# Example-1

Example-1
Consider the following equilibrium in a closed container: N2O4(g)  2NO2(g). At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant (Kp) and degree of dissociation (α)?
1. Neither Kp nor α changes
2. Both Kp and α change
3. Kp changes, but α does not change
4. Kp does not change, but α changes
Solution (D)
For the reaction: N2O4(g)  2NO2(g)

where a is initial mole of N2O4 in V liter vessel and α is the degree of dissociation of N2O4. If volume is reduced to V/2, the initial concentration of N2O4 becomes

An increase in concentration of reactants leads to the forward reaction, i.e., the decomposition of N2O4 has a constant value for Kp.

# Example-2

Example-2
For the following three reactions 1, 2, and 3, equilibrium constants are given:
1. CO(g) + H2O(g)  CO2(g) + H2(g)K1
2. CH4(g) + H2O(g)  CO(g) + 3H2(g)K2
3. CH4(g) + 2H2O(g)  CO2(g) + 4H2(g)K3.
Which of the following equations is correct?
1. K1
2. K2K3 = K1
3. K3 = K1 K2
4. K2 = K1 × K3 1
Solution (C)
Eq. (iii) = Eq. (i) + Eq. (ii)

Thus, K3 = K1 × K2

# Example-3

Example-3
The equilibrium constants KP1 and KP2 for the reactions X  2Y and Z  P + Q, respectively, are in the ratio of 1 : 9. If the degree of dissociations of X and Z are equal, then the ratio of total pressure at these equilibria is:
1. 1 : 36
2. 1 : 1
3. 1 : 3
4. 1 : 9
Solution (A)
=
∴

# Example-4

Example-4
Ammonia under a pressure of 15 atm at 27°C is heated to 347°C in a closed vessel in the presence of a catalyst. Under the conditions, NH3 is partially decomposed according to the equation, 2NH3  N2 + 3H2. The vessel is such that the volume remains effectively constant whereas pressure increases to 50 atm. Calculate the percentage of NH3 actually decomposed:
1. 65%
2. 61.3%
3. 62.5%
4. 64%
Solution (B)

Initial pressure of NH3 = 15 atm at 27°C

The pressure of a mole of NH3 = p atm at 347°C
∴

∴ p = 31 atm

At constant volume and at 347°C,
mole ∝ pressure
a ∝ 31 (before equilibrium)
∴ a + 2x ∝ 50 (after equilibrium)
∴
∴
∴ % of NH3 decomposed

# Example-5

Example-5
For the decomposition reaction: NH2COONH4(s)  2NH3(g) + CO2(g). The Kp = 2.9 × 10–5 atm2. The total pressure of gases at equilibrium when 1 mol of NH2COONH4(s) was taken to start with will be:
1. 0.0194 atm
2. 0.0388 atm
3. 0.0582 atm
4. 0.0766 atm
Solution (C)
NH2COONH4(s)

Kp = 2.9 × 10–5 atm3

Kp = (2P)2 × P = 2.9 × 10–5

or P = 0.0194

∴ Total pressure = 2P + P = 3P

= 3 × 0.0194 = 0.058 atm