Equilibrium JEE previous year questions with solutions are given here. Chemical equilibrium refers to the state of a system in which the concentration of the reactant and the concentration of the products do not change with time and the system does not display any further change in properties. In Ionic equilibrium, the ionic substance dissociates into their ions in polar solvents. The ions formed are always in equilibrium with its undissociated solute in the solution. Reactants and products coexist in equilibrium so that reactant conversion to product is always less than 100%. These reactions may involve the decomposition of a covalent (non-polar) reactant or ionization of ionic compounds into their ions in polar solvents.
As far as the JEE exam is concerned, Equilibrium is an important topic. The important concepts include electrolytes, non-electrolytes, Ostwald’s dilution law, ionic equilibrium formulas, homogeneous chemical equilibrium, heterogeneous chemical equilibrium, factors affecting chemical equilibrium, etc. This article gives an idea of what type of questions can be expected from this topic. BYJU’S provides accurate solutions prepared by our subject experts.
Students can easily download the questions and solutions in PDF format.
Download Equilibrium Previous Year Solved Questions PDF
JEE Main Previous Year Solved Questions on Equilibrium
1. Which of the following are Lewis acids?
(a) PH3 and BCl3
(b) AlCl3 and SiCl4
(c) PH3 and SiCl4
(d) BCl3 and AlCl3
Solution:
The compound which can accept a pair of electrons is known as Lewis acid. BCl3 and AlCl3 have vacant orbitals and their octet is not complete. Hence these can accept electron pairs and behave as Lewis acids.
Hence option (d) is the answer.
2. Species acting as both Bronsted acid and base is
(a) (HSO4) –
(b) Na2CO3
(c) NH3
(d) OH-
Solution:
A Bronsted acid is a substance that can donate a proton to any other substance and a Bronsted base is a substance that can accept a proton from any other substance. (HSO4)– can donate and accept a proton.
Hence option (a) is the answer.
3. What is the conjugate base of OH–?
(a) O2
(b) H2O
(c) O–
(d) O2-
Solution:
When acid gives H+ then the remaining of its part is called the conjugate base.
The conjugate base of OH– is O2-.
Hence option (d) is the answer.
4. Which one of the following substances has the highest proton affinity?
(a) H2S
(b) NH3
(c) PH3
(d) H2O
Solution:
The stability of the conjugate acid will give us the compound with the highest proton affinity.
Here ammonia has the highest proton affinity.
Hence option (b) is the answer.
5. When rain is accompanied by a thunderstorm, the collected rainwater will have a pH value
(a) slightly lower than that of rainwater without a thunderstorm
(b) slightly higher than that when the thunderstorm is not there
(c) uninfluenced by the occurrence of a thunderstorm
(d) which depends on the amount of dust in the air.
Solution:
The temperature increases due to the thunderstorm. As temperature increases, [H+] also increases, and thus pH decreases.
Hence option (a) is the answer.
6. For the reaction, 2SO2(g) + O2(g) ⇌ 2SO3(g), ∆H = -57.2 kJ mol–1 and Kc = 1.7 × 1016 Which of the following statements is incorrect?
(a) The equilibrium will shift in the forward direction as the pressure increases.
(b) The addition of inert gas at constant volume will not affect the equilibrium constant.
(c) The equilibrium constant is large, suggestive of reaction going to completion and so no catalyst is required.
(d) The equilibrium constant decreases as the temperature increases.
Solution:
The large value of Kc suggests that the reaction should go almost to completion. The oxidation of SO2 to SO3 is very slow. So the rate of reaction is increased by adding a catalyst. Statement c is wrong.
Hence option (c) is the answer.
7. 20 mL of 0.1 M H2SO4 solution is added to 30 mL of 0.2 M NH4OH solution. The pH of the resultant mixture is [pKb of NH4OH = 4.7]
(a) 9.4
(b) 9.0
(c) 5.0
(d) 5.2
Solution:
Given pKb of NH4OH = 4.7
20 mL of 0.1 M H2SO4 ⇒ nH+ = 4
30 ml 0.2 M NH4OH ⇒ nNH4OH = 6
Solution is basic buffer.
pOH = pKb + log [NH4+]/[NH4OH] = 4.7 + log (4/2)
= 4.7 + log 2
= 4.7+0.3
= 5
pH = 14-pOH
= 14-5 = 9
Hence option (b) is the answer.
