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NEET • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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331 Hard
331 of 408
Application +4 | -1
Equivalent amounts of H2  and  I2{\mathrm{H}_2}\;\mathrm{and}\;{\mathrm{I}_2} are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen can be converted to H,  the  Kc\mathrm{H},\;\mathrm{the}\;{K_c} at this temperature is
View answer and explanation
Correct option: A
64
Option A (64) is correct because the equilibrium constant is calculated based on the percentage of hydrogen converted to HI.
332 Hard
332 of 408
Understanding +4 | -1
The thermal dissociation of calcium carbonate showing heterogeneous equilibrium is
CaCO3( s)⇌CaO(s)+CO2( g){CaCO}_3({~s}) \rightleftharpoons {CaO}({s})+{CO}_2({~g})

For this reactions which of the following is/are true

(i) Kc′=[CO2( g)]{K}_{{c}}^{\prime}=\left[{CO}_2({~g})\right]

(ii) Kp=pCO2{Kp}={pCO}_2

(iii) [CaCO3\left[{CaCO}_3\right. (s)] and [CaO(s)][{CaO}({s})] are both constant

(iv) [CO2( g)]\left[{CO}_2({~g})\right] is constant
View answer and explanation
Correct option: B
(i), (ii) and (iii)
Option B ((i), (ii) and (iii)) is correct because for a heterogeneous equilibrium involving solids and gases, the equilibrium constant expressions only include the gaseous species, and the concentrations of solids are considered constant and incorporated into the equilibrium constant.
333 Hard
333 of 408
Application +4 | -1
The solubility of CaF2\mathrm{Ca}{\mathrm{F}_2} in pure water is 2.3×10−6  mol  dm−32.3 \times {10^{ - 6}}\;\mathrm{mol }\;\mathrm{d}{\mathrm{m}^{ - 3}}. Its solubility product will be
View answer and explanation
Correct option: B
48.66×10−1848.66 \times {10^{ - 18}}
Option D (48.66×10−1848.66 \times {10^{ - 18}}) is correct because it is the solubility product calculated from the given solubility of CaF2CaF_2.
334 Medium
334 of 408
Understanding +4 | -1
Conjugate base and conjugate acid of HSO4−HSO_4^ - respectively are
View answer and explanation
Correct option: C
 H2SO4 and SO42−\,{H_2}S{O_4}\,and\,SO_4^{2 - }
Option B is incorrect. The conjugate acid of HSO4−HSO_4^- is formed by adding a proton (H+H^+), resulting in H2SO4H_2SO_4. The conjugate base is formed by removing a proton, resulting in SO42−SO_4^{2-}.
335 Medium
335 of 408
Analyzing +4 | -1
Dissociation constants of two acids HA and HB are 4×10−104 \times {10^{ - 10}} and 1.8×10−51.8 \times {10^{ - 5}} resp. Whose PH{P^H} value will be higher for a given molarity
View answer and explanation
Correct option: A
HA
Option A (HA) is correct because HA has a smaller dissociation constant, indicating it's a weaker acid. Weaker acids have higher pH values at a given molarity.
336 Hard
336 of 408
Analyzing +4 | -1
For the reaction in equilibrium,

I. CO(g)+12O2(g)⇌CO2(g)\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{CO}_2(g)

II. 2CO(g)+O2(g)⇌2CO2(g)2 \mathrm{CO}(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{CO}_2(g)
Equilibrium  constantFree  energy  chargeK1ΔG1K2ΔG1\begin{array}{cccccccccccccccccccc} {{\text{Equilibrium}}\,\,{\text{constant}}}&{{\text{Free}}\,\,{\text{energy}}\,\,{\text{charge}}} \\ {{K_1}}&{\Delta {G_1}} \\ {{K_2}}&{\Delta {G_1}} \end{array}
Then
View answer and explanation
Correct option: C
K2=K12,ΔG2=2ΔG1K_2=K_1^2, \Delta G_2=2 \Delta G_1
Option C (K2=K12,ΔG2=2ΔG1K_2=K_1^2, \Delta G_2=2 \Delta G_1) is correct because doubling the reaction squares the equilibrium constant and doubles the free energy change.
337 Easy
337 of 408
Remembering +4 | -1
The solvent which is neither proton donor nor proton acceptor is called:
View answer and explanation
Correct option: C
Aprotic
Option C (Aprotic) is correct because aprotic solvents neither donate nor accept protons, unlike protic or amphoteric solvents.
338 Hard
338 of 408
Understanding +4 | -1
When NaNO3{NaNO}_3 is heated in a closed vessel, oxygen is liberated and NaNO2{NaNO}_2 is left behind, then which of the following statements is correct at equilibrium?

