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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Easy

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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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91 Easy
91 of 161
Understanding +4 | -1
Tribasic acid furnishes…type of anions.
View answer and explanation
Correct option: C
3
Option C (3) is correct because a tribasic acid like H3PO4 can furnish three types of anions by losing one, two, or three protons.
92 Easy
92 of 161
Remembering +4 | -1
The relationship between KP  and  KC{K_P}\mathrm{ \,\, and\,\, }{K_C} is given by
View answer and explanation
Correct option: B
KP=KC(RT)Δn{K_P} = {K_C}{(RT)^{\Delta n}}
Option B (KP=KC(RT)Δn{K_P} = {K_C}{(RT)^{\Delta n}}) is correct because this equation correctly relates the equilibrium constants Kp and Kc, where Δn is the change in the number of moles of gas.
93 Easy
93 of 161
Understanding +4 | -1
Equilibrium constants for some reactions are given. In which of the following case does the reaction go farthest to completion?
View answer and explanation
Correct option: A
K=102K=10^2
Option A (K=102K=10^2) is correct because a larger equilibrium constant (K) indicates that the reaction favors product formation and thus goes further to completion.
94 Easy
94 of 161
Application +4 | -1
The hydronium ion concentration in an aqueous solution of H2SO4\mathrm{H}_2 \mathrm{SO}_4 is 2.0×10−4M2.0 \times 10^{-4} \mathrm{M} at 25∘C.25^{\circ} \mathrm{C}. The hydroxide ion concentration in the solution is
View answer and explanation
Correct option: D
5×10−11M5 \times 10^{-11} \mathrm{M}
Option D (5×10−11M5 \times 10^{-11} \mathrm{M}) is correct because the hydroxide ion concentration is calculated using the ion product of water (KwK_w) and the given hydronium ion concentration.
95 Easy
95 of 161
Application +4 | -1
Calculate the  pOH\;\mathrm{pOH} of a solution at 25∘C25^\circ C that contains 1×10−10  M1 \times {10^{ - 10}}\;M of hydronium ions i.e.,    H3+Oi.e.,\;\;\mathrm{H}_3^ + \mathrm{O}:
View answer and explanation
Correct option: A
4
Option A (4) is correct because the pOH is calculated from the pH, which is derived from the hydronium ion concentration using the formula pH + pOH = 14.
96 Easy
96 of 161
Understanding +4 | -1
Statement-I: Acids and bases are conductors of electricity in aqueous solutions

Statement-II: Acids and bases ionise in water
View answer and explanation
Correct option: A
Statement I, II are correct
Option A (Statement I, II are correct) is correct because acids and bases conduct electricity due to ionization in water.
97 Easy
97 of 161
Analyzing +4 | -1
Consider the following reaction at equilibrium
N2(g)+3H2(g)⇌2NH3(g){N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g)
if small amount of Helium gas is added to the reaction mixture at constant pressure then
View answer and explanation
Correct option: B
Dissociation of NH3N{H_3} increases
The correct answer is B (Dissociation of NH3N{H_3} increases) because adding an inert gas at constant pressure increases the volume, shifting the equilibrium towards the side with more moles of gas.
98 Easy
98 of 161
Understanding +4 | -1
The oxo-acid of anhydride P2O5{\mathrm{P}_2}{\mathrm{O}_5} is:
View answer and explanation
Correct option: D
All of these
The correct answer is D (All of these) because P2O5{\mathrm{P}_2}{\mathrm{O}_5} can form H3PO4{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_4}, H4P2O7{\mathrm{H}_4}{\mathrm{P}_2}{\mathrm{O}_7}, and HPO3\mathrm{HP}{\mathrm{O}_3} upon hydration, depending on the conditions.
99 Easy
99 of 161
Analyzing +4 | -1
Strongest acid is
View answer and explanation
Correct option: B
CH3OH\mathrm{C}{\mathrm{H}_3}\mathrm{OH}
Option B (CH3OH) is the strongest acid because methanol is more acidic than the other hydrocarbons due to the electronegativity of oxygen.
100 Easy
100 of 161
Understanding +4 | -1
For the reaction, C2H4(g)+H2(g)⇋C2H6(g){C_2}{H_4}(g) + {H_2}(g)\underset{{}}{\overset{{}}{\leftrightharpoons}}{C_2}{H_6}\left( g \right) the equilibrium constant has the units
View answer and explanation
Correct option: B
dm3  mol−1  d{m^3}\,\,mo{l^{ - 1}}\,\,
Option B (dm3 mol−1, d{m^3}\\ \,mo{l^{ - 1}}\\ ,\,) is correct because the units of the equilibrium constant are derived from the change in the number of moles in the reaction.
101 Easy
101 of 161
Application +4 | -1
At 600 , Kp600\,,\,{K_p} for the following reaction is 1 atm.

