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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Medium

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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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Showing 31-60 of 401 questions
31 Medium
31 of 401
Analyzing +4 | -1
Statement-I: PH\mathrm{P}^{\mathrm{H}} of a solution changes from 6 to 7 when diluted by 10 times

Statement-II: If [H+]\left[\mathrm{H}^{+}\right]decreases 10 times, PH\mathrm{P}^{\mathrm{H}} increases by one unit .
View answer and explanation
Correct option: D
Statement I is incorrect, Statement II is correct
Option D is correct. Statement I is incorrect because diluting an acidic solution (pH 6) will increase the pH but it will not exceed 7. Statement II is correct because pH increases by one unit for every 10-fold decrease in [H+].
32 Medium
32 of 401
Understanding +4 | -1
The equilibrium, N2( g)+3H2( g)⇌2NH3( g)\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) is attained at 25∘C25^{\circ} \mathrm{C} in a closed container and an inert gas He\mathrm{He} is introduced. Choose the correct statement.
View answer and explanation
Correct option: D
Nothing happens
Option D (Nothing happens) is correct because adding an inert gas at constant volume does not shift the equilibrium.
33 Medium
33 of 401
Analyzing +4 | -1
In an aqueous solution of volume 500ml500 \mathrm{ml}, when the reaction of 2AgCu=Cu2++2Ag2 \mathrm{Ag} \mathrm{Cu}=\mathrm{Cu}^{2+}+2 \mathrm{Ag} reached equilibrium the [Cu2+]\left[\mathrm{Cu}^{2+}\right] was x\mathrm{x} M. When 500ml500 \mathrm{ml} of water is further added, at the equilibrium [Cu2+]\left[\mathrm{Cu}^{2+}\right] will be
View answer and explanation
Correct option: D
less than x/2M\mathrm{x} / 2 \mathrm{M}
Option D (less than x / 2 M) is correct because adding water shifts the equilibrium to the left, decreasing the concentration of Cu2+ to less than half its original value.
34 Medium
34 of 401
Application +4 | -1
A buffer solution contains 0.1 mole of sodium acetate dissolved in 1000 cm31000\,\mathrm{c}{\mathrm{m}^3} of 0.1 M acetic acid. To the above buffer solution, 0.1 mole of sodium acetate is further added and dissolved. The pH of the resulting buffer is
View answer and explanation
Correct option: D
pKa+log⁡2\mathrm{p}{K_\mathrm{a}} + \log 2
Option D (pKa+log⁡2\mathrm{p}{K_\mathrm{a}} + \log 2) is correct because adding more salt to the buffer changes the salt/acid ratio, affecting the pH as described by the Henderson-Hasselbalch equation.
35 Medium
35 of 401
Remembering +4 | -1
pKa\mathrm{p}{K_a} of a weak acid is defined as
View answer and explanation
Correct option: C
log⁡101Ka{\log _{10}}\frac{1}{{{K_a}}}
Option C (log⁡101Ka{\log _{10}}\frac{{1}}{{{{K_a}}}}) is correct because the pKapK_a is defined as the negative logarithm (base 10) of the acid dissociation constant (KaK_a), which is equivalent to the logarithm of the reciprocal of KaK_a.
36 Medium
36 of 401
Analyzing +4 | -1
Assertion :- pH of a buffer changes with temperature.

Reason :- Ionic product of water (Kw)(K_w) changes with temperature.
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 is correct because both the assertion and the reason are true, but the reason is not a correct explanation of the assertion. The pH of a buffer changes with temperature due to changes in the equilibrium constants of the buffer components, and the ionic product of water also changes with temperature.
37 Medium
37 of 401
Remembering +4 | -1
Calculate ionic product of H2O{H}_2 {O} at 50∘C50^{\circ} \mathrm{C}.
View answer and explanation
Correct option: A
4×10−144\times 10^{-14}
Option A (4×10−144\times 10^{-14}) is correct because the ionic product of water (KwK_w) increases with temperature, and at 50∘C50^{\circ} \mathrm{C}, it is approximately 4×10−144\times 10^{-14}.
38 Medium
38 of 401
Understanding +4 | -1
Aqueous solution is alkaline in case of
View answer and explanation
Correct option: A
Na2CO3\mathrm{Na}_2 \mathrm{CO}_3
Option A (Na2CO3\mathrm{Na}_2 \mathrm{CO}_3) is correct because it is a salt of a weak acid and a strong base, resulting in an alkaline solution due to hydrolysis.
39 Medium
39 of 401
Application +4 | -1
pH\mathrm{pH} of 0.01 M (NH4)2SO40.01 \text{ M } \left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 and 0.02 M NH4OH0.02 \mathrm{~M} ~\mathrm{NH}_4 \mathrm{OH} buffer (pKa[NH4+]=9.26)\left(\text{pK}_{\mathrm{a}}\left[\mathrm{NH}_4{ }^{+}\right]=9.26\right) is
View answer and explanation
Correct option: A
9.26
The correct answer is A (9.26) because at the given concentrations, the pOH equals pKb, and pH equals pKa.
40 Medium
40 of 401
Analyzing +4 | -1
2KClO3(s)⇌2KCl(s)+3O2(g)2 {KClO}_3(\mathrm{s}) \rightleftharpoons 2 {KCl}(\mathrm{s})+3 {O}_2(\mathrm{g})

