached
5. What if Qc is less than Kc ?
This can be analyzed in 4 steps:
(i) If Qc is less than Kc, it means that:
• In order for the Qc to become equal to Kc,
♦ The numerator has to increase
♦ And at the same time, denominator has to decrease
(ii) If numerator is to increase, more products should form
(iii) If denominator is to decrease, more reactants should be used up
(iv) From (iii) and (iv), it is clear that, the reaction should proceed in the forward direction
◼ So we can write:
If Qc is less than Kc, the reaction will proceed in the forward direction
6. We can write similar steps to prove the opposite also
• That is, we can prove that:
If Qc is greater than Kc, the reaction will proceed in the reverse direction
• Let us write those steps also:
(i) If Qc is greater than Kc, it means that:
• In order for the Qc to become equal to Kc,
♦ The numerator has to decrease
♦ And at the same time, denominator has to increase
(ii) If numerator is to decrease, more products should be used up
(iii) If denominator is to increase, more reactants should form
(iv) From (iii) and (iv), it is clear that, the reaction should proceed in the backward direction
◼ So we can write:
If Qc is greater than Kc, the reaction will proceed in the backward direction
Let us see an example. It can be written in steps:
1. Consider the reaction: H2(g) + I2 (g) ⇌ 2HI (g); Kc = 57.0 at 700 K
2. At time t, the following concentrations were measured:
[H2]t = 0.10 M, [I2]t = 0.20 M, [HI]t = 0.40 M
• So Qc = $\mathbf\small{\rm{\frac{0.40^2}{0.10 \times 0.20}}}$ = 8.0
3. We see that, Qc is less than Kc
• So Qc must increase
• For that, the numerator must increase. Also, denominators should decrease
• That means:
Concentration of product must increase. Also, concentrations of reactants must decrease
• Thus we can write:
The reaction will proceed in the forward direction
Now we will see some solved examples
Solved example 7.25
The value of Kc for the reaction 2A ⇌ B + C is 2 × 10-3. At a given time, the composition of the reaction mixture is: [A] = [B] = [C] = 3 × 10-4 M. In which direction the reaction will proceed?
Solution:
1. We have: Qc = $\mathbf\small{\rm{\frac{[B][C]}{[A]^2}=\frac{(3\times10^{-4})\times(3\times10^{-4})}{(3\times10^{-4})^2}}}$ = 1
2. We see that, Qc is greater than Kc
• So Qc must decrease
• For that, the numerators must decrease. Also, denominator must increase
• That means:
Concentration of products must decrease. Also, concentrations of reactants must increase
• Thus we can write:
The reaction will proceed in the reverse direction
Solved example 7.26
A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g) is 1.7 × 102 . Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Solution:
1. First we will find the molar concentrations:
• 1.57 mol N2 is present in 20 L
♦ So the concentration of N2 = No. of moles of N2 in 1 litre, [N2] = 1.57/20
• Similarly, [H2] = 1.92/20 and [NH3] = 8.13/20
2. We have: Qc =
$\mathbf\small{\rm{\frac{[NH_3]^2}{[N_2][H_2]^3}=\frac{(8.13/20)^2}{(1.57/20)\times(1.92/20)^3}}}$
= 2379.237
3. We see that, Qc is greater than Kc
• So the reaction is not at equilibrium
4. Predicting the direction:
• We see that, Qc is greater than Kc
• So Qc must decrease
• For that, the numerators must decrease. Also, denominator must increase
• That means:
Concentration of products must decrease. Also, concentrations of reactants must increase
• Thus we can write:
The reaction will proceed in the reverse direction
Solved example 7.27
Equilibrium constant, Kc for the reaction
N2 (g) + 3H2 (g) ⇌ 2NH3 (g) at 500 K is 0.061
At a particular time, the analysis shows that composition of the reaction mixture
is 3.0 mol L-1 N2, 2.0 mol L-1 H2 and 0.5 mol L-1 NH3 . Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?
Solution:
1. We have: Qc =
$\mathbf\small{\rm{\frac{[NH_3]^2}{[N_2][H_2]^3}=\frac{(0.5)^2}{(3.0)\times(2.0)^3}}}$
= 0.0104
2. We see that, Qc is less than Kc
So the reaction is not at equilibrium
3. Predicting the direction:
• We see that, Qc is less than Kc
• So Qc must increase
• For that, the numerators must increase. Also, denominator must decrease
• That means:
Concentration of products must increase. Also, concentrations of reactants must decrease
• Thus we can write:
The reaction will proceed in the forward direction
Calculating Equilibrium Concentrations
We have seen some problems where we calculate the concentrations at equilibrium. We solved them algebraically by putting 'x' as the unknown. Here we will see some more advanced problems:
Solved example 7.28
13.8 g of N2O4 was placed in a 1 L reaction vessel at 400K and allowed to attain
equilibrium
N2O4(g) ⇌ 2NO2(g)
The total pressure at equilibrium was found to be 9.15 bar. Calculate Kc, Kp and partial pressures at equilibrium.