NaNO3( s)⇌NaNO2( s)+12O2( g),ΔH=+ve{NaNO}_3({~s}) \rightleftharpoons {NaNO}_2({~s})+\frac{1}{2} {O}_2({~g}), \Delta {H}=+{ve}
View answer and explanation
Correct option: C
Increase in temperature favours forward reaction
Option C is correct because the reaction is endothermic (ΔH=+ve\Delta H = +ve), so increasing the temperature favors the forward reaction according to Le Chatelier's principle.
339 Hard
339 of 408
Analyzing +4 | -1
A solution is 0.1 M in Cl−C{l^ - } and 0.001 M in CrO42−CrO_4^{2 - }. Solid AgNO3AgN{O_3} is gradually added to it. Assuming that the addition does not change in volume and Ksp(AgCl)=1.7×10−10 M2{K_{sp}}(AgCl) = 1.7 \times {10^{ - 10}}\,{M^2} and Ksp(Ag2CrO4)=1.9×10−12 M3{K_{sp}}(A{g_2}Cr{O_4}) = 1.9 \times {10^{ - 12}}\,{M^3}
Select correct statement from the following
View answer and explanation
Correct option: D
AgClAgCl will precipitate first as the amount of Ag+A{g^ + } needed to precipitate is low.
Option D (AgClAgCl will precipitate first as the amount of Ag+A{g^ + } needed to precipitate is low) is correct because it requires a lower concentration of Ag+Ag^+ to reach the KspK_{sp} for AgClAgCl compared to Ag2CrO4Ag_2CrO_4.
340 Hard
340 of 408
Analyzing +4 | -1
Equilibrium constants K1{K_1} and K2{K_2}  for the following equilibria are related as :
NO(g)+12O2(g)⇋NO2(g);K1\mathrm{NO}\left( \mathrm{g} \right) + \frac{1}{2}{\mathrm{O}_2}\left( \mathrm{g} \right)\leftrightharpoons\mathrm{N}{\mathrm{O}_2}\left( \mathrm{g} \right);{K_1}
2NO2(g)⇋2NO(g)+O2(g);K22\mathrm{N}{\mathrm{O}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{NO}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right);{K_2}
View answer and explanation
Correct option: A
K2=1K12{K_2} = \frac{1}{{K_1^2}}
Option A (K₂ = 1/K₁²) is correct because the second reaction is the reverse of the first reaction multiplied by 2. Reversing a reaction inverts the equilibrium constant, and multiplying by a factor raises the equilibrium constant to that power.
341 Hard
341 of 408
Application +4 | -1
pH\mathrm{pH} of a mixture of 1M1 \mathrm{M} benzoic acid (pKa=4.20)\left(\mathrm{p} K_a=4.20\right) and 1M C6H5COONa1 \mathrm{M}\, {C}_6 {H}_5 {COONa} is 4.5 . In 300 mL300 \mathrm{~mL} buffer, benzoic acid is [log⁡2=0.3][\log 2=0.3]
View answer and explanation
Correct option: C
100 mL100 \mathrm{~mL}
Option C (100 mL100 \mathrm{~mL}) is correct because it correctly uses the Henderson-Hasselbalch equation and the given pH to determine the volume of benzoic acid in the buffer solution.
342 Medium
342 of 408
Application +4 | -1
In the reactions, A+2B⇋2CA + 2B\leftrightharpoons2C, if 2 moles of AA, 3.0 moles of BB and 2.0 moles of C are placed in a 2 L flask and the equilibrium concentration of CC is 0.5 mol/L, the equilibrium constant (Kc)\left( {{K_c}} \right) for the reactions is
View answer and explanation
Correct option: A
0.210.21
Option A (0.210.21) is correct because after calculating the equilibrium concentrations, the equilibrium constant Kc is found to be approximately 0.21.
343 Medium
343 of 408
Understanding +4 | -1
Any precipitate is formed when
View answer and explanation
Correct option: D
The value of ionic product is greater than the value of solubility product
Option D (The value of ionic product is greater than the value of solubility product) is correct because a precipitate forms when the solution is supersaturated, meaning the ionic product exceeds the solubility product.
344 Hard
344 of 408
Understanding +4 | -1
Blue colour disappears in which solution by passing SO2{SO}_2
View answer and explanation
Correct option: B
I2+{I}_2+ starch
Option B (I2+{I}_2+ starch) is correct because SO2{SO}_2 reduces iodine (I2{I}_2) to iodide (I−I^−), causing the blue color of the iodine-starch complex to disappear.
345 Hard
345 of 408
Understanding +4 | -1
Consider the equilibrium

CO(g)+3H2(g)⇌CH4(g)+H2O(g)CO(g) + 3H_2(g) \rightleftharpoons CH_4(g) + H_2O(g)

If the pressure applied over the system increases by
two fold at constant temperature then

(A) Concentration of reactants and products
increases.