In the equilibrium given above, the equilibrium constant for the reverse reaction increases as the temperature of the system increases. Which is the correct statement about the system:-
View answer and explanation
Correct option: B
The reaction is exothermic
Option B is correct because the equilibrium constant for the reverse reaction increases with temperature, indicating an exothermic forward reaction.
41 Medium
41 of 401
Analyzing +4 | -1
Statement (S): Cu+,Ag+\mathrm{Cu}^{+}, \mathrm{Ag}^{+}are noft acids whereas CN−,H−\mathrm{CN}^{-}, \mathrm{H}^{-}are soft bases.

Explanation (E): Soft acids do not possess noble configuration whereas soft bases have donor atom of easily polarised nature.
View answer and explanation
Correct option: C
Both S and E are correct and E is correct explanation for S
Option C is correct because both the statement and explanation are true, and the explanation correctly justifies the statement regarding soft acids and bases.
42 Medium
42 of 401
Application +4 | -1
For the reaction, PCl3(g)+Cl2(g)  ⇋  PCl5(g)\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right) the value of Kc{K_c} at 250∘C250^\circ C is 26. The value of Kp{K_p} at this temperature will be
View answer and explanation
Correct option: C
0.610.61
Option C (0.61) is correct because it is calculated using the relationship between Kp and Kc, considering the change in the number of moles of gas.
43 Medium
43 of 401
Analyzing +4 | -1
Statement I: Aqueous solutions of NaCl{NaCl} and CH3COONH4{CH}_3 {COONH}_4 are neutral (pH=7.0\left(\mathrm{pH}=7.0\right. at 25∘C).25^{\circ} \mathrm{C}).

Statement II: Both the salts do not undergo hydrolysis.
View answer and explanation
Correct option: C
If Statement I is CORRECT, but Statement II is INCORRECT
Option C is correct because while NaCl solutions are neutral due to not undergoing hydrolysis, CH3COONH4 solutions are also neutral because the hydrolysis of the cation and anion cancel each other out, but it DOES undergo hydrolysis.
44 Medium
44 of 401
Analyzing +4 | -1
STATEMENT-1 : Sparingly soluble salts ABA B and XY2X Y_2 with the same solubility product, will have different solubility.

STATEMENT-2 : Solubility of sparingly soluble salt depend upon solubility product.
View answer and explanation
Correct option: B
If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1
Option B is correct because both statements are true, but Statement-2 does not correctly explain Statement-1. Solubility depends on the solubility product and the stoichiometry of the salt.
45 Medium
45 of 401
Understanding +4 | -1
STATEMENT I : Solubility of sparingly soluble salt decreases due to common ion effect.

STATEMENT II : Solubility product concentare does not depend on common ion effect.
View answer and explanation
Correct option: A
Statement I, II are correct
Option A (Statement I, II are correct) is correct because the common ion effect reduces the solubility of sparingly soluble salts, but the solubility product remains constant at a given temperature.
46 Medium
46 of 401
Analyzing +4 | -1
Statement I: If equilibrium constant is N2+O2⇌2NO\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} is K\mathrm{K}, the equilibrium constant for 12N2+12O2⇌NO\frac{1}{2} N_2+\frac{1}{2} O_2 \rightleftharpoons N O is K2\mathrm{K} 2

