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NEET • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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61 Hard
61 of 408
Understanding +4 | -1
STATEMENT I : A solution containing mixture of acetic acid and sodium acetate resist pH\mathrm{pH} change on addition of small amount of acid or base.

STATEMENT II : A solution containing a mixture of acetic acid sodium acetate acts as buffer solution around pH.4.74
View answer and explanation
Correct option: A
Statement I, II are correct
Option A (Statement I, II are correct) is correct because a mixture of a weak acid (acetic acid) and its conjugate base (sodium acetate) forms a buffer solution that resists pH changes, and its buffering range is around its pKa value, which is approximately 4.74.
62 Hard
62 of 408
Understanding +4 | -1
(A): BaCO3\mathrm{BaCO}_3 is more soluble in HNO3\mathrm{HNO}_3 than in plain water.

(R): Carbonate is a weak base and reacts with H+\mathrm{H}^{+}from the strong acid, causing the barium salt to dissociate.
View answer and explanation
Correct option: A
Both (A) and (R) are true and (R) is the correct explanation of (A)
Option A (Both (A) and (R) are true and (R) is the correct explanation of (A)) is correct because the reaction of carbonate with H+ drives the dissolution of BaCO3.
63 Hard
63 of 408
Application +4 | -1
A+B ⇋  C+DA + B\,\leftrightharpoons\,\,C + D

Initially moles of A    and    BA\;\;\mathrm{and}\;\;B are equal. At equilibrium, moles of CC are three times that of A. the equilibrium constant of the reaction will be
View answer and explanation
Correct option: D
9
Option D (9) is correct because the equilibrium constant is calculated based on the changes in moles of reactants and products.
64 Easy
64 of 408
Analyzing +4 | -1
Which of the following statements about pH{pH} and H+{H}^{+}ion concentration is incorrect?
View answer and explanation
Correct option: B
A solution of the mixture of one equivalent of each of CH3COOH{CH}_3 {COOH} and NaOH{NaOH} has a pH{pH} of 7
Option B ( A solution of the mixture of one equivalent of each of CH3COOH{CH}_3 {COOH} and NaOH{NaOH} has a pH{pH} of 7) is correct because the resulting solution is a buffer, not a neutral solution with pH 7.
65 Hard
65 of 408
Understanding +4 | -1
Equilibrium constant K{K} is related to Ecell 0{E}_{\text {cell }}^0 and not Ecell {E}_{\text {cell }} because
View answer and explanation
Correct option: B
Ecell {E}_{\text {cell }} becomes zero at equilibrium point but Ecell 0{E}_{\text {cell }}^0 remains constant under all conditions
Option B is correct because Ecell becomes zero at equilibrium, while E°cell remains constant and is used to calculate the equilibrium constant.
66 Medium
66 of 408
Analyzing +4 | -1
An aqueous solution of hydrogen sulphide shows the equilibrium,

H2S⇋H++HS−{\mathrm{H}_2}\mathrm{S}\leftrightharpoons{\mathrm{H}^ + } + \mathrm{H}{\mathrm{S}^ - }

