# Solved Examples: Equilibrium Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE

## JEE: Solved Examples: Equilibrium Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE

The document Solved Examples: Equilibrium Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE is a part of the JEE Course Mock Test Series for JEE Main & Advanced 2022.
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Question 1: The Ksp of Ag2CrOis 1.1 ×10–12 at 298K. The solubility (in mol/L) of Ag2CrO4 in a 0.1M AgNO3 solution is    (IIT-JEE- 2013)
(a) 1.1×10-11
(b) 1.1×10-10
(c) 1.1×10-9
(d) 1.1×10-12
Solution:
Ag2+ is a common ion here. Hence, the contribution of Ag+ from AgNO3 would be ignored, and the solubility of Ag2CrOwill depend on CrO4-2 only.
Let the concentration of CrO4-2 in solution = x molL-1
Ksp = 1.1 ´10–12 = [Ag+][CrO4-2]
or
1.1´10–12 = [0.1]2[x]
or
X = 1.1×10-10
Hence, the correct option is B.

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

Solution:
For salt MX:
MX → M+ +X-
Ksp = [M+][X-]
Here [M+] = [X-]
So, Ksp =[M+]2
or
[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
or
[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
or
[X-]4 = Ksp/27 =(2.7×10-15)/27
or
[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
ΔG0f ( N2O4) = 100 KJ
ΔG0f ( 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)]/[P (N2O4)] = 100/10 =
Now,
ΔG= 2 ΔG0f ( NO2) - ΔG0f ( N2O4) = 2×50 -100 = 0
So,
ΔG = ΔG0 – 2.303RT log Qp = 0 – 2.303×8.314×298×log 10
= -5706 J = -5.706 kJ
(ii) Negative value of ΔG indicates that reaction is spontaneous one and it will shift in forward direction.

The document Solved Examples: Equilibrium Notes | Study Mock Test Series for JEE Main & Advanced 2022 - JEE is a part of the JEE Course Mock Test Series for JEE Main & Advanced 2022.
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