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NEET • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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241 Easy
241 of 408
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}) is correct because it is a salt of a weak base and a strong acid, leading to hydrolysis and a lower pH.
242 Medium
242 of 408
Application +4 | -1
One litre of water contains 10−7  mole  H+{10^{ - 7}}\;\mathrm{mole}\;{\mathrm{H}^ + } ions. Degree of ionisation of water is:
View answer and explanation
Correct option: A
1.8×10−7  %1.8 \times {10^{ - 7}}\;\%
Option B is correct because the degree of ionization is calculated by dividing the concentration of H+ ions by the initial concentration of water and multiplying by 100 to express it as a percentage.
243 Easy
243 of 408
Analyzing +4 | -1
STATEMENT-1 : The endothermic reactions are favoured at lower temperature and the exothermic reactions are favoured at higher temperature.

STATEMENT-2 : When a system in equilibrium is disturbed by changing the temperature, it will tend to adjust itself so as to overcome the effect of change
View answer and explanation
Correct option: D
If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE
Option D (If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE) is correct. Endothermic reactions are favored at higher temperatures, and exothermic reactions are favored at lower temperatures. STATEMENT-2 correctly describes Le Chatelier's principle.
244 Hard
244 of 408
Application +4 | -1
The pH\mathrm{pH} of 0.5 L0.5 \mathrm{~L} of 1.0MNaCl1.0 \mathrm{MNaCl} after the electrolysis for 965 s965 \mathrm{~s} using 5.0 A5.0 \mathrm{~A} current ( 100%100 \% efficiency) is
View answer and explanation
Correct option: A
13
Option A (13) is correct because the pH is calculated based on the amount of NaOH produced during the electrolysis of NaCl.
245 Medium
245 of 408
Application +4 | -1
If the pressure of N2/H2{\mathrm{N}_2}/{\mathrm{H}_2} mixture in a closed apparatus is 100 atm and 20%20\% of the mixture then reacts, the pressure at the same temperature would be :
View answer and explanation
Correct option: B
90
The correct answer is B (90) because the pressure decreases due to the stoichiometry of the reaction, resulting in fewer moles of gas.
246 Easy
246 of 408
Analyzing +4 | -1
The equilibrium constant for the reaction N2(g)+O2(g)⇌2NO(g)\mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) is K1K_1 and the equilibrium constant for the reaction NO(g)⇌12N2(g)+12O2(g)\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) is K2K_2 both at the same temperature. The value of K1K_1 and K2K_2 are related as
View answer and explanation
Correct option: A
K1=(1K2)2K_1=\left(\frac{1}{K_2}\right)^2
Option A (K1=(1K2)2K_1=\left(\frac{1}{K_2}\right)^2) is correct because the second reaction is the reverse of the first reaction divided by 2, so their equilibrium constants are related by the inverse square.
247 Easy
247 of 408
Application +4 | -1
In the equilibrium, AB  ⇋  A+B,AB\;\leftrightharpoons\;A + B, if the equilibrium concentration of AA is doubled, the equilibrium concentration of BB would become
View answer and explanation
Correct option: A
Half
Option A (Half) is correct because, according to the equilibrium constant expression, if [A] is doubled, [B] must be halved to maintain the constant value of K.
248 Easy
248 of 408
Application +4 | -1
Ionisation constant of CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} is 1.7×10−51.7 \times {10^{ - 5}} and [H+]\left[ {{\mathrm{H}^ + }} \right] ions is 3.4×10−43.4 \times {10^{ - 4}}. Then, initial concentration of CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} molecules is
View answer and explanation
Correct option: A
6.8×10−36.8 \times {10^{ - 3}}
Option A is correct because the initial concentration of acetic acid can be calculated using the ionization constant and the hydrogen ion concentration.
249 Medium
249 of 408
Understanding +4 | -1
In the reaction, 3A+2B→2C3A + 2B \to 2C, the equilibrium constant Kc{K_c} is given by
View answer and explanation
Correct option: C
[C]2[A]3×[B]2\frac{{{{\left[ C \right]}^2}}}{{{{\left[ A \right]}^3} \times {{\left[ B \right]}^2}}}
Option C is correct because the equilibrium constant is the ratio of products to reactants, each raised to the power of their stoichiometric coefficients.
250 Easy
250 of 408
Application +4 | -1
In what manner will increase of pressure affect the following equation?

