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JEE • CHEMISTRY • Equilibrium

Equilibrium practice · Medium

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Showing 1-30 of 401 questions
1 Medium
1 of 401
Analyzing +4 | -1
K1\mathrm{K}_1 and K2\mathrm{K}_2 are equilibrium constant for reactions (i) and (ii)

N2( g)+O2( g)⇌2NO(g)…\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\quad\ldots (i)

NO(g)⇌12 N2( g)+12O2( g)…\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\quad\ldots (ii)

Then,
View answer and explanation
Correct option: A
K1=(1 K2)2\mathrm{K}_1=\left(\frac{1}{\mathrm{~K}_2}\right)^2
Option A is correct because the equilibrium constant K1 is the inverse square of K2, derived from reversing and halving the reaction.
2 Medium
2 of 401
Understanding +4 | -1
Ka1,   Ka2   and   Ka3K{a_1},\,\,\,K{a_2}\,\,\,\mathrm{and}\,\,\,K{a_3} are the respective ionisation constants for the following reactions
H2S⇋H++HS−HS−⇋H++S2−H2S⇋2H++S2−\begin{array}{l} {H_2}S\leftrightharpoons{H^ + } + H{S^ - }\\ H{S^ - }\leftrightharpoons{H^ + } + {S^{2 - }}\\ {H_2}S\leftrightharpoons2{H^ + } + {S^{2 - }} \end{array}
The correct relationship between Ka1,  Ka2 ,Ka3K{a_1},\,\,K{a_2}\,,K{a_3} is
View answer and explanation
Correct option: B
Ka3=  Ka1 ×Ka2K{a_3} = \,\,K{a_1}\, \times K{a_2}
Option B (Ka3= Ka1 ×Ka2K{a_3} = \,K{a_1}\, \times K{a_2}) is correct because the overall ionization constant is the product of the stepwise ionization constants.
3 Medium
3 of 401
Application +4 | -1
The pH\mathrm{pH} of 0.1 M−N2H40.1 \mathrm{~M}-\mathrm{N}_2 \mathrm{H}_4 solution is (For N2H4,\mathrm{N}_2 \mathrm{H}_4, Kb1=3.6×10−6,Kb2=6.4×10−12,log⁡2=0.3K_{{b}_1}=3.6 \times 10^{-6}, K_{{b}_2}=6.4 \times 10^{-12}, \log 2=0.3, log⁡3=0.48\log 3=0.48)
View answer and explanation
Correct option: C
10.78
Option C (10.78) is correct because it accurately calculates the pH of the N2H4 solution considering the first base dissociation constant.
4 Medium
4 of 401
Understanding +4 | -1
Reaction BaO2(s)⇌BaO(s)+O2(g);ΔH=−ve.{\text{Ba}}{{\text{O}}_2}(s) \rightleftharpoons {\text{BaO}}(s) + {{\text{O}}_2}(g);\Delta H = - {\text{ve}}. In equilibrium condition pressure of O2\text{O}_2 depends on:
View answer and explanation
Correct option: C
Increase temperature on equilibrium
Option C (Increase temperature on equilibrium) is correct because the reaction is exothermic, and increasing the temperature will shift the equilibrium to the left, favoring the reactants and increasing the pressure of O2.
5 Medium
5 of 401
Understanding +4 | -1
A : At equilibrium concentration of the reactant and product does not change with time for a chemical reaction.
R: The rate of reaction is zero at equilibrium.
View answer and explanation
Correct option: C
If A is correct but R is incorrect
Option C is correct because at equilibrium, the concentrations of reactants and products do not change with time, but the rate of reaction is *not* zero; the forward and reverse rates are equal.
6 Medium
6 of 401
Understanding +4 | -1
Identify the reaction in which both products concentration and reactants concentration is appreciable
View answer and explanation
Correct option: B
X(g)+Y(g)⇌XY(g),Keq=54X(g)+Y(g) \rightleftharpoons X Y(g), {K}_{e q}=54
Option B (X(g)+Y(g)⇌XY(g),Keq=54X(g)+Y(g) \rightleftharpoons X Y(g), {K}_{e q}=54) is correct because a KeqK_{eq} value close to 1 (typically between 0.001 and 1000) indicates that both reactants and products are present in appreciable concentrations at equilibrium.
7 Medium
7 of 401
Analyzing +4 | -1
Solubility of Ni(OH)2\mathrm{Ni}(\mathrm{OH})_2 will be maximum in
View answer and explanation
Correct option: D
0.1M0.1 \mathrm{M} aq. solution of H2SO4\mathrm{H}_2 \mathrm{SO}_4
Option D is correct because the solubility of Ni(OH)2\mathrm{Ni}(\mathrm{OH})_2 will be maximum in 0.1M0.1 \mathrm{M} aq. solution of H2SO4\mathrm{H}_2 \mathrm{SO}_4 due to the acid-base reaction consuming hydroxide ions and shifting the equilibrium towards dissolution.
8 Medium
8 of 401
Application +4 | -1
Calculate the hydrolysis constant of the salt containing NO2−\mathrm{NO}_2^{-} ions. Ka\mathrm{K}_{\mathrm{a}} for HNO2=4.5×10−10\mathrm{HNO}_2=4.5 \times 10^{-10} and Kw=1×10−14\mathrm{K}_{\mathrm{w}}=1 \times 10^{-14}.
View answer and explanation
Correct option: A
2.2×10−52.2 \times 10^{-5}
Option A (2.2×10−52.2 \times 10^{-5}) is correct because it is the hydrolysis constant calculated using the formula Kh=KwKaK_h = \frac{{K_w}}{{K_a}}.
9 Medium
9 of 401
Application +4 | -1
At 3000 K3000 \mathrm{~K}, the equilibrium partial pressure of CO2,CO{CO}_2, {CO} and O2{O}_2 are 0.6, 0.4 and 0.2 atm respectively. Kp{K}_p for the reaction, 2CO2⇄2CO+O22 {CO}_2 \rightleftarrows 2 {CO}+{O}_2 is
View answer and explanation
Correct option: A
0.088
Option A (0.088) is correct because it is the result of correctly applying the equilibrium constant expression for the given reaction and partial pressures.
10 Medium
10 of 401
Understanding +4 | -1
Assertion :- H2OH_2O is amphoteric in nature.

