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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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541 Medium
541 of 680
Application +4 | -1
Equilibrium constant for the following equilibrium,

3ClO−(aq)⇌2Cl−(aq)+ClO3−(aq)3 \mathrm{ClO}^{-}(a q) \rightleftharpoons 2 \mathrm{Cl}^{-}(a q)+\mathrm{ClO}_3^{-}(a q)

is 10×102810 \times 10^{28} at a temperature at which RT=2500 J mol−1R T=2500 \mathrm{~J} \mathrm{~mol}^{-1}.
Gibbs free energy change (ΔG∘)\left(\Delta G^{\circ}\right) is
View answer and explanation
Correct option: C
−161.21 kJ mol−1-161.21 \mathrm{~kJ} \mathrm{~mol}^{-1}
Option C (-161.21 kJ mol⁻¹) is correct because it is calculated using the Gibbs free energy equation with the given equilibrium constant.
542 Easy
542 of 680
Understanding +4 | -1
Which can act as buffer:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the given mixtures (acetic acid/sodium acetate, ammonium chloride/ammonium hydroxide, and HCN/NaCN) are buffer solutions.
543 Medium
543 of 680
Application +4 | -1
The value of KspK_{s p} of different salts AB,AB2,A2BA B, A B_2, A_2 B and A2B3A_2 B_3 are 4×10−10,8×10−15,2×10−144 \times 10^{-10}, 8 \times 10^{-15}, 2 \times 10^{-14}, 5×10−185 \times 10^{-18} respectively. The most soluble salt is
View answer and explanation
Correct option: D
A2 B3{A}_2 {~B}_3
Option D (A2 B3{A}_2 {~B}_3) is correct because it has the highest solubility among the given salts, determined by comparing their solubility products.
544 Easy
544 of 680
Application +4 | -1
The degree of ionisation of an acid HAHA is 0.00001 at 0.1 M0.1 \, \mathrm{M} concentration. Its dissociation constant is
View answer and explanation
Correct option: B
10−1110^{-11}
The correct answer is 10−1110^{-11} because the dissociation constant is calculated using the formula Ka=Cα2K_a = C \alpha^2.
545 Medium
545 of 680
Understanding +4 | -1
The equilibrium constant in a reversible reaction at a given temperature
View answer and explanation
Correct option: A
Does not depend on the initial concentrations
Option A is correct. The equilibrium constant is a characteristic property of a reversible reaction at a given temperature and is independent of initial concentrations.
546 Easy
546 of 680
Analyzing +4 | -1
Which species acts as stronger acid than formic acid in aqueous solution?
View answer and explanation
Correct option: B
H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}
Option B is correct because sulfuric acid is a strong acid that almost completely ionizes in aqueous solution.
547 Medium
547 of 680
Analyzing +4 | -1
The equilibrium constant for A2(g)⇋ 2A(g){A_2}\left( g \right)\leftrightharpoons\,2A\left( g \right) at 400 K and 600 K are 1×10−8  and  1×10−21 \times {10^{ - 8}}\,\,\mathrm{ and }\,\,1 \times {10^{ - 2}} respectively. The reaction is
View answer and explanation
Correct option: B
Endothermic
Option B (Endothermic) is correct because the equilibrium constant increases with increasing temperature, indicating an endothermic reaction.
548 Medium
548 of 680
Application +4 | -1
A weak acid, HA{HA}, has a Ka{K}_{{a}} of 1.00×10−51.00 \times 10^{-5}. If 0.100 mole of this acid dissolved in one litre of water, the percentage of acid dissociated at equilbrium is closest to
View answer and explanation
Correct option: A
1.00%1.00 \%
Option A (1.00%1.00 \%) is correct because the percentage of acid dissociated is calculated using the square root of Ka/C.
549 Medium
549 of 680
Application +4 | -1
Ammonia under a pressure of 15atm15\mathrm{ }atm at 27∘C27^\circ C is heated to 347∘C347^\circ C in a closed vessel in the presence of catalyst. Under the conditions, NH3\mathrm{N}{\mathrm{H}_3} is partially decomposed according to the equation, 2NH3⇋N2+3H22\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\mathrm{N}_2} + 3{\mathrm{H}_2}. The vessel is such that the volume remains effectively constant whereas pressure increases to 50atm50\mathrm{ }atm. Calculate the percentage of NH3\mathrm{N}{\mathrm{H}_3} actually decomposed
View answer and explanation
Correct option: A
61.3%61.3\%
Option A (61.3%61.3\%) is correct because it is the percentage of ammonia decomposed calculated from the change in pressure and stoichiometry of the reaction.
550 Hard
550 of 680
Understanding +4 | -1
Assertion :- Boric acid behaves as a weak monobasic acid.

