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NEET • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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301 Hard
301 of 408
Understanding +4 | -1
Ionisation of weak acid can be calculated by the formula
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because both formulas are valid approximations for calculating the ionization of a weak acid.
302 Easy
302 of 408
Analyzing +4 | -1
Consider the following gaseous equilibrium with equilibrium constants K1{K_1} and K2{K_2} respectively
SO2(g)+12O2(g)⇋SO3(g)2SO3(g)⇋2SO2(g)+O2(g)\begin{array}{l} S{O_2}(g) + \frac{1}{2}{O_2}(g)\underset{{}}{\overset{{}}{\leftrightharpoons}}S{O_3}(g)\\ 2S{O_3}(g)\underset{{}}{\overset{{}}{\leftrightharpoons}}2S{O_2}(g) + {O_2}(g) \end{array}
The equilibrium constants are related as
View answer and explanation
Correct option: A
K12=1K2K_1^2 = \frac{1}{{{K_2}}}
Option A (K12=1K2K_1^2 = \frac{1}{{{K_2}}}) is correct because the second reaction is the reverse of the first reaction multiplied by 2, so the equilibrium constant is the inverse square of the first reaction's equilibrium constant.
303 Medium
303 of 408
Understanding +4 | -1
In the hydrolysis of a salt of weak acid and weak base, the hydrolysis constant Kh{K_h} is equal to
View answer and explanation
Correct option: C
KwKa.Kb\frac{{{K_w}}}{{{K_a}.{K_b}}}
Option C (KwKa.Kb\frac{{{K_w}}}{{{K_a}.{K_b}}}) is correct because the hydrolysis constant for a salt of a weak acid and weak base is given by this formula.
304 Medium
304 of 408
Application +4 | -1
Partial pressure of A,B,CA,B,C and DD on the basis of gaseous system, A+2B  ⇋  C+3D,A + 2B\;\leftrightharpoons\;C + 3D, are A=0.20,  B=0.10,  C=0.30A = 0.20,\;B = 0.10,\;C = 0.30 and D=0.50  atmD = 0.50\;\mathrm{atm}. The numerical value of equilibrium constant is
View answer and explanation
Correct option: B
18.7518.75
Option B (18.7518.75) is correct because it's the calculated value of the equilibrium constant KpK_p based on the given partial pressures.
305 Hard
305 of 408
Understanding +4 | -1
For a reaction in equilibrium :
View answer and explanation
Correct option: C
The rate of forward reaction is equal to the rate of backward reaction
Option C is correct because at equilibrium, the rate of the forward reaction is equal to the rate of the backward reaction.
306 Hard
306 of 408
Application +4 | -1
0.150.15 mole of pyridinium chloride has been added into 500 mL500 \mathrm{~mL} of 0.2 M0.2 \mathrm{~M} pyridine solution. Find the pH\mathrm{pH} and [OH−]\left[\mathrm{OH}^{-}\right] ion concentration in the resulting solution. Kb=1.5×10−9M\mathrm{K}_{\mathrm{b}}=1.5 \times 10^{-9} \mathrm{M} (assuming no change in volume).
View answer and explanation
Correct option: A
10−910^{-9}
Option C (10−1010^{-10}) is correct because it correctly calculates the hydroxide ion concentration in the buffer solution using the given Kb value and the concentrations of pyridine and pyridinium chloride.
307 Medium
307 of 408
Understanding +4 | -1
For the reaction, H2(g)+I2(g)→2HI(g){\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right) \to 2\mathrm{HI}\left( \mathrm{g} \right), the equilibrium constant Kp{K_p} changes with
View answer and explanation
Correct option: D
Temperature
Option D (Temperature) is correct because the equilibrium constant Kp{K_p} is temperature-dependent. Changing the temperature affects the equilibrium position and thus the value of Kp{K_p}.
308 Hard
308 of 408
Analyzing +4 | -1
N2O4(g)⇌2NO2(g){N}_2 {O}_4(\mathrm{g}) \rightleftharpoons 2 {NO}_2(\mathrm{g})