Reason :- H2OH_2O can accept a proton to form H3O+H_3O^+ ion and can donate a proton to form OH–{OH}^–
ion.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because water can act as both an acid and a base.
11 Medium
11 of 401
Application +4 | -1
If ΔG0\Delta {G^0} for the reaction given below is 1.7  kJ1.7\;\mathrm{ }kJ; the equilibrium constant of the reaction, 2HI(g)  ⇋  H2(g)+I2(g)2\mathrm{HI}\left( \mathrm{g} \right)\;\leftrightharpoons\;{\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right) at 25∘C25^\circ C is :
View answer and explanation
Correct option: D
0.50.5
Option D (0.50.5) is correct because it is calculated using the relationship between Gibbs free energy change and the equilibrium constant.
12 Medium
12 of 401
Understanding +4 | -1
A chemical system is in equilibrium. Addition of a catalyst would result in:
View answer and explanation
Correct option: C
A new reaction path way to reaction
Option C (A new reaction path way to reaction) is correct because a catalyst provides an alternative reaction pathway with a lower activation energy, thus speeding up the reaction without changing the equilibrium.
13 Medium
13 of 401
Understanding +4 | -1
The addition of a catalyst to the reaction system
View answer and explanation
Correct option: D
Increases the rate of forward as well as backward reaction equally
Option D (Increases the rate of forward as well as backward reaction equally) is correct because a catalyst lowers the activation energy for both forward and reverse reactions equally, thus increasing both rates.
14 Medium
14 of 401
Application +4 | -1
A buffer solution is prepared in which the concentration of NH3\mathrm{N}{\mathrm{H}_3} is 0.30  M0.30\;M and the concentration of NH4+\mathrm{NH}_4^ + is 0.20  M0.20\;M. If the equilibrium constant, Kb{K_b} for NH3\mathrm{N}{\mathrm{H}_3} equals 1.8×10−51.8 \times {10^{ - 5}} what is the pH of this solution?
(log⁡2.7=0.43)(\log 2.7 = 0.43)
View answer and explanation
Correct option: C
9.439.43
Option C (9.439.43) is correct because it is calculated using the Henderson-Hasselbalch equation for a basic buffer solution.
15 Medium
15 of 401
Application +4 | -1
If equilibrium constant is 2.6×1082.6 \times 10^8 then find the value of ΔG∘\Delta {G}^{\circ}.
View answer and explanation
Correct option: A
−43.36 kJ-43.36 \mathrm{~kJ}
Option A (−43.36 kJ-43.36 \mathrm{~kJ}) is correct because it is calculated using the Gibbs free energy equation with the given equilibrium constant.

