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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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391 Hard
391 of 680
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}}
The correct answer is (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}}} because it correctly represents the equilibrium constant Kp in terms of degree of dissociation and total pressure.
392 Medium
392 of 680
Understanding +4 | -1
pHpH for the solution of salt undergoing anionic hydrolysis (say CH3COONa\mathrm{C}{\mathrm{H}_3}\mathrm{COONa} ) is given by:
View answer and explanation
Correct option: A
pH=12[pKw+pKa+log⁡c]\mathrm{pH} = \frac{1}{2}\left[ {\mathrm{p}{K_w} + \mathrm{p}{K_a} + \log c} \right]
Option A is correct because it represents the correct formula for calculating the pH of a salt undergoing anionic hydrolysis, derived from the hydrolysis constant and water's ionic product.
393 Hard
393 of 680
Application +4 | -1
For weak electrolyte, AB,\mathrm{AB}, the degree of ionization would be (V=V= volume of solution having 1 mole1 \mathrm{~mole} of electrolyte and KK is the ionization constant of the electrolyte)
View answer and explanation
Correct option: D
K⋅V\sqrt{K \cdot V}
Option D (K⋅V\sqrt{K \cdot V}) is correct because the degree of ionization for a weak electrolyte is inversely proportional to the square root of the concentration, and concentration is inversely proportional to volume.
394 Medium
394 of 680
Application +4 | -1
The solubility of AgCl\mathrm{AgCl} in 0.2 M NaCl0.2 \mathrm{~M} \mathrm{~NaCl} is