For the reaction, if the percentage dissociation of N2O4{N}_2 {O}_4 are 25%,50%,75%25 \%, 50 \%, 75 \% and 100%100 \%, then the sequence of observed vapour densities will be (d1, d2, d3\left({d}_1, {~d}_2, {~d}_3\right. and d4{d}_4 are vapour density at 25%,50%25 \%, 50 \%, 75%75 \% and 100%100 \% dissociation respectively)
View answer and explanation
Correct option: A
d1>d2>d3>d4d_1>d_2>d_3>d_4
Option A (d1>d2>d3>d4d_1>d_2>d_3>d_4) is correct because as the percentage dissociation increases, the number of moles increases, leading to a decrease in vapor density.
309 Easy
309 of 408
Analyzing +4 | -1
Which of the following has lowest PH{P^H} value
View answer and explanation
Correct option: D
NH4ClN{H_4}Cl
Option D (NH4ClN{H_4}Cl) is correct because it is a salt of a strong acid and a weak base, resulting in an acidic solution with the lowest pH.
310 Medium
310 of 408
Remembering +4 | -1
The pHpH of tears coming out of a person’s eye is:
View answer and explanation
Correct option: A
7.47.4
Option A (7.4) is correct because the pH of tears is approximately 7.4, which is slightly basic and similar to the pH of blood.
311 Easy
311 of 408
Understanding +4 | -1
The stronger Bronsted base is:
View answer and explanation
Correct option: A
ClO−\mathrm{Cl}{\mathrm{O}^ - }
Option A (ClO−\mathrm{Cl}{\mathrm{O}^ - }) is the strongest Bronsted base because it is the conjugate base of the weakest acid (HClO) among the given options.
312 Easy
312 of 408
Analyzing +4 | -1
Which of the following is least likely to behave as Lewis acid?
View answer and explanation
Correct option: D
BF3\mathrm{B}{\mathrm{F}_3}
The correct answer is A (OH−\mathrm{O}{\mathrm{H}^ - }) because it has a negative charge and a lone pair of electrons, making it a strong nucleophile and base rather than a Lewis acid.
313 Easy
313 of 408
Application +4 | -1
For the gaseous reaction, C2H4+H2⇋C2H6,  ΔH=−130  kJ  mol−1{\mathrm{C}_2}{\mathrm{H}_4} + {\mathrm{H}_2}\leftrightharpoons{\mathrm{C}_2}{\mathrm{H}_6},\;\Delta H = - 130\;\mathrm{kJ}\;\mathrm{ mo}{\mathrm{l}^{ - 1}} carried in a closed vessel, the equilibrium concentration of the C2H6{\mathrm{C}_2}{\mathrm{H}_6} can definitely be increased by
View answer and explanation
Correct option: B
Decreasing temperature and increasing pressure
Option B (Decreasing temperature and increasing pressure) is correct because the reaction is exothermic (ΔH<0\Delta H < 0) and involves a decrease in the number of moles of gas, both of which favor the formation of C2H6.
314 Hard
314 of 408
Analyzing +4 | -1
Given, pKbp K_b values
SpeciespKbNH34.74A−4.00B−5.00C−6.00\begin{array}{cccccccccccccccccccc} {{\text{Species}}}&{p{K_b}} \\ {N{H_3}}&{4.74} \\ {{A^ - }}&{4.00} \\ {{B^ - }}&{5.00} \\ {{C^ - }}&{6.00} \end{array}
queous solutions of 0.01M0.01 \mathrm{M} salts of HX\mathrm{HX}

(X=A,B,C)(X=A, B, C) and NH4OH\mathrm{NH}_4 \mathrm{OH} have pH\mathrm{pH} in the order
View answer and explanation
Correct option: B
NH4C<NH4B<NH4A\mathrm{NH}_4 {C} < \mathrm{NH}_4 B < \mathrm{NH}_4 A
Option B (NH4C<NH4B<NH4A\mathrm{NH}_4 {C} < \mathrm{NH}_4 B < \mathrm{NH}_4 A) is correct because pH is inversely proportional to the strength of the conjugate acid, which is related to the pKbpK_b values.
315 Hard
315 of 408
Analyzing +4 | -1
A buffer solution is 0.25 M−CH3COOH+0.15 M−CH3COONa,0.25 \mathrm{~M}-\mathrm{CH}_3 \mathrm{COOH}+0.15 \mathrm{~M}-\mathrm{CH}_3 \mathrm{COONa}, saturated in H2S(0.1 M)\mathrm{H}_2 \mathrm{S}(0.1 \mathrm{~M}) and has [Mn2+]=0.04 M,Ka(CH3COOH)=2.0×10−5,\left[\mathrm{Mn}^{2+}\right]=0.04 \mathrm{~M}, K_{\mathrm{a}}\left(\mathrm{CH}_3 \mathrm{COOH}\right)=2.0 \times 10^{-5}, Ka(H2S)=1.0×10−21K_{\mathrm{a}}\left(\mathrm{H}_2 \mathrm{S}\right)=1.0 \times 10^{-21} and Ksp(MnS)=2.5×10−13.K_{\mathrm{sp}}(\mathrm{MnS})=2.5 \times 10^{-13}.
Which buffer component should be increased in concentration and to which minimum value to just start precipitation of MnS?\mathrm{MnS}?
View answer and explanation
Correct option: C
CH3COONa,1.25 M\mathrm{CH}_3 \mathrm{COONa}, 1.25 \mathrm{~M}
Option C (CH3COONa,1.25 M\mathrm{CH}_3 \mathrm{COONa}, 1.25 \mathrm{~M}) is correct because increasing the concentration of CH3COONa shifts the equilibrium to reduce the sulfide ion concentration, preventing MnS precipitation until a certain concentration is reached.
316 Medium
316 of 408
Application +4 | -1
For the reaction equilibrium,

