# Example-1

Example-1

Consider the following equilibrium in a closed container: N

_{2}O_{4(g) }2NO_{2(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 (*K*) and degree of dissociation (_{p}*α*)?- Neither
*K*nor_{p}*α*changes - Both
*K*and_{p}*α*change *K*changes, but_{p}*α*does not change*K*does not change, but_{p}*α*changes

Solution (D)

For the reaction: N

_{2}O_{4(g)}2NO_{2(g)}where

*a*is initial mole of N_{2}O_{4}in*V*liter vessel and α is the degree of dissociation of N_{2}O_{4}. If volume is reduced to*V*/2, the initial concentration of N_{2}O_{4}becomesAn increase in concentration of reactants leads to the forward reaction, i.e., the decomposition of N

_{2}O_{4}has a constant value for*K*._{p}# Example-2

Example-2

For the following three reactions 1, 2, and 3, equilibrium constants are given:

- CO
_{(g)}+ H_{2}O_{(g)}CO_{2(g)}+ H_{2(g)};*K*_{1} - CH
_{4(g)}+ H_{2}O_{(g)}CO_{(g)}+ 3H_{2(g)};*K*_{2} - CH
_{4(g)}+ 2H_{2}O_{(g)}CO_{2(g)}+ 4H_{2(g)};*K*_{3}.

Which of the following equations is correct?

*K*_{1}*K*_{2}*K*_{3}=*K*_{1}*K*_{3}=*K*_{1}*K*_{2}*K*_{2}=*K*_{1 }×*K*_{3}_{1}

Solution (C)

Eq. (iii) = Eq. (i) + Eq. (ii)

Thus,

*K*_{3}= K_{1}× K_{2}# Example-3

Example-3

The equilibrium constants

*K*_{P}_{1}and*K*_{P}_{2}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 : 36
- 1 : 1
- 1 : 3
- 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, NH

_{3}is partially decomposed according to the equation, 2NH_{3}N_{2}+ 3H_{2}. The vessel is such that the volume remains effectively constant whereas pressure increases to 50 atm. Calculate the percentage of NH_{3}actually decomposed:- 65%
- 61.3%
- 62.5%
- 64%

Solution (B)

Initial pressure of NH

_{3}= 15 atm at 27°CThe pressure of

*a*mole of NH_{3}= 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 NH

_{3}decomposed# Example-5

Example-5

For the decomposition reaction: NH

_{2}COONH_{4(s) }2NH_{3(g)}+ CO_{2(g)}. The*K*= 2.9 × 10_{p}^{–5}atm^{2}. The total pressure of gases at equilibrium when 1 mol of NH_{2}COONH_{4(s)}was taken to start with will be:- 0.0194 atm
- 0.0388 atm
- 0.0582 atm
- 0.0766 atm

Solution (C)

NH

_{2}COONH_{4(s)}*K*= 2.9 × 10

_{p}^{–5}atm

^{3}

*K*= (2

_{p}*P*)

^{2}×

*P*= 2.9 × 10

^{–5}

or

*P*= 0.0194∴ Total pressure = 2

*P*+*P*= 3*P*= 3 × 0.0194 = 0.058 atm