Reason :- Boric acid contains hydrogen bonds in its structure.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because boric acid is a weak monobasic acid and contains hydrogen bonds, but the hydrogen bonds don't directly explain its weak acidity.
551 Medium
551 of 680
Understanding +4 | -1
Addition of sodium acetate to 0.1  M0.1\;M acetic acid will cause
View answer and explanation
Correct option: A
Increase in pHpH
Option A (Increase in pHpH) is correct because the addition of sodium acetate, a salt of a weak acid, causes a common ion effect, suppressing acetic acid ionization and increasing pHpH.
552 Hard
552 of 680
Application +4 | -1
What mass of 95% pure CaCO3CaC{O_3} will be required to neutralise 50 mL of 0.5 m HCl solution according to the following reaction?
CaCO3+2HCl(aq)→CaCl2(aq)+CO2(g)+2H2O(l)CaC{O_3} + 2HC{l_{(aq)}} \to CaC{l_{2(aq)}} + C{O_{2(g)}} + 2{H_2}{O_{(l)}}
[Calculate upto second place of decimal point]
View answer and explanation
Correct option: C
1.32 g
Option C (1.32 g) is correct because it accounts for the purity of the CaCO3CaCO_3 required to neutralize the HCl solution.
553 Easy
553 of 680
Application +4 | -1
For the reaction, H2+I2  ⇋  2HI{\mathrm{H}_2} + {\mathrm{I}_2}\;\leftrightharpoons\;2\mathrm{HI}, the equilibrium concentration of H2,I2  and    HI{\mathrm{H}_2},{\mathrm{I}_2}\;\mathrm{and}\;\;\mathrm{ HI} are 8.0,  3.0  and  28.0  mol/L8.0,\;3.0\mathrm{ }\;and\mathrm{ }\;28.0\mathrm{ }\;mol/L respectively. The equilibrium constant is
View answer and explanation
Correct option: B
32.6632.66
Option B (32.66) is correct because it is the equilibrium constant calculated from the given equilibrium concentrations using the equilibrium expression.
554 Medium
554 of 680
Analyzing +4 | -1
1 mol1 \mathrm{~mol} of helium is added to the dissociation equilibrium N2O4(g)⇌2NO2(g)−x kJ\mathrm{N}_2 \mathrm{O}_4(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{g})-x \mathrm{~kJ} set up in a cylinder fitted with a piston in such a way that the piston is moved outwards to keep the total equilibrium pressure constant
View answer and explanation
Correct option: A
The equilibrium will shift in the forward direction
Option A (The equilibrium will shift in the forward direction) is correct because adding an inert gas at constant pressure increases the volume, favoring the side with more moles of gas.
555 Easy
555 of 680
Understanding +4 | -1
In the heterogeneous equilibrium:
CaCO3(s)⇌NH3(g);Kp=4.0×104{CaCO}_3({s}) \rightleftharpoons {NH}_3({g}) ; {K}_{{p}}=4.0 \times 10^4
what would be the effect of addition of CaCO3{CaCO}_3 on the equilibrium concentration of CO2?{CO}_2?
View answer and explanation
Correct option: D
Remains unaffected
Option D (Remains unaffected) is correct because the addition of a solid reactant or product does not affect the equilibrium concentration of gaseous species in a heterogeneous equilibrium.
556 Medium
556 of 680
Application +4 | -1
For preparing a buffer solution of pH=7.0\mathrm{pH}=7.0, which buffer system you will choose?
View answer and explanation
Correct option: B
H2PO4−,HPO42−\mathrm{H}_2 \mathrm{PO}_4^{-}, \mathrm{HPO}_4^{2-}