Statement II : Equilibrium constant does not depend on the stoichiometry form of equation
View answer and explanation
Correct option: A
Statement I, II are correct
Option A (Statement I, II are correct) is correct because both statements accurately describe the relationship between equilibrium constants and stoichiometry.
47 Medium
47 of 401
Application +4 | -1
Solubility product of the salt, AxByA_x B_y will be represented most suitably, if the solubility is represented by SS
View answer and explanation
Correct option: C
Ksp=xxyy(S)x+yK_{s p}=x^x y^y(S)^{x+y}
Option C (Ksp=xxyy(S)x+yK_{s p}=x^x y^y(S)^{x+y}) is correct because it represents the solubility product of the salt AxBy in terms of its solubility S.
48 Medium
48 of 401
Understanding +4 | -1
Assertion (A): If the equation for a reaction is reversed, the equilibrium constant is inverted and if the equation is multiplied by 2, the equilibrium constant is squared.

Reason (R): The numerical value of an equilibrium constant depends on the way the equation for the reaction is written.
View answer and explanation
Correct option: A
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because the equilibrium constant's value is directly dependent on how the reaction is written.
49 Medium
49 of 401
Understanding +4 | -1
Statement-I: The pNp^N of a buffer solution containing equal moles of acetic ncid and sodium ncetate is 4.8

 ( Pka of of acetic acid is 4.8 ) \text { ( } P^{k_a} \text { of of acetic acid is } 4.8 \text { ) }

Statement-II: The ionic product of water at 25∘C25^{\circ} \mathrm{C} is 10−14 mol2lit−210^{-14} \mathrm{~mol}^2 \mathrm{lit}^{-2}
View answer and explanation
Correct option: A
Statement I, II are correct
Option A (Statement I, II are correct) is correct because both the pH calculation and the ionic product of water are accurate.
50 Medium
50 of 401
Application +4 | -1
If 10 ml10\,ml of 0.4 m CH3COOHC{H_3}COOH solution is mixed with 40 mol40\,mol of 0.1 m NaOHNaOH solution then the pH{p^H} of the resultant solution becomes (given pka of CH3COOH=4.75{p^{ka}}\,of\,C{H_3}COOH = 4.75)
View answer and explanation
Correct option: B
8.8
Option B (8.8) is correct because it's the calculated pH of the buffer solution formed after mixing the weak acid and strong base.
51 Medium
51 of 401
Analyzing +4 | -1
For the reaction, N2    +  3H2  ⇋  2NH3;{N_2}\;\; + \;3{H_2}\;\leftrightharpoons\;2N{H_3}; in a vessel, after the addition of equal number of mole of N2{N_2} and H2{H_2}, equilibrium state is formed. Which of the following is correct?
View answer and explanation
Correct option: B
[H2]<[N2]\left[ {{\mathrm{H}_2}} \right] < \left[ {{\mathrm{N}_2}} \right]
Option B ([H₂] < [N₂]) is correct because hydrogen is consumed three times faster than nitrogen in the given reaction, leading to a lower concentration of hydrogen at equilibrium.
52 Medium
52 of 401
Application +4 | -1
Given KSP(Ag I)=8.5×10−17{K_{SP}}\left( {Ag\,I} \right) = 8.5 \times {10^{ - 17}}. The solubility of AgIAgI in 0.1M0.1 M KIKI solution is
View answer and explanation
Correct option: B
8.5×10−16 M8.5 \times {10^{ - 16}}\,M
Option B (8.5×10−16 M8.5 \times {10^{ - 16}}\,M) is correct because the solubility is calculated by considering the common ion effect of KI on the solubility of AgI.
53 Medium
53 of 401
Analyzing +4 | -1
Statement-I: In the reaction, I2+I−→I3−,I2I_2+\mathrm{I}^{-} \rightarrow \mathrm{I}_3^{-}, \mathrm{I}_2 acts as Lewis base.