X(g)   ⇋   Y(g)+Z(g)X\left( \mathrm{g} \right)\,\,\,\leftrightharpoons\,\,\,Y\left( \mathrm{g} \right) + Z\left( \mathrm{g} \right)

At equilibrium, 50 %50\,\% of X(g)X\left( \mathrm{g} \right) is dissociated. The total pressure of the equilibrium system is pp atm. What is the partial pressure (in atm) of X(g)X\left( \mathrm{g} \right) at equilibrium?
View answer and explanation
Correct option: A
1
Option A (1) is correct because the partial pressure of X at equilibrium is calculated to be 1 atm using the given Kp and dissociation percentage.
102 Easy
102 of 161
Understanding +4 | -1
The conjugate base of H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} in the following reaction is:

H2SO4+H2O  ⇋  H3O++HSO4−{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} + {\mathrm{H}_2}\mathrm{O}\;\leftrightharpoons\;{\mathrm{H}_3}{\mathrm{O}^ + } + \mathrm{HSO}_4^ -
View answer and explanation
Correct option: B
HSO4−\mathrm{HSO}_4^ -
Option B (HSO4−\mathrm{HSO}_4^ -) is the correct conjugate base because it is formed when H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} donates a proton.
103 Easy
103 of 161
Application +4 | -1
The pH{pH} of a 10−3 M HCl10^{-3} \,{M} \,{HCl} solution at 25∘C25^{\circ} {C} if it is diluted 1000 times, will be-
View answer and explanation
Correct option: C
5.98
Option C (5.98) is correct because after dilution, the concentration of HCl becomes very low, and the contribution of water's autoionization to the [H+] cannot be neglected. The pH is calculated considering both HCl and water's contribution.
104 Easy
104 of 161
Analyzing +4 | -1
Which of the following has concentration dependent pH\mathrm{pH} ?
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Correct option: A
CH3COONa\mathrm{CH}_3 \mathrm{COONa}
Option A (CH3COONa\mathrm{CH}_3 \mathrm{COONa}) is correct because it's a salt of a weak acid and a strong base, leading to concentration-dependent hydrolysis and pH.
105 Easy
105 of 161
Remembering +4 | -1
The indicator used in the titration of sodium carbonate with sulphuric acid is
View answer and explanation
Correct option: C
Methyl orange
Option C (Methyl orange) is correct because methyl orange is a suitable indicator for titrations involving a strong acid and a weak base.
106 Easy
106 of 161
Understanding +4 | -1
According to Le-Chatelier principle, adding heat a solid and liquid in equilibrium will cause the
View answer and explanation
Correct option: A
Amount of solid to decrease
Option A (Amount of solid to decrease) is correct because the fusion of a solid is an endothermic process, and adding heat favors the forward reaction (solid to liquid), thus decreasing the amount of solid.
107 Easy
107 of 161
Application +4 | -1
For the reaction, N2(g)+3H2(g) ⇋ 2NH3(g), {N_2}(g) + 3{H_2}(g)\,\leftrightharpoons\,2N{H_3}(g),\, If the value of equilibrium constant is K,K, then the equilibrium constant for the reaction, NH3(g) ⇋ 12N2(g)+32H2(g) N{H_3}(g)\,\leftrightharpoons\,\frac{1}{2}{N_2}(g) + \frac{3}{2}{H_2}(g)\, will be
View answer and explanation
Correct option: B
1K\frac{1}{{\sqrt K }}
Option B (1K\frac{1}{{\sqrt K }}) is correct because reversing the reaction and halving the stoichiometric coefficients results in taking the reciprocal of the square root of the original equilibrium constant.
108 Easy
108 of 161
Application +4 | -1
In the reaction, N2O4  ⇋  2NO2{\mathrm{N}_2}{\mathrm{O}_4}\;\leftrightharpoons\;2\mathrm{N}{\mathrm{O}_2}, α\mathrm{\alpha } is that part of N2O4{\mathrm{N}_2}{\mathrm{O}_4} which dissociates, then the number of moles at equilibrium will be
View answer and explanation
Correct option: C
(1+α)\left( {1 + \mathrm{\alpha }} \right)
The correct answer is C ((1+α)\left( {1 + \mathrm{\alpha }} \right)) because it accurately calculates the total number of moles at equilibrium based on the dissociation of N2O4.
109 Easy
109 of 161
Application +4 | -1
For the reaction, PCl3(g)+Cl2(g)PCl5(g)PC{l_3}(g) + C{l_2}(g)PC{l_5}(g) at 2500C{250^0}C, the value of KC{K_C} is 26, then the value of KP{K_P} at the same temperature will be
View answer and explanation
Correct option: A
0.61
Option A (0.61) is correct because it's the calculated Kp value using the given Kc and the relationship Kp = Kc(RT)^Δn.
110 Easy
110 of 161
Remembering +4 | -1
Tetramine copper (II) sulphate is an example of:
View answer and explanation
Correct option: C
Complex salt
Option C (Complex salt) is correct because tetramine copper(II) sulphate, [Cu(NH3)4]SO4, contains a complex ion, [Cu(NH3)4]2+, which does not dissociate into simple ions in solution.
111 Easy
111 of 161
Analyzing +4 | -1
Aqueous solution of which salt has the lowest pHpH?
View answer and explanation
Correct option: B
NH4Cl\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}
Option B (NH4Cl\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}) has the lowest pH because it is an acidic salt due to the hydrolysis of NH4+\mathrm{NH}_4^+.
112 Easy
112 of 161
Analyzing +4 | -1
Statement I: For the reaction A(g)⇌B(g),\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g}), equilibrium moles of A\text{A} and B\text{B} are, respectively, 'aa' and ' bb ' in 1 L1 \mathrm{~L} container. If 5 moles of A\mathrm{A} and 3 moles of B\mathrm{B} are added, then reaction must move in forward direction.

