Skip to main content
Back to Equilibrium

Your Progress

0 / 146 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • CHEMISTRY • Equilibrium

Equilibrium practice · Medium

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 31-60 of 146 questions
31 Medium
31 of 146
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.
32 Medium
32 of 146
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.
33 Medium
33 of 146
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}.
34 Medium
34 of 146
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.
35 Medium
35 of 146
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.
36 Medium
36 of 146
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.
37 Medium
37 of 146
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.
38 Medium
38 of 146
Application +4 | -1
When 10−8{10^{ - 8}} mole of HCl\mathrm{HCl} is dissolved in one litre of water, the pH\mathrm{pH} of the solution will be:
View answer and explanation
Correct option: D
Below 7
Option D is correct because adding a small amount of acid to water will lower the pH, but since the amount is very small, the pH will be slightly below 7.
39 Medium
39 of 146
Application +4 | -1
A physician wishes to prepare a buffer solution at pH=3.58\mathrm{pH} = 3.58 that efficiently resists a change in pH\mathrm{pH} yet contains only small conc. of the buffering agents. Which one of the following weak acid together with its sodium salt would be best to use?
View answer and explanation
Correct option: D
Acetoacetic acid (pKa=3.58)\left( {\mathrm{p}{K_a} = 3.58} \right)
The correct answer is D (Acetoacetic acid (pKa = 3.58)) because the best buffer is one where the pKa is closest to the desired pH.
40 Medium
40 of 146
Understanding +4 | -1
In which of the following cases, does not reaction go farthest to completion?
View answer and explanation
Correct option: B
K=10−2K = {10^{ - 2}}
Option B is correct because a smaller equilibrium constant (K) indicates that the reaction does not proceed far towards completion.
41 Medium
41 of 146
Application +4 | -1
When 20  g20\;g of CaCO3\mathrm{CaC}{\mathrm{O}_3} were put into 10  litre10\;\mathrm{ }litre flask and heated to 800∘C800^\circ C, 35%35\% of CaCO3\mathrm{CaC}{\mathrm{O}_3} remained unreacted at equilibrium. Kp{K_p} for decomposition of CaCO3\mathrm{CaC}{\mathrm{O}_3} is :
View answer and explanation
Correct option: A
1.145  atm1.145\mathrm{ }\;atm
Option A is correct because it calculates the Kp for the decomposition of CaCO3 using the given percentage of unreacted CaCO3 at equilibrium.
42 Medium
42 of 146
Understanding +4 | -1
For the gaseous phase reaction, 2NO⇋N2+O22NO\leftrightharpoons{N_2} + \mathrm{ }{O_2}, ΔH∘=−43.5  k  cal  mol−1,\Delta {H^ \circ } = \mathrm{ } - \mathrm{ }43.5\mathrm{ }\;k\;cal\mathrm{ }\;mo{l^{ - 1}},  which statement is correct for, N2(g)+O2(g)⇋2NO(g)?{N_2}\left( g \right)\mathrm{ } + \mathrm{ }{O_2}\left( g \right)\mathrm{ }\leftrightharpoons\mathrm{ }2NO\left( g \right)?
View answer and explanation
Correct option: C
KK decreases as temperature decreases
Option C (KK decreases as temperature decreases) is correct because the forward reaction is exothermic, so decreasing the temperature will favor the reactants, thus decreasing the equilibrium constant for the reverse reaction.
43 Medium
43 of 146
Analyzing +4 | -1
For the following Assertion and Reason the correct option is
Assertion: The pH of water increases with increase in temperature.
Reason: The dissociation of water into H+{H^ + } and OH−O{H^ - } is an exothermic reaction.
View answer and explanation
Correct option: C
Both Assertion and Reason are false
Option C (Both Assertion and Reason are false) is correct because the pH of water decreases with an increase in temperature, and the dissociation of water is an endothermic reaction.
44 Medium
44 of 146
Analyzing +4 | -1
The following reactions are known to occur in the body,