Option B (H2PO4−,HPO42−\mathrm{H}_2 \mathrm{PO}_4^{-}, \mathrm{HPO}_4^{2-}) is correct because this buffer system has a pKa value closest to 7, making it suitable for maintaining a pH of 7.
557 Medium
557 of 680
Analyzing +4 | -1
1M solution of CH3COOHC{H_3}COOH should be diluted to…….. times so that pH is doubled.
View answer and explanation
Correct option: B
5.55×1045.55 \times {10^4} times
Option B (5.55×1045.55 \times {10^4} times) is correct because it is the dilution factor required to double the pH of a 1M acetic acid solution.
558 Easy
558 of 680
Analyzing +4 | -1
Nature of 0.1 M solution of sodium bisulphite is
View answer and explanation
Correct option: B
Alkaline
Option B (Alkaline) is correct because sodium bisulphite (NaHSO3NaHSO_3) forms bisulphite ions (HSO3−HSO_3^−) in solution, which can act as a weak acid, but the hydrolysis of SO32−SO_3^{2-} makes the solution alkaline.
559 Medium
559 of 680
Application +4 | -1
Equal volumes of two solutions of a strong acid having PH=3 and PH=4{P^H} = 3\,and\,{P^H} = 4 are mixed together. The PH{P^H} of the resulting solution is [log⁡0.5=−0.3]\left[ {\log 0.5 = - 0.3} \right]
View answer and explanation
Correct option: B
3.26
Option B (3.26) is correct because it's the calculated pH of the mixture of two strong acid solutions using the average hydrogen ion concentration.
560 Medium
560 of 680
Understanding +4 | -1
Which of the following salts when dissolved in water will hydrolyze?
View answer and explanation
Correct option: C
NH4Cl{NH}_4 {Cl}
Option C (NH4Cl{NH}_4 {Cl}) is correct because ammonium chloride is a salt of a weak base (ammonium hydroxide) and a strong acid (hydrochloric acid), and therefore undergoes hydrolysis in water.
561 Medium
561 of 680
Application +4 | -1
Some solid ammonium carbonate NH4COONH4\mathrm{NH}_4 \mathrm{COONH}_4 is take in a evacuated vessel and the vessel is sealed after a very long time a constant pressure of 0.6 atm0.6 \mathrm{~atm} is observed due to dissociation of the solid into NH3\mathrm{NH}_3 and CO2\mathrm{CO}_2 gas. Calculate the dissociation constant of solid.
View answer and explanation
Correct option: A
0.032 atm30.032 \mathrm{~atm}^3
Option A (0.032 atm30.032 \mathrm{~atm}^3) is correct because it is calculated by determining the partial pressures of the products and then using them to calculate the equilibrium constant Kp.
562 Hard
562 of 680
Application +4 | -1
The silver ion in a solution is precipitated by the addition of
chloride ion.
Ag+(aq)+Cl−(aq)⇌AgCl(s),K=1×1010\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s), K=1 \times 10^{10}.
The final volume of the solution is 500 mL500 \mathrm{~mL}. At a point when not more than 0.10 mg0.10 \mathrm{~mg} of Ag+\mathrm{Ag}^{+}ion remains unprecipitated [Cl−][\mathrm{Cl^-}] ion is
View answer and explanation
Correct option: A
5.4×10−5M5.4 \times 10^{-5} \mathrm{M}
Option C (5.4×10−4M5.4 \times 10^{-4} \mathrm{M}) is correct because it calculates the chloride ion concentration using the solubility product (KspK_{sp}) and the given silver ion concentration.
563 Medium
563 of 680
Analyzing +4 | -1
XeF6+H2O⇌XeOF4+2HF;\mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{XeOF}_4+2 \mathrm{HF}; equilibrium constant =K1.=K_1.