Statement II: Even if the amount of reactant added to a system at equilibrium is more than the amount of product added at the same time, the equilibrium can shift in any direction.
View answer and explanation
Correct option: D
If Statement I is INCORRECT, but Statement II is CORRECT
The correct answer is D (If Statement I is INCORRECT, but Statement II is CORRECT) because the direction of equilibrium shift depends on the relative values of a and b, and Statement II correctly states that the equilibrium can shift in any direction.
113 Easy
113 of 161
Application +4 | -1
What is the pH\mathrm{pH} of the resulting solution when equal volumes of 0.1 m NaOH and 0.01 M HCl0.1\,m\,NaOH\,and\,0.01\,M\,HCl are mixed?
View answer and explanation
Correct option: C
12.65\mathrm{12}\mathrm{.65}
Option C (12.65) is correct because after mixing equal volumes of NaOH and HCl, the resulting solution is basic, and the pH is calculated based on the excess hydroxide ions.
114 Easy
114 of 161
Understanding +4 | -1
For the hypothetical reaction, A(g)⇌B(g)+C(g)A(g) \rightleftharpoons B(g)+C(g) the correct relation between Kp{Kp} and Kc{Kc} at 12.18 Kelvin12.18 \mathrm{~Kelvin} is
View answer and explanation
Correct option: A
Kp=Kc{Kp}={Kc}
Option A (Kp=Kc{Kp}={Kc}) is correct because the change in the number of moles of gas (Δng\Delta n_g) is zero, making Kp{Kp} equal to Kc{Kc}.
115 Easy
115 of 161
Application +4 | -1
The pH\mathrm{pH} of M100Ca(OH)2\frac{\mathrm{M}}{100} \mathrm{Ca}(\mathrm{OH})_2 is
View answer and explanation
Correct option: C
12.30112.301
Option C (12.30112.301) is correct because the pOH is calculated from the hydroxide concentration, and then pH is found using pH + pOH = 14.
116 Easy
116 of 161
Application +4 | -1
The pH\mathrm{pH} of 0.1  M  HCl0.1\;M\;\mathrm{HCl} is approximately 1. The approximate pH\mathrm{pH} of 0.05  M  H2SO40.05\;M\;{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} is:
View answer and explanation
Correct option: C
1
Option C (1) is correct because the normality of H2SO4 is 0.1, resulting in a [H+] concentration of 0.1 and a pH of 1.
117 Easy
117 of 161
Understanding +4 | -1
For the reaction: NH4HS(s)⇌NH3(g)+H2S(g)\mathrm{NH}_4 \mathrm{HS{(s})} \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_2 \mathrm{S{(g})}
The unit of Kp\mathrm{Kp} for the reactions is
View answer and explanation
Correct option: C
atm2\mathrm{atm}^{2}
The correct answer is C (atm²) because the unit of Kp is (atm)^Δng, and Δng = 2 for this reaction.
118 Easy
118 of 161
Understanding +4 | -1
The KC{K_C} for H2(g)+I2(g)2HI(g){H_2}(g) + {I_2}(g)2HI(g) is 64, if the volume of the container is reduced to one-half of its
original volume, the value of equilibrium constant will be
View answer and explanation
Correct option: B
64
Option B (64) is correct because the equilibrium constant, Kc, is independent of volume changes for the given reaction.
119 Easy
119 of 161
Remembering +4 | -1
The Henderson’s equation used to calculate the pOH\mathrm{pOH} of basic buffer is:
View answer and explanation
Correct option: A
pOH=pKb+log⁡[Conjugate  acid][  Base]\mathrm{pOH} = \mathrm{p}{K_b} + \log \frac{{\left[ {\mathrm{Conjugate\; acid}} \right]}}{{\left[ {\mathrm{\;Base}} \right]}}
Option A is correct. The Henderson-Hasselbalch equation for a basic buffer is pOH=pKb+log⁡[Conjugate  acid][  Base]\mathrm{pOH} = \mathrm{p}{K_b} + \log \frac{{\left[ {\mathrm{Conjugate\; acid}} \right]}}{{\left[ {\mathrm{\;Base}} \right]}}
120 Easy
120 of 161
Understanding +4 | -1
Which addition would not change the pH\mathrm{pH} of 10 mL10\,\mathrm{mL} of dilute hydrochloric acid?
View answer and explanation
Correct option: A
20 mL20\,\mathrm{mL} of the same dilute hydrochloric acid
Option A (20 mL20\,\mathrm{mL} of the same dilute hydrochloric acid) is correct because adding more of the same concentration of HCl doesn't change the pH.
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