CO2+H2O⇋H2CO3⇋H++HCO3−\mathrm{C}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O}\leftrightharpoons{\mathrm{H}_2}\mathrm{C}{\mathrm{O}_3}\leftrightharpoons{\mathrm{H}^ + } + \mathrm{HCO}_3^ -

If CO2\mathrm{C}{\mathrm{O}_2} escapes from the system, then:
View answer and explanation
Correct option: B
Hydrogen ion concentration will diminish
Option B (Hydrogen ion concentration will diminish) is correct because if CO2 escapes, the equilibrium shifts to the left, reducing H2CO3 and consequently H+ concentration.
45 Medium
45 of 146
Application +4 | -1
pH of 0.2 M aqueous solution of sodium benzoate will be (PKaP{K_a} of benzoic acid = 4.2)
View answer and explanation
Correct option: B
8.758.75
Option B (8.75) is correct because it is calculated using the hydrolysis of salt of weak acid and strong base formula.
46 Medium
46 of 146
Application +4 | -1
The concentration of CO2\mathrm{C}{\mathrm{O}_2} be in equilibrium with 2.5×10−2  mol  litre−1  of  CO  at  100∘C2.5 \times {10^{ - 2}}\;\mathrm{mol}\;\mathrm{ litr}{\mathrm{e}^{ - 1}}\;\mathrm{of }\;\mathrm{CO}\;\mathrm{at}\,\,100^\circ C for the reaction : FeO(s)  +  CO(g)   ⇋  Fe(s)+CO2(g);  Kc=5.0\mathrm{FeO}\left( s \right)\; + \,\,\mathrm{CO}\left( \mathrm{g} \right)\;\,\leftrightharpoons\;\mathrm{Fe}\left( s \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right);\;{K_c} = 5.0
View answer and explanation
Correct option: D
0125  M0125\;M
Option D (0.125 M) is correct because it is calculated using the equilibrium constant expression and the given concentration of CO.
47 Medium
47 of 146
Analyzing +4 | -1
STATEMENT I : pH value of acidic buffer solution change, k if bufler solution is diluted upto very larger extent.

STATEMENT II : [H+]\left[H^{+}\right] decrease due to change in concentration as well as α\alpha and decrease in concentration is more as compare to increase in α\alpha.
View answer and explanation
Correct option: A
Statement I, II are correct
Option A (Statement I, II are correct) is correct because upon significant dilution of a buffer, the pH can change because the concentrations of the acid and conjugate base become very low, affecting the buffer capacity and the degree of dissociation.
48 Medium
48 of 146
Application +4 | -1
100  mL  of  0.01M100\;\mathrm{ }mL\;\mathrm{ }of\;\mathrm{ }0.01\mathrm{ }M solution of NaOH\mathrm{NaOH} is diluted to 1  dm31\;\mathrm{d}{\mathrm{m}^3}. What is the pH of the diluted solution?
View answer and explanation
Correct option: B
11
Option B (11) is correct because diluting the NaOH solution decreases the hydroxide concentration, leading to a pOH of 3 and therefore a pH of 11.
49 Medium
49 of 146
Analyzing +4 | -1
The decreasing order of strength of the bases,