2NOBr(g)  ⇋  2NO(g)+Br2(g),2\mathrm{NOBr}\left( \mathrm{g} \right)\;\leftrightharpoons\;2\mathrm{NO}\left( \mathrm{g} \right) + \mathrm{B}{\mathrm{r}_2}\left( \mathrm{g} \right), if PBr2=P9{P_{\mathrm{B}{\mathrm{r}_2}}} = \frac{P}{9} at equilibrium and PP is total pressure. The ratio Kp/P{K_p}/P is equal to:
View answer and explanation
Correct option: B
1/811/81
Option B is correct because the Kp/P ratio can be calculated from the given partial pressure of Br2 and the stoichiometry of the reaction.
317 Medium
317 of 408
Analyzing +4 | -1
Which of the following salts is the most basic in aqueous solution?
View answer and explanation
Correct option: B
CH3COOKC{H_3}COOK
Option B (CH3COOKCH_3COOK) is the most basic because it is a salt of a weak acid (CH3COOHCH_3COOH) and a strong base (KOH).
318 Easy
318 of 408
Application +4 | -1
If the equilibrium constant for N2(g) + O2(g)   ⇋   2NO(g) {\mathrm{N}_\mathrm{2}}\left( \mathrm{g} \right)\,\mathrm{ + }\,{\mathrm{O}_\mathrm{2}}\left( \mathrm{g} \right)\,\,\,\leftrightharpoons\,\,\,\mathrm{2NO}\left( \mathrm{g} \right)\, is k, the equilibrium constant for 12N2(g) + 12 O2(g)  ⇋  NO(g)\frac{\mathrm{1}}{\mathrm{2}}{\mathrm{N}_\mathrm{2}}\left( \mathrm{g} \right)\,\mathrm{ + }\,\frac{\mathrm{1}}{\mathrm{2}}\,{\mathrm{O}_\mathrm{2}}\left( \mathrm{g} \right)\,\,\leftrightharpoons\,\,\mathrm{NO}\left( \mathrm{g} \right) will be:
View answer and explanation
Correct option: C
 K12\,{\mathrm{K}^{\frac{1}{2}}}
Option C (K\sqrt{{K}}) is correct because when the reaction is halved, the equilibrium constant is raised to the power of 1/2.
319 Hard
319 of 408
Application +4 | -1
Calculate the partial pressure of carbon monoxide from the following datas

CaCO3C→ΔCaO(g)+CO2  ↑  Kp=8×10−2\mathrm{CaC}{\mathrm{O}_3}\mathrm{C}\xrightarrow{\Delta }\mathrm{CaO}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{O}_2}\; \uparrow \;{K_p} = 8 \times {10^{ - 2}}

O2(g)+C(s)→2CO(g),      Kp=2{\mathrm{O}_2}\left( \mathrm{g} \right) + \mathrm{C}\left( s \right) \to 2\mathrm{CO}\left( \mathrm{g} \right),\;\;\;{K_p} = 2
View answer and explanation
Correct option: B
0.40.4
Option B (0.4) is correct because it is calculated by combining the two equilibrium expressions and solving for the partial pressure of CO.
320 Medium
320 of 408
Analyzing +4 | -1
Statement I :SO2(g)+12O2(g)⇌SO2(g)\mathrm{SO}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightleftharpoons S \mathrm{O}_{2(g)}+heat. Forward rection is favoured at high temperature and low pressure.