8. The increase of pressure on ice water system at constant temperature will lead to
(a) no effect on that equilibrium
(b) a decrease in the entropy of the system
(c) a shift of the equilibrium in the forward direction
(d) an increase in the Gibbs energy of the system.
Solution:
On increasing the pressure on this system in equilibrium, the equilibrium tends to shift in a direction in which volume decreases, i.e., in the forward direction.
Hence option (c) is the answer.
9. The pH of rain water is approximately
(a) 7.5
(b) 6.5
(c) 7.0
(d) 5.6
Solution:
The pH of rainwater is approximately 5.6.
Hence option (d) is the answer.
10. A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is
(a) 1.8 atm
(b) 3 atm
(c) 0.3 atm
(d) 0.18 atm
Solution:
Given total pressure = 0.8 atm
CO2(g) + C(s) ⇌ 2CO(g)
Total pressure = 0.5-P +2P = 0.8
P = 0.8-0.5 = 0.3
KP = P2CO/PCO2 = (2P)2/(0.5-P)
= (0.6)2/0.2
= 0.36/0.2
= 1.8 atm
Hence option (a) is the answer.
11. Which of the following is a Lewis acid?
(a) NaH
(b) NF3
(c) PH3
(d) B(CH3)3
Solution:
The compound which can accept a pair of electrons is known as Lewis acid. B(CH3)3
Is a Lewis acid.
Hence option (d) is the answer.
12. The exothermic formation of ClF3 is represented by the equation:
Cl2(g) + 3F2(g) ⇌ 2ClF3(g); ∆H = –329 kJ Which of the following will increase the quantity of ClF3 in an equilibrium mixture of Cl2, F2 and ClF3?
(a) Increasing the temperature
(b) Removing Cl2
(c) Increasing the volume of the container
(d) Adding F2
Solution:
The addition of reactants or removal of the product will favour the forward reaction.
So adding F2 will increase the quantity of ClF3.
Hence option (d) is the answer.
13. Among the following acids which have the lowest pKa value?
(a) CH3COOH
(b) (CH3)2CH-COOH
(c) HCOOH
(d) CH3CH2COOH
Solution:
Higher the pKa value, weaker is the acid. So the strongest acid has the lowest pKa value.
Hence option (c) is the answer.
14. The correct relationship between free energy change in a reaction and the corresponding equilibrium constant Kc is
(a) ∆G = RT ln Kc
(b) –∆G = RT ln Kc
(c) ∆G° = RT ln Kc
(d) –∆G° = RT ln Kc
Solution:
∆G = ∆G° + 2.303 RT logKc
At equilibrium, ∆G = 0
So ∆G° = –2.303 RT logKc
Hence option (d) is the answer.
15. The molar solubility of Cd(OH)2 is 1.84 × 10–5 M in water. The expected solubility of Cd(OH)2 in a buffer solution of pH = 12 is
(a) 1.84 × 10–9 M
(b) 2.49 × 10–10 M
(c) (2.49/ 1.84)× 10–9 M
(d) 6.23 × 10–11 M
Solution:
Given molar solubility, s = 1.84 × 10–5
Ksp = 4s3
= 4( 1.84 × 10–5)3
Cd(OH2) ⇌ Cd2+ + 2OH-.
s’ represents the solubility in buffer solution
pH = 12
pOH = 2
[OH–] = 10-2s’ ×(10-2)2 = 4(1.84×10-5)3
So, s’ = 2.492 ×10-10 moles L-1
Hence option (b) is the answer.
Also read:-
Ionic equilibrium – Ostwald Dilution Law
Introduction To Chemical Equilibrium
Chemical Equilibrium
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