[KspAgCl=1.8×10−10]\left[\mathrm{K}_{\mathrm{sp}} \mathrm{AgCl}=1.8 \times 10^{-10}\right]
View answer and explanation
Correct option: B
9.0×10−10M9.0 \times 10^{-10} \mathrm{M}
Option B (9.0×10−10M9.0 \times 10^{-10} \mathrm{M}) is correct because it accurately calculates the solubility of AgCl in the presence of a common ion (Cl-) using the solubility product constant.
395 Medium
395 of 680
Application +4 | -1
An aqueous solution of aniline of concentration 0.24 M0.24 \mathrm{~M} is prepared. What is concentration of sodium hydroxide is needed in this solution so that anilinium ion concentration remains at 1×10−8M1 \times 10^{-8} \mathrm{M} ? ( KbK_b for C6H5NH3+=2.4×10−5M)\left.\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3{ }^{+}=2.4 \times 10^{-5} \mathrm{M}\right)
View answer and explanation
Correct option: A
10−2M10^{-2} \mathrm{M}
Option A (10−2M10^{-2} \mathrm{M}) is correct because the concentration of sodium hydroxide is calculated using the equilibrium expression for the aniline solution, considering the suppression of aniline dissociation due to NaOH.
396 Medium
396 of 680
Understanding +4 | -1
Solubility product principle can be applied when
View answer and explanation
Correct option: C
Any ionic compound is sparingly soluble in a liquid
Option C is correct because the solubility product principle applies specifically to sparingly soluble ionic compounds in a liquid, where an equilibrium exists between the solid and its ions in solution.
397 Hard
397 of 680
Analyzing +4 | -1
A container of capacity V LV \mathrm{~L} contains an equilibrium mixture that consists of 2 moles each of PCl5,PCl3{PCl}_5, {PCl}_3 and Cl2{Cl}_2 (all as gases). The pressure is 30.3975 kPa30.3975 \mathrm{~kPa} and temperature is T K.T \mathrm{~K}. A certain amount ('xx' mole) of Cl2(g){Cl}_2(\mathrm{g}) is now introduced keeping the pressure and temperature constant, until the equilibrium volume becomes 2V L.2 V \mathrm{~L}. The value of '90x90 x' is
View answer and explanation
Correct option: A
600
Option A (600) is correct because after adding chlorine and re-establishing equilibrium, the value of 90x is calculated to be 600.
398 Medium
398 of 680
Analyzing +4 | -1
Assertion :-For a reaction A(g)+B(g)⇄AB(g)A\left( g \right) + B\left( g \right) \rightleftarrows AB_{\left( g \right)} if inert gas is added in a container at
constant volume. The equilibrium shifts to left side.
Reason :- Because partial pressure of A,BA, B and ABAB decreases.
View answer and explanation
Correct option: D
If both Assertion & Reason are False
The correct answer is D because adding an inert gas at constant volume does not shift the equilibrium, as the partial pressures of the reactants and products remain unchanged.
399 Easy
399 of 680
Application +4 | -1
A buffer solution is prepared by mixing 10  mL10\;mL of 1.0  M1.0\;M acetic and 20  mL20\;mL of 0.5  M0.5\;M sodium acetate and then diluted to 100  mL100\;mL with distilled water. If the pKa\mathrm{p}{\mathrm{K}_a} of CH3COOH\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} is 4.764.76, what is the pH of the buffer solution prepared?
View answer and explanation
Correct option: B
4.764.76
Option B (4.764.76) is correct because, in this case, the concentrations of the acid and conjugate base are equal, making pH equal to pKa.
400 Easy
400 of 680
Understanding +4 | -1
A white substance having alkaline nature in solution is
View answer and explanation
Correct option: C
Na2CO3\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}
Option C (Na2CO3\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}) is correct because sodium carbonate hydrolyzes in water to produce a strong base (NaOH) and a weak acid (H2CO3H_2CO_3), resulting in an alkaline solution.
401 Medium
401 of 680
Application +4 | -1
The KspK_{s p} of PbCrO4\mathrm{PbCrO}_4 is 1.0×10−161.0 \times 10^{-16}. Then the molar solubility of PbCrO4\mathrm{PbCrO}_4 is:
View answer and explanation
Correct option: D
1.0×10−81.0 \times 10^{-8}
Option D (1.0×10−81.0 \times 10^{-8}) is correct because it is the square root of the Ksp value.
402 Medium
402 of 680
Application +4 | -1
The value of KPK_{\mathrm{P}} for the process: CuSO4⋅5H2O(s)⇌CuSO4⋅3H2O(s)+2H2O(g)\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}(\mathrm{s})\rightleftharpoons \mathrm{CuSO}_4 \cdot 3 \mathrm{H}_2 \mathrm{O}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) is 1.21×10−4 atm21.21 \times 10^{-4} \mathrm{~atm}^2 at certain temperature. If aqueous tension at that temperature is 40 torr, then at what relative humidity of air will CuSO4⋅5H2O\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O} effloresces?
View answer and explanation
Correct option: D
Below 20.9%20.9 \%
Option A ([Co(NH3)6]3+) is correct because it forms a low-spin, diamagnetic octahedral complex due to the strong field ligand NH3.
403 Easy
403 of 680
Application +4 | -1
An acid solution of pH=5\mathrm{pH}=5 is diluted 1000 times, the pH\mathrm{pH} of final solution becomes :-
View answer and explanation
Correct option: C
6.956.95
Option C (6.956.95) is correct because after dilution, the pH is calculated considering the contribution of hydrogen ions from both the acid and water.
404 Easy
404 of 680
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.
405 Medium
405 of 680
Remembering +4 | -1
The strength of an acid depends on its tendency to
View answer and explanation
Correct option: B
Donate protons
Option B (Donate protons) is correct because, according to the Bronsted-Lowry definition, an acid is a substance that donates protons.
406 Medium
406 of 680
Application +4 | -1
A saturated solution of Mg(OH)2\mathrm{Mg}{\left( {\mathrm{OH}} \right)_2} in water at 25∘C25^\circ C contains 0.11  g  Mg(OH)20.11\;\mathrm{g}\;\mathrm{Mg}{\left( {\mathrm{OH}} \right)_2} per litre of solution. The solubility product of Mg(OH)2\mathrm{Mg}{\left( {\mathrm{OH}} \right)_2} is:
View answer and explanation
Correct option: D
4×(0.11)3/(58)34 \times {\left( {0.11} \right)^3}/{\left( {58} \right)^3}
Option D (4×(0.11)3/(58)34 \times {\left( {0.11} \right)^3}/{\left( {58} \right)^3}) is correct because it correctly calculates the solubility product using the given solubility and molar mass.
407 Medium
407 of 680
Analyzing +4 | -1
The concentration of KI  and  KCl\mathrm{KI }\;\mathrm{and}\;\mathrm{ KCl} in a certain solution containing both is 0.001 M each. If 20 mL of this solution is added to 20 mL of a saturated solution of AgI in water. What will happen?