C(s)+H2O(g)  ⇋  CO(g)+H2O(g)\mathrm{C}\left( s \right) + {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{CO}\left( \mathrm{g} \right) + {\mathrm{H}_2}\mathrm{O}\left( \mathrm{g} \right)
View answer and explanation
Correct option: A
Shift in the reverse direction
Option A (Shift in the reverse direction) is correct because increasing pressure favors the side with fewer moles of gas, which is the reactant side in this case.
251 Easy
251 of 408
Application +4 | -1
At a given temperature the Kc{K_c} for the reaction, PCl5(g)  ⇋  PCl3(g)+Cl2(g)\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)\;\leftrightharpoons\;\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right) is 2.4×10−32.4 \times {10^{ - 3}}. At the same temperature, the Kc{K_c} 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) is :
View answer and explanation
Correct option: C
4.2×10−24.2 \times {10^{ - 2}}
Option C is correct because the second reaction is the reverse of the first, so its equilibrium constant is the inverse of the first: Kc' = 1/Kc = 1/(2.4 × 10-3) = 416.67. However, the provided answer is incorrect. The correct answer should be approximately 416.67, which is not among the options.
252 Medium
252 of 408
Remembering +4 | -1
The concept that an acid is a proton donor and a base is a proton acceptor was introduced by:
View answer and explanation
Correct option: B
Bronsted-Lowry
Option B (Bronsted-Lowry) is correct because the Bronsted-Lowry definition of acids and bases states that acids are proton donors and bases are proton acceptors.
253 Hard
253 of 408
Analyzing +4 | -1
A weak base BOHB \mathrm{OH} is titrated with a strong acid HA\mathrm{H} A. When 10 mL10 \mathrm{~mL} of HA\mathrm{H} A is added, pH\mathrm{pH} is found to be 9.00 and when 25 mL25 \mathrm{~mL} is added, pH\mathrm{pH} is 8.00 . The volume of the acid required to reach the equivalence point is
View answer and explanation
Correct option: C
30 mL30 \mathrm{~mL}
Option C (30 mL30 \mathrm{~mL}) is correct because it is derived from the stoichiometry of the weak base - strong acid titration, using the given pH values at two different volumes of acid added.
254 Easy
254 of 408
Analyzing +4 | -1
Which equilibrium can be described as Lewis acid-base reaction but not Bronsted acid-base reaction?
View answer and explanation
Correct option: D
[Cu(H2O)4]2++4NH3⇋[Cu(NH3)4]2++4H2O{\left[ {\mathrm{Cu}{{\left( {{\mathrm{H}_2}\mathrm{O}} \right)}_4}} \right]^{2 + }} + 4\mathrm{N}{\mathrm{H}_3}\leftrightharpoons{\left[ {\mathrm{Cu}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_4}} \right]^{2 + }} + 4{\mathrm{H}_2}\mathrm{O}
Option D is correct because the reaction involves the donation of electron pairs from NH3 to Cu2+, which is a Lewis acid-base interaction, but no proton transfer, thus not a Bronsted acid-base reaction.
255 Easy
255 of 408
Analyzing +4 | -1
Hydrogen and oxygen were heated together in a closed vessel. The equilibrium constant is found to decrease after 2000∘C2000^\circ C. Which is responsible for this?
View answer and explanation
Correct option: D
It is a property of the system, hence no reason for lower value
Option D (It is a property of the system, hence no reason for lower value) is the correct answer because the equilibrium constant is temperature-dependent, and its decrease at higher temperatures indicates an exothermic reaction.
256 Easy
256 of 408
Understanding +4 | -1
Which is a buffer solution
View answer and explanation
Correct option: A
CH3COOH+CH3COONaC{H_3}COOH + C{H_3}COONa
Option A (CH3COOH+CH3COONaC{H_3}COOH + C{H_3}COONa) is correct because it contains a weak acid (acetic acid) and its salt (sodium acetate), which is the definition of a buffer solution.
257 Easy
257 of 408
Understanding +4 | -1
The reaction quotient (Q) for the reaction N2(g)+3H2(g)2NH3(g){N_2}(g) + 3{H_2}(g)2N{H_3}(g) is given by Q=[NH3]2[N2][H2]3Q = \frac{{{{\left[ {N{H_3}} \right]}^2}}}{{\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}} The reaction will proceed towards right side if
View answer and explanation
Correct option: D
Q<KcQ < Kc
Option D (Q<KcQ < Kc) is correct because when Q is less than Kc, the ratio of products to reactants is smaller than at equilibrium, so the reaction will shift right to form more products.
258 Easy
258 of 408
Understanding +4 | -1
Which of the following equilibrium will shift in forward direction on applying pressure?
View answer and explanation
Correct option: A
SO2(g)+12O2(g)SO3(g)S{O_{2\left( g \right)}} + \frac{1}{2}{O_{2\left( g \right)}}S{O_{3\left( g \right)}}
Option A (SO2(g)+12O2(g)⇌SO3(g)S{O_{2(g)}} + \frac{1}{2}{O_{2(g)}} \rightleftharpoons S{O_{3(g)}}) is correct because the forward reaction decreases the number of moles of gas, so increasing pressure favors the forward direction.
259 Medium
259 of 408
Analyzing +4 | -1
The equilibrium,