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16 Medium
16 of 401
Analyzing +4 | -1
Which of the following when mixed, will give a solution with
pH greater than 7?
View answer and explanation
Correct option: B
100 mL100 \mathrm{~mL} of 0.1 M CH3COOH+100 mL0.1 \mathrm{~M} \mathrm{~CH}_3 \mathrm{COOH}+100 \mathrm{~mL} of 0.1 M KOH0.1 \mathrm{~M} \mathrm{~KOH}
The correct answer is B (100 mL100 \mathrm{~mL} of 0.1 M CH3COOH+100 mL0.1 \mathrm{~M} \mathrm{~CH}_3 \mathrm{COOH}+100 \mathrm{~mL} of 0.1 M KOH0.1 \mathrm{~M} \mathrm{~KOH}) because it forms a basic buffer solution after the reaction.
17 Medium
17 of 401
Analyzing +4 | -1
A vessel contains A(g)\mathrm{A}(\mathrm{g}) and B(g)\mathrm{B}(\mathrm{g}) at 2 atm2 \mathrm{~atm} and 4 atm4 \mathrm{~atm} respectively at TK\mathrm{TK}, the mixture is allowed to attain equilibrium at TK\mathrm{TK}, according to the reaction

8B(g)⇌8A(g)+C(s)8 \mathrm{B}(\mathrm{g}) \rightleftharpoons 8 \mathrm{A}(\mathrm{g})+\mathrm{C}(\mathrm{s})

At equilibrium, (nAnB)eq =(nBnA)initial \left(\frac{\mathrm{n_A}}{\mathrm{n_B}}\right)_{\text {eq }}=\left(\frac{\mathrm{n_B}}{\mathrm{n_A}}\right)_{\text {initial }}
Find the value of y\mathrm{y} if Kc=2y\mathrm{K_c}=2^\mathrm{y}
View answer and explanation
Correct option: A
8
Option A (8) is correct because the equilibrium constant Kc is calculated to be 28, thus y=8.
18 Medium
18 of 401
Application +4 | -1
In which of the following cases, pH is greater than 7?
View answer and explanation
Correct option: C
50 mL of 0.1 M CH3COOHC{H_3}COOH + 50 mL of 0.1 M KOH
Option C (50 mL of 0.1 M CH3COOHC{H_3}COOH + 50 mL of 0.1 M KOH) is correct because it involves the reaction of a weak acid (CH3COOHCH_3COOH) with a strong base (KOH), resulting in a salt of a weak acid and strong base, which hydrolyzes to give a pH greater than 7.
19 Medium
19 of 401
Analyzing +4 | -1
A: HCl,HNO3{HCl}, {HNO}_3 and H2SO4{H}_2 {SO}_4 are equally strong acids in water.
R: Water is a stronger acid than alcohols.
View answer and explanation
Correct option: B
If both A and R are correct, but R is not the correct explanation of A
Option B is correct because both the assertion and the reason are correct statements, but the reason does not explain the assertion. The leveling effect explains the assertion.
20 Medium
20 of 401
Analyzing +4 | -1
The degree of hydrolysis of which of the
following salt is independent of the
concentration of salt solution?
View answer and explanation
Correct option: D
NaCl{NaCl}
The correct answer is D (NaCl{NaCl}) because it is a salt of a strong acid and a strong base, and therefore does not undergo hydrolysis. Its degree of hydrolysis is zero and independent of concentration.
21 Medium
21 of 401
Application +4 | -1
If the concentration of the salt (AB)(\mathrm{AB}) in the aqueous solution is 5.0×10−5 M.Ksp5.0 \times 10^{-5}\, \mathrm{M} . \mathrm{K}_{\mathrm{sp}} for the salt is 3.6×10−9 M23.6 \times 10^{-9}\, \mathrm{M}^2. Then which sentence describe the solution most correctly.
View answer and explanation
Correct option: B
The solution is unsaturated
Option B (The solution is unsaturated) is correct because the Ionic Product (IP) is less than the solubility product (Ksp), indicating that more salt can dissolve in the solution.
22 Medium
22 of 401
Understanding +4 | -1
Assertion :- There is no effect on equilibrium constant if inert gas is added to the reaction.
A(g)+B(g)⇌C(g){A_{\left( g \right)}} + {B_{\left( g \right)}} \rightleftharpoons {C_{\left( g \right)}}
Reason :- Equilibrium constant changes only with temperature.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the equilibrium constant is indeed affected only by temperature, and adding an inert gas at constant volume does not shift the equilibrium.
23 Medium
23 of 401
Application +4 | -1
3.2 moles of hydrogen iodide were heated in a sealed bulb at 444∘C444^\circ C till the equilibrium state was reached. Its degree of dissociation at this temperature was found to be 22%.22\% . The number of moles of hydrogen iodide present at equilibrium are
View answer and explanation
Correct option: B
2.496
Option B (2.496) is correct because it is the number of moles of HI remaining at equilibrium after dissociation.
24 Medium
24 of 401
Application +4 | -1
The pH of a buffer solution containing equal molal concentration of a weak base and its chloride (Kb  for  weak  base  =2×10−5)({K_b}\;\mathrm{for}\;\mathrm{ weak}\;\mathrm{ base}\; = 2 \times {10^{ - 5}}) is
View answer and explanation
Correct option: D
9.39.3
Option D (9.39.3) is correct because it is calculated using the Henderson-Hasselbalch equation for a basic buffer.
25 Medium
25 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.
26 Medium
26 of 401
Analyzing +4 | -1
Which of the hydrates species can exist?