Statement II : Reaction is endothermic.
View answer and explanation
Correct option: B
Statement I, II are incorrect
Option B (Statement I, II are incorrect) is correct because the forward reaction is favored at low temperature and high pressure, and the reaction is exothermic, not endothermic.
321 Medium
321 of 408
Understanding +4 | -1
A solution of FeCl3\mathrm{FeC}{\mathrm{l}_3} in water acts as acidic due to:
View answer and explanation
Correct option: C
Hydrolysis of Fe3+\mathrm{F}{\mathrm{e}^{3 + }}
Option C (Hydrolysis of Fe3+\mathrm{F}{\mathrm{e}^{3 + }}) is correct because the ferric ion undergoes hydrolysis, releasing H+{\mathrm{H}^ + } ions and making the solution acidic.
322 Medium
322 of 408
Application +4 | -1
The first and second dissociation constants of an acid
H2A are  1.0×10−5 and  5.0×10−10{\mathrm{H}_2}A\,\mathrm{are}\,\,1.0 \times {10^{ - 5}}\,\mathrm{and}\,\,5.0 \times {10^{ - 10}} respectively. The overall dissociation constant of the acid will be
View answer and explanation
Correct option: C
5.0×10−155.0 \times {10^{ - 15}}
Option C (5.0×10−155.0 \times {10^{ - 15}}) is correct because the overall dissociation constant is the product of the individual dissociation constants.
323 Medium
323 of 408
Remembering +4 | -1
If KcK_c is in the range of _____ appreciable concentrations of both reactants and products are present.
View answer and explanation
Correct option: B
10−310^{-3} to 10310^3
Option B (10−310^{-3} to 10310^3) is correct because it indicates that the concentrations of reactants and products are neither overwhelmingly large nor small, allowing for a measurable equilibrium.
324 Medium
324 of 408
Analyzing +4 | -1
Which of the following will have maximum pH?
View answer and explanation
Correct option: C
M10  NaOH\frac{\mathrm{M}}{{10}}\;\mathrm{NaOH}
Option C (M10  NaOH\frac{\mathrm{M}}{{10}}\;\mathrm{NaOH}) is correct because it represents the most concentrated strong base, leading to the highest pH.
325 Easy
325 of 408
Application +4 | -1
20  mL\mathrm{20}\;\mathrm{mL} of a 0.1  N  HCl0.1\;N\;\mathrm{HCl} is mixed with 20  mL\mathrm{20}\;\mathrm{mL} of a 0.1  N  KOH0.1\;N\;\mathrm{KOH} solution. The pH\mathrm{pH} of the solution would be:
View answer and explanation
Correct option: B
7
Option B is correct because equal volumes and concentrations of a strong acid and strong base will neutralize each other, resulting in a neutral solution with a pH of 7.
326 Medium
326 of 408
Application +4 | -1
For the reaction C(s)+CO2(g)2  CO(g)C(s) + C{O_2}(g)2\,\,CO(g) the partial pressure of CO2C{O_2} and are 2.0 and 4.0 atm resp, at equilibrium. The Kp{K_p} for the reaction is
View answer and explanation
Correct option: C
3.03.0
Option D (32.0) is correct because Kp is calculated as the ratio of the partial pressure of products to reactants, each raised to the power of their stoichiometric coefficients.
327 Easy
327 of 408
Analyzing +4 | -1
In which of the following reactions, increases in the volume at constant temperature do not affect the number of moles at equilibrium?
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because in all given reactions, the number of moles of gaseous reactants and products are different, so a change in volume will affect the equilibrium.
328 Easy
328 of 408
Application +4 | -1
pHpH of 0.05    M  Mg(OH)20.05\;\;M\;\mathrm{Mg}{\left( {\mathrm{OH}} \right)_2} is:
View answer and explanation
Correct option: A
13
Option A (13) is correct because Mg(OH)2 is a strong base that dissociates to give 2 OH- ions, leading to a high pOH and thus a pH close to 13.
329 Medium
329 of 408
Understanding +4 | -1
Which of the following equilibria will shift to right side on increasing the temperature
View answer and explanation
Correct option: C
H2O(g)H2(g)+12O2(g){H_2}O(g){H_2}(g) + \frac{1}{2}{O_2}(g)
Option C (H2O(g)H2(g)+12O2(g){H_2}O(g){H_2}(g) + \frac{1}{2}{O_2}(g)) is correct because increasing the temperature will shift the equilibrium to the right side only if the forward reaction is endothermic.
330 Easy
330 of 408
Application +4 | -1
At 600 , Kp600\,,\,{K_p} for the following reaction is 1 atm.

X(g)   ⇋   Y(g)+Z(g)X\left( \mathrm{g} \right)\,\,\,\leftrightharpoons\,\,\,Y\left( \mathrm{g} \right) + Z\left( \mathrm{g} \right)

At equilibrium, 50 %50\,\% of X(g)X\left( \mathrm{g} \right) is dissociated. The total pressure of the equilibrium system is pp atm. What is the partial pressure (in atm) of X(g)X\left( \mathrm{g} \right) at equilibrium?
View answer and explanation
Correct option: A
1
Option A is correct because, at equilibrium, the partial pressure of X is equal to 1 atm, given the total pressure and the degree of dissociation.
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