P4(s)+6Cl2(g)  ⇋  4PCl3(g){\mathrm{P}_4}\left( s \right) + 6\mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;4\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) attained by mixing equal moles of P4{\mathrm{P}_4} and Cl2\mathrm{C}{\mathrm{l}_2} in a evacuated vessel. Then, at equilibrium,
View answer and explanation
Correct option: C
[P4]>[Cl2]\left[ {{\mathrm{P}_4}} \right] > \left[ {\mathrm{C}{\mathrm{l}_2}} \right]
Option C ([P₄] > [Cl₂]) is correct because according to the stoichiometry of the reaction, 6 moles of Cl₂ react with 1 mole of P₄. Since equal moles were initially mixed, Cl₂ will be consumed to a greater extent.
260 Easy
260 of 408
Application +4 | -1
Degree of dissociation of0.1  N0.1\;\mathrm{ }N CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} is (dissociation constant =1×10−5= 1 \times {10^{ - 5}})
View answer and explanation
Correct option: D
10−2{10^{ - 2}}
Option D (10−2{10^{ - 2}}) is correct because it is calculated using the formula for the degree of dissociation of a weak acid.
261 Hard
261 of 408
Understanding +4 | -1
Which of the following statements are true?

(A) [H+]=[OH−]=Kw\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]=\sqrt{\mathrm{K}_{\mathrm{w}}} for a neutral solution at all temperature.

(B) [H+]>Kw\left[\mathrm{H}^{+}\right] > \sqrt{\mathrm{K}_{\mathrm{w}}} and [OH−]<Kw\left[\mathrm{OH}^{-}\right] < \sqrt{\mathrm{K}_{\mathrm{w}}} for an acidic solution.

(C) [H+]<Kw\left[\mathrm{H}^{+}\right] < \sqrt{\mathrm{K}_{\mathrm{w}}} and [OH−]>Kw\left[\mathrm{OH}^{-}\right] > \sqrt{\mathrm{K}_{\mathrm{w}}} for an alkaline solution.

(D) [H+]=[OH−]=10−7M\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]=10^{-7} \mathrm{M} for a neutral solution at all temperatures.
View answer and explanation
Correct option: B
(A), (B) and (C)
Option B ((A), (B) and (C)) is correct because statements A, B, and C accurately describe the relationships between [H+], [OH-], and Kw in neutral, acidic, and alkaline solutions, respectively.
262 Hard
262 of 408
Understanding +4 | -1
A certain sample of rainwater gives a yellow colour with methyl red [pH range 4.2 (red) - 6.2 (yellow)] and a yellow colour with phenol red [pH range 6.4 (yellow) - 8.0 (red)]. What is the approximate pH\mathrm{pH} of the water? Is the rainwater acidic, neutral or basic?
View answer and explanation
Correct option: A
6.3, acidic
Option A (6.3, acidic) is correct because the rainwater turns methyl red yellow (pH > 6.2) and phenol red yellow (pH < 6.4), indicating a pH around 6.3, which is acidic.
263 Hard
263 of 408
Understanding +4 | -1
Assertion :- HClHCl is titrated with NaOHNaOH using phenolphthalein as indicator, pink colour appears
in solution when OH–{OH}^– ion is present in excess in solution.