If dilute hydrochloric acid is added to an aqueous solution of hydrogen sulphide without any change in temperature, then:
View answer and explanation
Correct option: D
The concentration of HS−\mathrm{H}{\mathrm{S}^ - } will decreases
Option D is correct because adding HCl increases the concentration of H+, which shifts the equilibrium to the left, decreasing the concentration of HS- according to Le Chatelier's principle.
67 Hard
67 of 408
Understanding +4 | -1
A solute undergoes complex formation with ions of sparingly soluble salt, the solubility of salt:
View answer and explanation
Correct option: A
Increases
Option A (Increases) is correct because complex formation removes ions from the solution, shifting the equilibrium towards dissolution and increasing solubility.
68 Medium
68 of 408
Analyzing +4 | -1
The correct order of increasing [H3O+]\left[ {{\mathrm{H}_3}{\mathrm{O}^ + }} \right] in the following aqueous solutions is:
View answer and explanation
Correct option: C
0.01  M  NaNO2<0.01  M  NaCl<0.01  M  H2S<0.01  M  H2SO40.01\;M\;\mathrm{NaN}{\mathrm{O}_2} < 0.01\;M\;\mathrm{NaCl} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}
Option C (0.01  M  NaNO2<0.01  M  NaCl<0.01  M  H2S<0.01  M  H2SO40.01\;M\;\mathrm{NaN}{\mathrm{O}_2} < 0.01\;M\;\mathrm{NaCl} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}) is correct because it correctly orders the solutions from least to greatest acidity based on their chemical properties.
69 Medium
69 of 408
Analyzing +4 | -1
Which of the following has highest pH?
View answer and explanation
Correct option: D
M4Ca(OH)2\frac{\mathrm{M}}{4}\mathrm{Ca}{\left( {\mathrm{OH}} \right)_2}
Option D (M4Ca(OH)2\frac{\mathrm{M}}{4}\mathrm{Ca}{\left( {\mathrm{OH}} \right)_2}) is correct because it provides the highest hydroxide ion concentration due to being a strong base and having two hydroxide ions per formula unit.
70 Easy
70 of 408
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 contains a complex ion, specifically the tetramine copper (II) ion, making it a complex salt.
71 Medium
71 of 408
Understanding +4 | -1
For the reactions, A+B+Q  ⇋  C+DA + B + Q\;\leftrightharpoons\;C + D&nbsp; if the temperature is increased then concentration of the products will
View answer and explanation
Correct option: A
Increase
Option A is correct because increasing the temperature favors the endothermic reaction, which in this case, is the forward reaction, thus increasing the concentration of products.
72 Hard
72 of 408
Application +4 | -1
The standard Gibbs energy change at 300 K for the reaction, 2A⇋B+C2A\underset{{}}{\overset{{}}{\leftrightharpoons}}B + C is 2494.2J.2494.2 J. At a given time, the composition of the reaction mixture is [A]=12,   [B]=2     and    [C]=12[A] = \frac{1}{2},\,\,\,[B] = 2\,\,\,\,\,\mathrm{and}\,\,\,\,[C] = \frac{1}{2} The reaction proceeds in the [R = 8.314 JK/mole, e = 2.718]
View answer and explanation
Correct option: B
Reverse direction because Q>KCQ > {K_C}
Option B (Reverse direction because Q>KCQ > {K_C}) is correct because the reaction quotient (Q) is greater than the equilibrium constant (Kc), indicating that the reaction will proceed in the reverse direction to reach equilibrium.
73 Medium
73 of 408
Understanding +4 | -1
The equilibrium constant for the reaction,N2(g)+O2(g)⇋2NO(g){\mathrm{N}_2}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{NO}\left( \mathrm{g} \right) is 4×10−44 \times {10^{ - 4}} at 2000 K. In presence of a catalyst the equilibrium is attained ten times faster. Therefore, the equilibrium constant, in present of the catalyst, at 2000 K is:
View answer and explanation
Correct option: B