(Ksp  AgCl=10−10,  Ksp  AgI=10−16)({K_{\mathrm{sp}}}\;\mathrm{AgCl} = {10^{ - 10}},\;{K_{\mathrm{sp}}}\;\mathrm{AgI} = {10^{ - 16}})
View answer and explanation
Correct option: A
AgI\mathrm{AgI} will be precipitated
Option A (AgI will be precipitated) is correct because the addition of KI increases the iodide ion concentration, causing the ionic product of AgI to exceed its solubility product, leading to precipitation.
408 Medium
408 of 680
Understanding +4 | -1
The tendency of BF3, BCl3   and   BBr3B{F_3},\,BC{l_3}{\rm\,\,\,{ and}}\,\,\,\mathrm{ }BB{r_3} to behaves as Lewis acid decreases in the sequence
View answer and explanation
Correct option: B
BBr3>BCl3>BF3BB{r_3} > BC{l_3} > B{F_3}
Option B (BBr3>BCl3>BF3BB{r_3} > BC{l_3} > B{F_3}) is correct because the Lewis acidity decreases as the extent of pπ-pπ backbonding increases, which is greatest in BF3.
409 Medium
409 of 680
Analyzing +4 | -1
For the given reaction, 2 A( s)+B(g)=C(g)+2D(s)+E(s)2 \mathrm{~A}(\mathrm{~s})+\mathrm{B}(\mathrm{g})=\mathrm{C}(\mathrm{g})+2 \mathrm{D}(\mathrm{s})+\mathrm{E}(\mathrm{s}) the degree of dissociation of B\mathrm{B} was found to be 20%20 \% at 300 K300 \mathrm{~K} and 24%24 \% at 500 K500 \mathrm{~K}. The rate of forward reaction
View answer and explanation
Correct option: C
depends on temperature only and increases with increase in temperature.
Option C is correct because, with Δng=0\Delta n_g = 0, pressure has no effect, and the forward reaction rate depends only on temperature, increasing with temperature due to the endothermic nature.
410 Hard
410 of 680
Understanding +4 | -1
For the reactions

I2(aq)⇌I2\mathrm{I}_2(\mathrm{aq}) \rightleftharpoons \mathrm{I}_2 (oil) Equilibrium constant is K1\mathrm{K}_1

I2\mathrm{I}_2 (oil) ⇌I2\rightleftharpoons \mathrm{I}_2 (ether) Equilibrium constant is K2\mathrm{K}_2

for the reaction

I2(aq)⇌I2\mathrm{I}_2(\mathrm{aq}) \rightleftharpoons \mathrm{I}_2 (ether) Equilibrium constant is K3\mathrm{K}_3