Statement-II: In this reaction I' donates an electron pair for sharing with iodine
View answer and explanation
Correct option: D
Statement I is incorrect, Statement II is correct
Option D is correct because in the reaction, I2 accepts an electron pair from I- to form I3-, acting as a Lewis acid, not a Lewis base.
54 Medium
54 of 401
Application +4 | -1
The following reaction was allowed to come to equilibrium: A(g)+2 B(g)⇌C(g).\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{g}) \rightleftharpoons \mathrm{C}(\mathrm{g}). The initial amounts of reactants placed into a 5 L5 \mathrm{~L} in both vessels were 1 mol1 \mathrm{~mol} of A\mathrm{A} and 1.8 mol1.8 \mathrm{~mol} of B.\mathrm{B}. After the reaction reached equilibrium, 1 mol1 \mathrm{~mol} of B\text{B} was found. Calculate the value of KCK_C for this reaction.
View answer and explanation
Correct option: C
17
Option C (17) is correct because the equilibrium constant Kc is calculated based on the equilibrium concentrations of the reactants and products.
55 Medium
55 of 401
Understanding +4 | -1
Read the following statements carefully and choose the correct answer

(i) Water and water vapour remain in equilibrium position at atmospheric pressure ( 1.013 bar)1.013 \,{bar}) and at 100∘C100^{\circ} {C} in a closed vessel.

(ii) The boiling point of water is 100∘C100^{\circ} {C} at 1.013 bar pressure

(iii) Boiling point of the liquid depends on the atmospheric pressure.

(iv) Boiling point depends on the altitude of the place; at high altitude the boiling point increases.
View answer and explanation
Correct option: C
(i), (ii) and (iii) are correct
Option C ((i), (ii) and (iii) are correct) is correct because statements (i), (ii), and (iii) accurately describe the equilibrium of water and water vapor, the boiling point of water, and the dependence of boiling point on atmospheric pressure, while statement (iv) is incorrect.
56 Medium
56 of 401
Analyzing +4 | -1
Assertion :- The precipitate the cations of fourth group in qualitative analysis, medium is made alkaline before passing H2S\mathrm{H}_2 \mathrm{S} gas.

Reason :- This is done to suppress the ionization of H2S.\mathrm{H}_2 \mathrm{S}.
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
The correct answer is C (If Assertion is True but the Reason is False) because making the medium alkaline before passing H2S gas does help precipitate the cations of the fourth group, but it's done to increase the sulfide ion concentration, not to suppress the ionization of H2S.
57 Medium
57 of 401
Understanding +4 | -1
Which of the following solutions cannot act as a buffer?
View answer and explanation
Correct option: C
HCl+NH4Cl\mathrm{HCl} + \mathrm{N}{\mathrm{H}_4}\mathrm{Cl}
Option C (HCl+NH4Cl\mathrm{HCl} + \mathrm{N}{\mathrm{H}_4}\mathrm{Cl}) is correct because it consists of a strong acid (HCl) and its salt with a strong base (NH4ClNH_4Cl), which does not form a buffer solution.
58 Medium
58 of 401
Application +4 | -1
The [H+]\left[\mathrm{H}^{+}\right]of a resulting solution that is 0.01M0.01 \mathrm{M} acetic acid (Ka=1.8×10−5)\left(K_a=1.8 \times 10^{-5}\right) and 0.01M0.01 \mathrm{M} in benzoic acid (Ka=6.3×10−5)\left(K_a=6.3 \times 10^{-5}\right) :
View answer and explanation
Correct option: A
9×10−49 \times 10^{-4}
Option A (9×10−49 \times 10^{-4}) is correct because the total [H+] is the square root of the sum of the products of Ka and concentration for each acid.
59 Medium
59 of 401
Application +4 | -1
A2B\mathrm{A}_2 \mathrm{B} dissociates as A2B(s)⇌2 A(g)+B(g)\mathrm{A}_2 \mathrm{B}(\mathrm{s}) \rightleftharpoons 2 \mathrm{~A}(\mathrm{g})+\mathrm{B}(\mathrm{g}).

At equilibrium total pressure is PP, then KpK_p for the reaction is
View answer and explanation
Correct option: C
427P3\frac{4}{27} P^3
Option C (427P3\frac{4}{27} P^3) is correct because the KpK_p is calculated using the partial pressures of the gaseous products at equilibrium.
60 Medium
60 of 401
Analyzing +4 | -1
(A): The addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate AgBr{AgBr} rather than AgCl{AgCl}

(R): The value of KspK_{s p} of AgCl<Ksp{AgCl} < K_{s p} of AgBr{AgBr}
View answer and explanation
Correct option: C
(A) is true but (R) is false
Option C ((A) is true but (R) is false) is correct because AgBrAgBr precipitates first due to its lower solubility product (KspK_{sp}), meaning KspK_{sp} of AgClAgCl is greater than AgBrAgBr.
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