4×10−44 \times {10^{ - 4}}
Option B (4×10−44 \times {10^{ - 4}}) is correct because a catalyst only affects the rate of reaction, not the equilibrium constant.
74 Easy
74 of 408
Application +4 | -1
For the reaction CO(g)+12O2(g)CO2(g), KP/KC+ \frac{1}{2}{O_2}\left( g \right)C{O_2}(g),\,{K_P}/{K_C} will be
View answer and explanation
Correct option: C
(RT)−1/2{\left( {RT} \right)^{ - 1/2}}
Option C ((RT)−1/2{\left( {RT} \right)^{ - 1/2}}) is correct because the relationship between Kp and Kc is Kp = Kc(RT)^Δ n, and in this case, Δ n = -1/2.
75 Medium
75 of 408
Analyzing +4 | -1
The equilibrium constant for N2(g)+O2(g)⇋2NO(g){N_2}(g) + {O_2}(g)\leftrightharpoons2NO\left( g \right) is K1{K_1} and that for NO(g)⇋12N2(g)+12O2(g)NO(g)\leftrightharpoons\frac{1}{2}{N_2}(g) + \frac{1}{2}{O_2}(g) is K2{K_2}, then K1{K_1} and K2{K_2} related as
View answer and explanation
Correct option: A
K1=(1K2)2{K_1} = {\left( {\frac{1}{{{K_2}}}} \right)^2}
The correct answer is A (K1=(1K2)2{K_1} = {(\frac{1}{{{K_2}}})^2}) because it correctly relates the equilibrium constants of the forward and reverse reactions.
76 Medium
76 of 408
Understanding +4 | -1
In a lime kiln, to get higher yield of CO2\mathrm{C}{\mathrm{O}_2}, the measure that can be taken is
View answer and explanation
Correct option: D
To pump out CO2\mathrm{C}{\mathrm{O}_2}
Option D is correct because removing CO2 shifts the equilibrium to the right, increasing the yield of CO2 according to Le Chatelier's principle.
77 Hard
77 of 408
Analyzing +4 | -1
Phenolphthalein can’t be used in the titration of
View answer and explanation
Correct option: B
CH3COOH\mathrm{CH}_3 \mathrm{COOH} with NaOH\mathrm{NaOH}
Option D is correct. Phenolphthalein is unsuitable for titrations involving a weak acid and a weak base because the pH change near the equivalence point is gradual, making it difficult to determine the endpoint accurately.
78 Medium
78 of 408
Analyzing +4 | -1
A white salt is readily soluble in water and gives a colourless solution with a pH of about 9. The salt would be:
View answer and explanation
Correct option: B
CH3COONa\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}
Option B (CH3COONa\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}) is correct because it is a salt of a weak acid and a strong base, which will hydrolyze in water to produce OH- ions, resulting in a pH greater than 7.
79 Medium
79 of 408
Application +4 | -1
In a reaction PCl5PCl3+Cl2PC{l_5}PC{l_3} + C{l_2}, degree of dissociation is 30%, If initial moles of PCl5PC{l_5} is one then total moles at equilibrium is
View answer and explanation
Correct option: A
1.31.3
Option A (1.31.3) is correct because at equilibrium, the total number of moles is calculated by considering the degree of dissociation of PCl5PC{l_5}.
80 Medium
80 of 408
Application +4 | -1
0.1 mole of N2O4(g){\mathrm{N}_2}{\mathrm{O}_4}\left( \mathrm{g} \right) was sealed in a tube under one atmospheric conditions at 25∘C25^\circ C. Calculate the number of moles of NO2(g)\mathrm{N}{\mathrm{O}_2}\left( \mathrm{g} \right) present, if the equilibrium N2O4(g)  ⇋  2NO2(g)(Kp=0.14){\mathrm{N}_2}{\mathrm{O}_4}\left( \mathrm{g} \right)\;\leftrightharpoons\;2\mathrm{N}{\mathrm{O}_2}\left( \mathrm{g} \right)\left( {{K_p} = 0.14} \right) is reached after some times
View answer and explanation
Correct option: A
0.036
Option A is correct because it calculates the number of moles of NO2 at equilibrium using the given Kp value and initial conditions.
81 Hard
81 of 408
Understanding +4 | -1
Which of the following relation represents correct relation between standard electrode potential and equilibrium constant?