I.H5O2+,II.H3O+\mathrm{I.H}_5 \mathrm{O}_2^{+} , \mathrm{II.H}_3 \mathrm{O}^{+}III. H3O2−,\mathrm{H}_3 \mathrm{O}_2^{-} , IV. H5O3−\mathrm{H}_5 \mathrm{O}_3^{-}

Select the correct alternate.
View answer and explanation
Correct option: D
I, II, III and IV
Option D (I, II, III and IV) is correct because both H+ and OH- can be hydrated, forming the given hydrated species.
27 Medium
27 of 401
Application +4 | -1
A certain buffer solution contains equal concentration of X−{X^ - } and HX.\mathrm{H}X. The Ka{K_a} for HX.is10.The\mathrm{H}X. is 10. ThepH$ of the buffer is:
View answer and explanation
Correct option: B
8
Option B (8) 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, which is 8 in this case.
28 Medium
28 of 401
Analyzing +4 | -1
Which of the given salt solution will has minimum pH\mathrm{pH} ?
View answer and explanation
Correct option: C
NH4Cl\mathrm{NH}_4 \mathrm{Cl}
Option C is correct because NH4Cl is a salt of a weak base and strong acid, resulting in an acidic solution with a pH less than 7.
29 Medium
29 of 401
Analyzing +4 | -1
STATEMENT-1 : For the reaction A(g)⇌B(g)+C(g),Kp=1 atmA(g) \rightleftharpoons B(g)+C(g), K_p=1 \mathrm{~atm}. If we start with equal moles of all gases at 9 atm9 \mathrm{~atm} of initial pressure, then at equilibrium partial pressure of AA increases.

STATEMENT-2: Reaction quotient Qp>KpQ_p>K_p hence equilibrium shifts in backward direction.
View answer and explanation
Correct option: A
If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1
Option A (If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1) is correct because the reaction quotient is greater than Kp, causing the equilibrium to shift backward and increase the partial pressure of A.
30 Medium
30 of 401
Application +4 | -1
100 ml100 \,\mathrm{ml} of 0.2 M HCl0.2 \,\mathrm{M} \,{HCl} is reacted with 400 ml400 \,\mathrm{ml} of 0.1 M NaOH0.1 \,\mathrm{M}\, {NaOH} solution. The pH{pH} of the resultant solution becomes
View answer and explanation
Correct option: B
12.6
Option B (12.6) is correct because after reacting 100 mL of 0.2 M HCl with 400 mL of 0.1 M NaOH, the pH of the resulting solution is 12.6.
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