Solution:
1. Given that, 13.8 grams of N2O4 is taken. We have to convert it into number of moles
• One mole of N2O4 is 92 grams. So 1 gram of N2O4 is 1/92 moles
• Thus we get: 13.8 grams of N2O4 = 138/92 = 0.15 mol N2O4
2. Let x mol NO2 be present at equilibrium
• Then x/2 mol N2O4 would have been used up
• So the number of moles of N2O4 at equilibrium = (0.15 - x/2)
3. So the equilibrium constant Kc will be given by:
$\mathbf\small{\rm{K_c=\frac{x^2}{0.15-{\frac{x}{2}}}=\frac{2x^2}{0.30-x}}}$
4. Next we will find the similar expression for Kp. The following steps from (5) to (11) will help us to obtain the expression
5. Total number of moles at equilibrium = (0.15 - x/2 + x) = (0.15 + x/2)
6. Number of moles of N2O4 at equilibrium = (0.15 - x/2)
• So mol fraction of N2O4 = $\mathbf\small{\rm{\frac{0.15-{\frac{x}{2}}}{0.15+{\frac{x}{2}}}=\frac{0.3-x}{0.3+x}}}$
7. Number of moles of NO2 at equilibrium = x
• So mol fraction of N2O4 = $\mathbf\small{\rm{\frac{x}{0.15+{\frac{x}{2}}}=\frac{2x}{0.3+x}}}$
8. Partial pressure of N2O4
= Total pressure × Mole fraction of N2O4
= 9.15 × $\mathbf\small{\rm{\frac{0.3-x}{0.3+x}}}$
9. Partial pressure of NO2
= Total pressure × Mole fraction of NO2
= 9.15 × $\mathbf\small{\rm{\frac{2x}{0.3+x}}}$
10. Consider the results in (8) and (9)
• $\mathbf\small{\rm{\frac{9.15}{0.3+x}}}$ is common. Will will put it as 'U'
• So the results become:
(i) Partial pressure of N2O4 = $\mathbf\small{\rm{(0.3-x)U}}$
(ii) Partial pressure of NO2 = $\mathbf\small{\rm{2xU}}$
11. Now we can write the expression for Kp:
$\mathbf\small{\rm{K_p=\frac{(2xU)^2}{(0.3-x)U}=\frac{4x^2U}{0.3-x}}}$
12. We know the relation between Kp and Kc: $\mathbf\small{\rm{K_p=K_c(RT)^{\Delta n}}}$
• In our present case, Δn = (2-1) = 1
• So from (3) and (11), we get: $\mathbf\small{\rm{\frac{4x^2U}{0.3-x}=\frac{2x^2}{0.3-x}(RT)^1}}$
⇒ 2U = RT
13. Substituting for U from (10), we get:
$\mathbf\small{\rm{\frac{2 \times 9.15}{0.3+x}}}$ = RT
⇒ $\mathbf\small{\rm{\frac{2 \times 9.15}{0.3+x}}}$ = (0.083 × 400)
x = 0.2512
14. Using this value of x in (3), we get: $\mathbf\small{\rm{K_c=\frac{2(0.2512)^2}{0.30-0.2512}}}$ = 2.5861
15. Using this value of x in (11), we get:
Kp = $\mathbf\small{\rm{\frac{4x^2U}{0.3-x}=[\frac{4x^2}{0.3-x}\times \frac{9.15}{0.3+x}]}}$
= $\mathbf\small{\rm{[\frac{4(0.2512)^2 \times 9.15}{(0.3-0.2512)(0.3+0.2512)}]}}$ = 85.86
16. The partial pressure of N2O4 at equilibrium can be calculated using the result in (8)
• By putting x = 0.2512 in (8), we get: pN2O4 = 0.813 bar
17. The partial pressure of NO2 at equilibrium can be calculated using the result in (9)
• By putting x = 0.2512 in (9), we get: pNO2 = 8.34 bar
Links to three more solved examples is given below:
•
In the next section, we will see the relation between equilibrium constant, reaction coefficient and Gibb's free energy
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