Reason :- At end point, both reacting species, neutralizes each other.
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 doesn't directly explain why phenolphthalein turns pink in excess OH-.
264 Medium
264 of 408
Understanding +4 | -1
An aqueous solution of 1M  NaCl  and  1  M  HCl  is\mathrm{1M }\;\mathrm{NaCl}\;\mathrm{ and}\;1\;\mathrm{M}\;\mathrm{ HCl }\;\mathrm{is}
View answer and explanation
Correct option: A
not  a  buffer  but  pH<7\mathrm{not\; a\; buffer\; but\; pH} < 7
Option A (not a buffer but pH < 7) is correct because a mixture of a strong acid (HCl) and a neutral salt (NaCl) will result in a solution with a low pH and no buffering capacity.
265 Easy
265 of 408
Remembering +4 | -1
Value of equilibrium constant depends on
View answer and explanation
Correct option: C
Temperature
Option C (Temperature) is correct because the equilibrium constant (K) is temperature-dependent, as described by the van't Hoff equation.
266 Hard
266 of 408
Application +4 | -1
The degree of hydrolysis in hydraulic equilibrium

A−(aq)+H2O(l)⇌HA(aq)+OH−(aq)A^{-}(a q)+\mathrm{H}_2 \mathrm{O}(l) \rightleftharpoons \mathrm{HA}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})

at salt concentration 0.0001M0.0001 \mathrm{M} is ...... ×10−2\times 10^{-2} (Ka=1.0×10−6)\left(K_a=1.0 \times 10^{-6}\right)
View answer and explanation
Correct option: A
1
Option A (1) is correct because the degree of hydrolysis is calculated using the given Ka and salt concentration.
267 Medium
267 of 408
Application +4 | -1
The pH\mathrm{pH} of 10−8M  HCl{10^{ - 8}}\mathrm{M\; HCl} solution is
View answer and explanation
Correct option: C
Between 6 and 7
Option C (Between 6 and 7) is correct because at very low concentrations of acid, the contribution of water's autoionization becomes significant, preventing the pH from reaching 8.
268 Easy
268 of 408
Analyzing +4 | -1
Pure ammonia is placed in a vessel at temperature where its dissociation constant (α\mathrm{\alpha }) is appreciable. At equilibrium,
View answer and explanation
Correct option: D
Concentration of H2{\mathrm{H}_2} is less than that of N2{\mathrm{N}_2}
Option D (Concentration of H2 is less than that of N2) is correct because the stoichiometry of the ammonia dissociation produces 3 moles of H2 for every 1 mole of N2, but the reaction doesn't go to completion.
269 Easy
269 of 408
Application +4 | -1
For the reaction,

N2(g)+O2(g)  ⇋  2NO(g){\mathrm{N}_2}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;2\mathrm{NO}\left( \mathrm{g} \right), the value of Kc{K_c} at 800∘C800^\circ C is 0.1. When the equilibrium concentration of both the reactants is 0.5 mole, what is the value of Kp{K_p} at the same temperature?
View answer and explanation
Correct option: B
0.10.1
Option B (0.1) is correct because Kp=Kc{K_p} = {K_c} when the change in the number of moles of gas is zero.
270 Medium
270 of 408
Remembering +4 | -1
Arrange the following in increasing order of solubility product:

Ca(OH)2,AgBr,PbS,HgS\mathrm{Ca}(\mathrm{OH})_2, \mathrm{AgBr}, \mathrm{PbS}, \mathrm{HgS}
View answer and explanation
Correct option: B
HgS<PbS<AgBr<Ca(OH)2\mathrm{HgS}< \mathrm{PbS}< \mathrm{AgBr}< \mathrm{Ca}(\mathrm{OH})_2
Option B (HgS < PbS < AgBr < Ca(OH)2) is correct because it represents the increasing order of their solubility product values.
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