Chemical Equilibrium Solved Examples

Question: 1) The Ksp of Ag₂CrO₄ is 1.1 ×10^–12 at 298K. The solubility (in mol/L) of Ag₂CrO₄in a 0.1M AgNO₃ solution is (IIT-JEE- 2013)

1) 1.1×10^-11
2) 1.1×10^-10
3) 1.1×10^-9
4) 1.1×10^-12

Answer: B

Solution:

Ag²⁺is a common ion here. Hence, the contribution of Ag⁺ from AgNO₃would be ignored, and the solubility of Ag₂CrO₄will depend on CrO₄^-2only.

Let the concentration of CrO₄^-2in solution = x molL^-1

K_sp = 1.1 ´10^–12= [Ag⁺][CrO₄^-2]

1.1´10^–12= [0.1]²[x]

X = 1.1×10^-10

Hence, the correct option is B.

Question 2) Solubility product constant (K_SP) of salts of type MX, MX₂ and M₃X at temperature “T” are 4.0×10^-8, 3.2×10^-14and 2.7×10^-15, respectively. Solubilities (mol dm^-3) of the salts at temperature ‘T’ are (IIT JEE -2008)

1) MX> MX₂ > M₃X
2) M₃X> MX₂> MX
3) MX₂>M₃X> MX
4) MX >M₃X> MX₂

Answer: D

Solution:

For salt MX:

MX → M⁺ +X^-

Ksp = [M⁺][X^-]

Here [M⁺] = [X^-]

So, Ksp =[M⁺]²

[M⁺] =[X^-] =√Ksp = √(4.0×10^-8) = 2×10^-4

For salt MX₂:

MX₂ → M⁺ +2X^-

Ksp = [M⁺][X^-]²

Here 2[M⁺] = [X^-]

So, Ksp =4[M⁺]³

[M⁺] =?Ksp/4 = ? [(3.2×10^-14)/4] = 2×10^-5

For salt M₃X,

M₃X → 3M⁺ +X^-

Ksp = [M⁺]³[X^-]

Here [M⁺] = 3[X^-]

So, Ksp =27[X^-]⁴

[X^-]⁴ = Ksp/27 =(2.7×10^-15)/27

[X^-] = 10^-4

So, order of the solubilies is

MX >M₃X> MX₂

Hence, the correct option is D.

Question: 3) In the following equilibrium, N₂O₄(g)↔ NO₂(g) When 5 moles of each are taken, the temperature is kept at 298 K the total pressure was found to be 20 bar. Given that

ΔG⁰_f( N₂O₄) = 100 KJ

ΔG⁰_f( NO₂) = 50 KJ

(i) Find ΔG of the reaction

(ii) The direction of the reaction in which the equilibrium shifts. (IIT IEE -2004)

Solution:

(i) Given that, P total = P (N₂O₄) = P (NO₂) = 20 bar

But at equilibrium, P (N₂O₄) = P (NO₂) ,

So, P (N₂O₄) = P (NO₂) = 10 bar

Now

For the equilibrium reaction,

N₂O₄(g)↔ 2NO₂(g)

Q = [P (NO₂)]²/[P (N₂O₄)] = 100/10 =