XeO4+XeF6⇌XeOF4+XeO3F2;\mathrm{XeO}_4+\mathrm{XeF}_6 \rightleftharpoons \mathrm{XeOF}_4+\mathrm{XeO}_3 \mathrm{F}_2; equilibrium constant =K2.=K_2.

The equilibrium constant for the following reaction will be:

XeO4+2HF⇌XeO3F2+H2O\mathrm{XeO}_4+2 \mathrm{HF} \rightleftharpoons \mathrm{XeO}_3 \mathrm{F}_2+\mathrm{H}_2 \mathrm{O}
View answer and explanation
Correct option: C
K2K1\frac{K_2}{K_1}
Option C is correct because the equilibrium constant for the overall reaction is obtained by dividing the equilibrium constant of the second reaction by that of the first.
564 Hard
564 of 680
Application +4 | -1
In a 10.0 L10.0 \mathrm{~L} container, an equilibrium was established between SO3,SO2\mathrm{SO}_3, \mathrm{SO}_2 and O2\mathrm{O}_2 gases, by starting with SO3\mathrm{SO}_3 only. The density of equilibrium mixture was found to be 16 g/litre16 \mathrm{~g/litre} at a temperature of 9000.0821 K.\frac{900}{0.0821} \mathrm{~K}. If the degree of dissociation of SO3\mathrm{SO}_3 is 40%,40 \%, then the value of KPK_{{P}} (in atm) for the reaction 2SO3(g)⇌2SO2(g)+O2(g)2 \mathrm{SO}_3(\mathrm{g})\rightleftharpoons 2 \mathrm{SO}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) is
View answer and explanation
Correct option: A
16 atm16 \mathrm{~atm}
The correct answer is 16 atm. The equilibrium constant KPK_P is calculated using partial pressures derived from the degree of dissociation and total pressure.
565 Easy
565 of 680
Analyzing +4 | -1
Assertion : The value of KK gives us a relative idea about the extent to which a reaction proceeds.

Reason : The value of KK is independent of the stochiometry of reactants and products at the point of equilibrium.
View answer and explanation
Correct option: C
If assertion is true but reason is false.
Option C is correct. The assertion is true because K indicates the extent of a reaction. The reason is false because K is dependent on the stoichiometry of the balanced chemical equation.
566 Easy
566 of 680
Analyzing +4 | -1
Which one of the following statement is not true?
View answer and explanation
Correct option: C
The  pH  of  1×10−8  M  HCl  is  8\mathrm{The}\;\mathrm{ pH}\;\mathrm{ of}\;1 \times {10^{ - 8}}\;\mathrm{M}\;\mathrm{ HCl}\;\mathrm{ is}\;8
Option C is correct because the pH of 1 × 10-8 M HCl is not 8. Since HCl is an acid, the pH must be less than 7. The pH will be slightly less than 7 due to the contribution of H+ from water.
567 Medium
567 of 680
Understanding +4 | -1
What is the best description of the change that occurs when Na2O(s)\mathrm{N}{\mathrm{a}_2}\mathrm{O}\left( s \right) is dissolved in water?
View answer and explanation
Correct option: B
Oxide ion accepts sharing in a pair of electrons
Option C (Oxide ion donates a pair of electrons) is correct because when Na₂O dissolves in water, the oxide ion (O²⁻) acts as a Lewis base and donates a pair of electrons to a water molecule, forming hydroxide ions (OH⁻).
568 Medium
568 of 680
Understanding +4 | -1
Favorable condition for product formation in the given reaction.
SO2+1/2O2→SO3(g)\mathrm{SO}_2+1 / 2 \mathrm{O}_2 \rightarrow \mathrm{SO}_3(\mathrm{g})
View answer and explanation
Correct option: A
High pressure
Option A (High pressure) is correct because the forward reaction decreases the number of moles of gas, so high pressure favors product formation according to Le Chatelier's principle.
569 Easy
569 of 680
Remembering +4 | -1
The pH{pH} value of ordinary water is :
View answer and explanation
Correct option: C
5.3
Option B (6.5) is incorrect, Option C (5.3) is the closest to the actual pH of ordinary water which is slightly acidic due to dissolved CO2, but Option A (7) is the pH of pure water, and Option D (7.8) is slightly alkaline. The correct answer is B (6.5)
570 Easy
570 of 680
Understanding +4 | -1
The strongest Lewis base in the following is
View answer and explanation
Correct option: A
CH3−\mathrm{CH}_3^ -
Option A (CH3−\mathrm{CH}_3^ -) is correct because CH3−\mathrm{CH}_3^ - is the conjugate base of the very weak acid CH4\mathrm{CH}_4, making it the strongest base among the given options.
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