(B) Equilibrium will shift in forward direction.

(C) Equilibrium constant increases since concentration of products increases.

(D) Equilibrium constant remains unchanged as concentration of reactants and products remain same.

Choose the correct answer from the options given below:
View answer and explanation
Correct option: A
(A) and (B) only
Option A ((A) and (B) only) is correct because increasing the pressure increases the concentration of both reactants and products, and the equilibrium shifts in the forward direction due to the decrease in the number of moles.
346 Easy
346 of 408
Application +4 | -1
The pHpH of an aqueous solution of CH3COONa\mathrm{C}{\mathrm{H}_3}\mathrm{COONa} of concentration C(M)C\left( \mathrm{M} \right) is given by
View answer and explanation
Correct option: D
12pKw+12pKa+12log⁡C\frac{1}{2}p{K_w} + \frac{1}{2}p{K_a} + \frac{1}{2}\log C
Option D is correct because the pH of a salt of a weak acid and a strong base is given by the formula pH = 1/2 pKw + 1/2 pKa + 1/2 log C.
347 Medium
347 of 408
Application +4 | -1
At constant temperature in one litre vessel, when the reaction,

2SO3(g)  ⇋  2SO2(g)+O2(g)2\mathrm{S}{\mathrm{O}_3}\left( \mathrm{g} \right)\;\leftrightharpoons\;2\mathrm{S}{\mathrm{O}_2}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right) is at equilibrium, the SO2\mathrm{S}{\mathrm{O}_2} concentration is 0.6  M0.6\;M, initial concentration of SO3\mathrm{S}{\mathrm{O}_3} is 1M1M. The equilibrium constant is :
View answer and explanation
Correct option: D
0.6750.675
Option D is correct because it accurately calculates the equilibrium constant based on the given concentrations and stoichiometry of the reaction.
348 Hard
348 of 408
Analyzing +4 | -1
Equilibrium constant (K)({K}) for the reaction Ni(s)+4CO(g)⇌Ni(CO)4( g){Ni}({s})+4 {CO}({g}) \rightleftharpoons {Ni}({CO})_4({~g}) can be written in terms of

(1) Ni(s)+2CO2( g)+2C(s)⇌Ni(CO)4( g){Ni}({s})+2 {CO}_2({~g})+2 {C}(\mathrm{s}) \rightleftharpoons {Ni}({CO})_4({~g}); equilibrium constant =K1={K}_{{1}}.

(2) CO2( g)+C(s)⇌2CO(g){CO}_2({~g})+{C}({s}) \rightleftharpoons 2 {CO}({g}); equilibrium constant =K2={K}_2.