The relation between K1,K2,K3\mathrm{K}_1, \mathrm{K}_2, \mathrm{K}_3 is
View answer and explanation
Correct option: B
K3=K1 K2\mathrm{K}_3=\mathrm{K}_1 \mathrm{~K}_2
Option B (K3=K1 K2\mathrm{K}_3=\mathrm{K}_1 \mathrm{~K}_2) is correct because when adding reactions, the equilibrium constants are multiplied.
411 Medium
411 of 680
Application +4 | -1
The solubility of AgBr\mathrm{AgBr} having solubility product 5×10−135 \times10^{-13} in 0.1 M CaBr20.1 \mathrm{~M} \mathrm{~CaBr}_2 solution will be
View answer and explanation
Correct option: B
2.5×10−12M2.5 \times 10^{-12} \mathrm{M}
Option B (2.5×10−12M2.5 \times 10^{-12} \mathrm{M}) is correct because the solubility of AgBr is reduced due to the common ion effect from the Br−Br^- ions in the CaBr2CaBr_2 solution.
412 Easy
412 of 680
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.
413 Medium
413 of 680
Application +4 | -1
pH\mathrm{pH} of the buffer solution formed by mixing acetic acid and sodium acetate taken in equimolar concentration will be [Given : pKa\mathrm{p} K_a value for CH3COOH=4.76\mathrm{CH}_3 \mathrm{COOH}=4.76 ]
View answer and explanation
Correct option: C
=4.76= 4.76
Option C (= 4.76) is correct because, according to the Henderson-Hasselbalch equation, when the concentrations of the acid and its conjugate base are equal, the pH equals the pKa.
414 Medium
414 of 680
Application +4 | -1
What would be the pH\mathrm{pH} of an ammonia solution if the pH\mathrm{pH} of acetic acid solution of same strength is 3.2? The dissociation constants of ammonia and acetic acid are same.
View answer and explanation
Correct option: D
10.8
Option D (10.8) is correct because for a solution of a salt of weak acid and weak base, if the dissociation constants are the same, pH + pOH = 14.
415 Medium
415 of 680
Application +4 | -1
The KaK_a of a weak monobasic acid is 1×10−51 \times 10^{-5}. The percentage of ionization in a decimolar acid solution is:
View answer and explanation
Correct option: D
1
Option D (1) is correct because the percentage of ionization is calculated using the formula derived from the equilibrium constant Ka and concentration.
416 Medium
416 of 680
Analyzing +4 | -1
STATEMENT I: pH\mathrm{pH} of 10−7M10^{-7} \mathrm{M} HCl{HCl} is less than 7 at 25∘C25^{\circ} \mathrm{C}
STATEMENT II : At very low concentration of HCl{HCl}, contribution of H+\mathrm{H}^{+}from water considerable.
View answer and explanation
Correct option: A
Statement I, II are correct
The correct answer is 3 because the complex has Fe in the +1 oxidation state with a 3d7 configuration, resulting in three unpaired electrons.
417 Medium
417 of 680
Analyzing +4 | -1
The equilibrium constants for A2(g)⇌2A(g)\mathrm{A}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{A}(\mathrm{g}) at 400 K400 \mathrm{~K} and 600 K600 \mathrm{~K} are 1×10−81 \times 10^{-8} and 1×10−21 \times 10^{-2} respectively. The reaction is
View answer and explanation
Correct option: B
Endothermic
The correct answer is Endothermic because the equilibrium constant increases with increasing temperature, indicating that heat is absorbed, favoring product formation.
418 Easy
418 of 680
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.
419 Easy
419 of 680
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.
420 Hard
420 of 680
Application +4 | -1
In the presence of excess of anhydrous SrCl2{SrCl}_2, the amount of water taken up is governed by Kp=1012 atm−4K_p=10^{12} {~atm}^{-4} for the following reaction at 273 K273 {~K}
SrCl2⋅2H2O(s)+4H2O(g)⇌SrCl2⋅6H2O(s){SrCl}_2 \cdot 2 {H}_2 {O}({s})+4 {H}_2 {O}({g}) \rightleftharpoons {SrCl}_2 \cdot 6 {H}_2 {O}({s})

What is equilibrium vapour pressure (in torr) of water in a closed vessel that contains SrCl2⋅2H2O(s){SrCl}_2 \cdot 2 {H}_2 {O}(s) ?
View answer and explanation
Correct option: C
0.76 torr0.76\, torr
Option C (0.76 torr0.76\, torr) is correct because the equilibrium vapor pressure is calculated from the given KpK_p value.
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