I. log⁡K=nFE∘2.303RT\log {K}=\frac{{nFE}^{\circ}}{2.303 {RT}}

II. K=enFE∘RT{K}={e}^{\frac{{nFE}^{\circ}}{{RT}}}

III. log⁡K=−nFE∘2.303RT\log {K}=\frac{-{nFE}^{\circ}}{2.303 {RT}}

IV. log⁡K=0.4342−nFE∘RT\log {K}=0.4342 \frac{-{nFE}^{\circ}}{{RT}}

Choose the correct statement(s).
View answer and explanation
Correct option: C
I, II and IV are correct
Option C is correct because relations I, II, and IV correctly represent the relationship between standard electrode potential and equilibrium constant.
82 Hard
82 of 408
Application +4 | -1
A saturated solution of Ag2SO4  is  2.5×10−2  M\mathrm{A}{\mathrm{g}_2}\mathrm{S}{\mathrm{O}_4}\;\mathrm{is}\;2.5 \times {10^{ - 2}}\;\mathrm{M} The value of its solubility product is
View answer and explanation
Correct option: A
62.5×10−662.5 \times {10^{ - 6}}
Option B (Both (A) and (R) are true but (R) is not the explanation for (A)) is correct because both the assertion and reason are true, but the reason does not fully explain the speed of the reaction.
83 Hard
83 of 408
Analyzing +4 | -1
For the equilibrium A(g)⇌nB(g),{A}(\mathrm{g}) \rightleftharpoons {nB}(\mathrm{g}), the equilibrium constant KPK_{{P}} is related with the degree of dissociation α,\alpha, and the total pressure of gases at equilibrium PP is
View answer and explanation
Correct option: A
(nα)n⋅Pn−1(1−α)⋅[1+(n−1)α]n−1\frac{(n \alpha)^n \cdot P^{n-1}}{(1-\alpha) \cdot[1+(n-1) \alpha]^{n-1}}
Option A is correct. The equilibrium constant Kp is related to the degree of dissociation α and total pressure P by the formula derived from partial pressures at equilibrium.
84 Easy
84 of 408
Remembering +4 | -1
The auto protonation constant of H2O{\mathrm{H}_2}\mathrm{O} is:
View answer and explanation
Correct option: B
3.23×10−183.23 \times {10^{ - 18}}
Option B (3.23×10−183.23 \times {10^{ - 18}}) is correct because the auto-protonation constant of water, also known as the autoionization constant, is a specific value at a given temperature.
85 Medium
85 of 408
Analyzing +4 | -1
Which is decreasing order of strength of bases?
O_H,N_H2,HC≡C−andCH3CH2−\mathop \mathrm{O}\limits^\_ \mathrm{H},\mathop \mathrm{N}\limits^\_ {\mathrm{H}_2},\mathrm{HC} \equiv {\mathrm{C}^ - }\mathrm{and C}{\mathrm{H}_3}\mathrm{CH}_2^ -
View answer and explanation
Correct option: A
H3CCH2−>NH2−>HC≡C−>OH−{\mathrm{H}_3}\mathrm{CCH}_2^ - > N\mathrm{H}_2^ - > HC \equiv {\mathrm{C}^ - } > O{\mathrm{H}^ - }
Option A is correct. The order of base strength is determined by the stability of the conjugate acid. The weaker the conjugate acid, the stronger the base.
86 Hard
86 of 408
Analyzing +4 | -1
N2( g)+3H2( g)⇌2NH3( g);K1\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}) ; \mathrm{K}_1

 N2( g)+O2( g)⇌2NO(g);K2\mathrm{~N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) ; \mathrm{K}_2

H2( g)+12O2( g)⇌H2O(g);K3\mathrm{H}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons \mathrm{H}_2 \mathrm{O}(\mathrm{g}) ; \mathrm{K}_3

The equilibrium constant for 2NH3( g)+52O2( g)⇌2NO(g)+3H3O(g)2 \mathrm{NH}_3(\mathrm{~g})+\frac{5}{2} \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+3 \mathrm{H}_3 \mathrm{O}(\mathrm{g}) will be
View answer and explanation
Correct option: D
K2 K33 K1\frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_1}
The correct answer is D (K2 K33 K1\frac{\mathrm{K}_2 \mathrm{~K}_3^3}{\mathrm{~K}_1}) because it represents the equilibrium constant obtained by manipulating the given reactions to match the target reaction.
87 Easy
87 of 408
Remembering +4 | -1
Which of the following is Lewis acid?
View answer and explanation
Correct option: B
H+{H}^{+}
Option B (H+{H}^{+}) is correct because a Lewis acid is an electron pair acceptor, and H+{H}^{+} has an empty orbital and can accept an electron pair.
88 Medium
88 of 408
Analyzing +4 | -1
For the reaction
2SO2(g)+O2(g)→2SO3(g),ΔH=−57.2kJ mol−1andKc=1.7×10162S{O_2}(g) + {O_2}(g) \to 2S{O_3}(g),\Delta H = - 57.2\mathrm{kJ}\,\mathrm{mo}{\mathrm{l}^{ - 1}}\mathrm{ and }{K_c} = 1.7 \times {10^{16}}
Which of the following statement is incorrect?
View answer and explanation
Correct option: D
The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required
The correct answer is D because a large equilibrium constant suggests the reaction goes to completion, but a catalyst is still needed to increase the rate of reaction.
89 Medium
89 of 408
Understanding +4 | -1
The equilibrium which remains unaffected by change in pressure of the reactants is
View answer and explanation
Correct option: A
N2(g)+O2(g)⇋2NO(g){\mathrm{N}_2}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{NO}\left( \mathrm{g} \right)
Option A is correct because the number of moles of gaseous reactants equals the number of moles of gaseous products, so pressure changes have no effect on the equilibrium.
90 Easy
90 of 408
Understanding +4 | -1
A precipitate is formed when
View answer and explanation
Correct option: C
The ionic product exceeds the solubility product
Option C (The ionic product exceeds the solubility product) is correct because a precipitate forms when the solution is supersaturated, meaning the ionic product is greater than the solubility product.
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