What is the relation between K,K2{K}, {K}_2 and K2{K}_2 ?
View answer and explanation
Correct option: A
K=(K1)/(K2)2{K}=\left({K}_1\right) /\left({K}_2\right)^2
Option A (K=(K1)/(K2)2{K}=\left({K}_1\right) /\left({K}_2\right)^2) is correct because it represents the correct relationship between the equilibrium constants when manipulating the given reactions to obtain the target reaction.
349 Hard
349 of 408
Analyzing +4 | -1
How much water should be added to 300 mL300 \mathrm{~mL} of 0.2M0.2 \mathrm{M} solution of acetic acid (Ka=1.8×10−5)\left(K_a=1.8 \times 10^{-5}\right) to increase its degree of dissociation two times?
View answer and explanation
Correct option: C
0.9 litre
Option C (0.9 litre) is correct because it involves calculating the dilution required to double the degree of dissociation of a weak acid using the Ostwald's dilution law.
350 Medium
350 of 408
Remembering +4 | -1
Urea is
View answer and explanation
Correct option: A
Monoacidic base
Option A (Monoacidic base) is correct because urea can accept one proton due to the presence of two nitrogen atoms, each with a lone pair of electrons.
351 Easy
351 of 408
Understanding +4 | -1
Which of the following is a wrong statement about equilibrium state?
View answer and explanation
Correct option: C
Catalysts increase value of equilibrium constant
Option C is correct because catalysts affect the rate of reaction but do not change the equilibrium constant.
352 Easy
352 of 408
Understanding +4 | -1
Which salt will give maximum pH in aqueous solution :
View answer and explanation
Correct option: C
CH3COONaC{H_3}COONa
Option C (CH3COONaC{H_3}COONa) is correct because it is a salt of a weak acid and a strong base, which will undergo hydrolysis to produce a basic solution with a higher pH compared to the other options.
353 Easy
353 of 408
Understanding +4 | -1
If 1  M  CH3COONa1\;M\;\mathrm{C}{\mathrm{H}_3}\mathrm{COONa} is added to 1  M  CH3COOH1\;M\;\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}
View answer and explanation
Correct option: A
pH\mathrm{pH} of the solution increases
Option A (pH\mathrm{pH} of the solution increases) is correct because adding CH3COONa, a salt of a weak acid, to CH3COOH, a weak acid, creates a buffer solution. The addition of the salt increases the concentration of the conjugate base, shifting the equilibrium and increasing the pH.
354 Easy
354 of 408
Remembering +4 | -1
Methyl orange is an indicator in acid-alkali titration. It gives:
View answer and explanation
Correct option: D
Yellow colour in alkaline medium and red colour in acid medium
Option D (Yellow colour in alkaline medium and red colour in acid medium) is correct because methyl orange exhibits different colors in acidic and alkaline solutions.
355 Hard
355 of 408
Application +4 | -1
The percentage error in [H+]\left[ {{\mathrm{H}^ + }} \right] made by neglecting the ionisation of water in 1.0×10−6  M  NaOH1.0 \times {10^{ - 6}}\;M\;\mathrm{NaOH} is:
View answer and explanation
Correct option: A
1%1\mathrm{\% }
Option A (1%) is correct because the percentage error is calculated by comparing the H+ concentration from water to the total H+ concentration.
356 Hard
356 of 408
Application +4 | -1
The Ksp{K_{sp}} of PbCO3\mathrm{PbC}{\mathrm{O}_3} and MgCO3\mathrm{MgC}{\mathrm{O}_3} are 1.5×10−151.5 \times {10^{ - 15}} and 1×10−151 \times {10^{ - 15}} respectively at 298  K298\;K. The concentration of Pb2+\mathrm{P}{\mathrm{b}^{2 + }} ions in a saturated solution containing MgCO3\mathrm{MgC}{\mathrm{O}_3} and PbCO3\mathrm{PbC}{\mathrm{O}_3} is:
View answer and explanation
Correct option: B
3×10−8  M3 \times {10^{ - 8}}\;M
Option D (Ka3=Ka1×Ka2\mathrm{K}_{\mathrm{a}_3}=\mathrm{K}_{\mathrm{a}_1} \times \mathrm{K}_{\mathrm{a}_2}) is correct because the overall ionization constant is the product of the stepwise ionization constants.
357 Easy
357 of 408
Remembering +4 | -1
Ostwald’s dilution law is applicable for
View answer and explanation
Correct option: A
Weak electrolyte
Option A (Weak electrolyte) is correct because Ostwald's dilution law is specifically applicable to weak electrolytes due to their partial dissociation in solution.
358 Medium
358 of 408
Understanding +4 | -1
When CaCO3\mathrm{CaC}{\mathrm{O}_3} is heated at a constant temperature in a closed container, the pressure due to CO2\mathrm{C}{\mathrm{O}_2} produced will:
View answer and explanation
Correct option: C
Remain constant so long as temperature is constant
The correct answer is C because at a constant temperature, the equilibrium pressure of CO2 will remain constant regardless of the amount of CaCO3 or the container size.
359 Medium
359 of 408
Application +4 | -1
2  ICl→I2+Cl2;Kc=0.142\,\,ICl \to {I_2} + C{l_2};Kc = 0.14 Initial concentration of ICl is 0.6 M, then equilibrium concentration of I2{I_2} is
View answer and explanation
Correct option: B
0.128 M
Option B (0.128 M) is correct because it represents the equilibrium concentration of I2 calculated using the equilibrium constant expression.
360 Hard
360 of 408
Application +4 | -1
A mixture of H2\mathrm{H}_2 and I2\mathrm{I}_2 (vapour) in molecular proportion 2 : 3 was heated at 440∘C440^{\circ} \mathrm{C} till the reaction

H2( g)+l2( g)⇌2H(g)\mathrm{H}_2(\mathrm{~g})+\mathrm{l}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{H}(\mathrm{g})

reached equilibrium state. Calculate the percentage of I2\mathrm{I}_2 converted into HI. (Kc\left(\mathrm{K}_{\mathrm{c}}\right. at 440∘C440^{\circ} \mathrm{C} is 0.02 and x\mathrm{x} is small compared to unity)
View answer and explanation
Correct option: B
5.77%5.77 \%
Option D (45717) is incorrect. This question seems to have an error. The calculated Kp value is much smaller than any of the options. There might be a typo in the question or options.
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