OH−,NH2−,H−C≡C−  and  CH3−CH2−,  is\mathrm{O}{\mathrm{H}^ - },\mathrm{NH}_2^ - ,\mathrm{H}-\mathrm{C} \equiv {\mathrm{C}^ - }\;and\;\mathrm{C}{\mathrm{H}_3}-\mathrm{CH}_2^ - ,\;is
View answer and explanation
Correct option: A
CH3−CH2−>NH2−>H−C≡C−>OH−\mathrm{C}{\mathrm{H}_3}-\mathrm{CH}_2^ - > N\mathrm{H}_2^ - > H-C \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.
50 Medium
50 of 146
Remembering +4 | -1
The internal and external indicator among the following respectively are:
View answer and explanation
Correct option: A
KMnO4\mathrm{KMnO}_4, phenolphthalein
Option B (Phenolphthalein, KMnO4\mathrm{KMnO}_4) is correct because phenolphthalein is an internal indicator and KMnO4\mathrm{KMnO}_4 can act as a self-indicator.
51 Medium
51 of 146
Analyzing +4 | -1
Arrange NH4+,H2O,H3O+,HF  and  OH−\mathrm{NH}_4^ + ,{\mathrm{H}_2}\mathrm{O},{\mathrm{H}_3}{\mathrm{O}^ + },\mathrm{HF }\;\mathrm{and}\;\mathrm{ O}{\mathrm{H}^ - } in increasing order of acidic nature
View answer and explanation
Correct option: C
OH−<H2O<NH4+<HF<H3O+\mathrm{O}{\mathrm{H}^ - } < {\mathrm{H}_2}\mathrm{O} < N\mathrm{H}_4^ + < HF < {\mathrm{H}_3}{\mathrm{O}^ + }
Option C is correct. The acidity increases as follows: OH- < H2O < NH4+ < HF < H3O+. This is based on the stability of the conjugate base and electronegativity of the central atom.
52 Medium
52 of 146
Application +4 | -1
4 moles of A is mixed with 4 moles of B, at equilibrium for the reaction A+BC+DA + BC + D, 2 moles of C and D is formed. The equilibrium constant for the reaction will be
View answer and explanation
Correct option: C
1
Option C (1) is correct because the equilibrium constant is calculated as the ratio of products to reactants, and in this case, the change in moles is equal for all species.
53 Medium
53 of 146
Analyzing +4 | -1
In a reaction at equilibrium ′X′'X' mole of the reactant AA decompose to give 1  mole1\;mole each of C  and  DC\;and\;D . If the fraction of AA decomposed at equilibrium is independent of initial concentration of AA, then the value of ′X′'X' is :
View answer and explanation
Correct option: C
2
Option C (2) is correct because for the fraction of A decomposed to be independent of the initial concentration, the reaction must be first order, implying X=2.
54 Medium
54 of 146
Application +4 | -1
Which of the following has PH{P^H} is equal to near about one
View answer and explanation
Correct option: D
75mℓ  M5  HCℓ+25ml  M5  NaOH75m\ell \,\,\frac{M}{5}\,\,HC\ell + 25ml\,\,\frac{M}{5}\,\,NaOH
Option D (75mℓM5\HCℓ+25mlM5\NaOH75m\ell \frac{M}{5}\HC\ell + 25ml\frac{M}{5}\NaOH) is correct because it results in an excess of HCl that gives a pH near 1 after calculation.
55 Medium
55 of 146
Remembering +4 | -1
What is equilibrium expression for the reaction, P4(s)+5O2(g)⇋P4O10(s)  ?{P_4}(s) + 5{O_2}(g)\leftrightharpoons{P_4}{O_{10}}(s)\,\,?
View answer and explanation
Correct option: B
Kc=1[O2]5{K_c} = \frac{1}{{{{\left[ {{O_2}} \right]}^5}}}
Option B (Kc=1[O2]5{K_c} = \frac{1}{{{{\left[ {{O_2}} \right]}^5}}}) is correct because solids are not included in the equilibrium expression.
56 Medium
56 of 146
Remembering +4 | -1
For the system;&nbsp;3A+2B⇋  C3A + 2B\leftrightharpoons\;C,the expression for equilibrium constant is
View answer and explanation
Correct option: B
[C][A]3[B]2\frac{{\left[ C \right]}}{{{{\left[ A \right]}^3}{{\left[ B \right]}^2}}}
Option B is correct because the equilibrium constant is the ratio of product concentration to reactant concentrations, each raised to the power of their stoichiometric coefficients.
57 Medium
57 of 146
Understanding +4 | -1
A cylinder fitted with a movable piston contains liquid water in equilibrium with water vapour at 250C. Which operation result in a decrease in the equilibrium vapour pressure?
View answer and explanation
Correct option: D
Dissolving salt in the water
Option D is correct because dissolving salt in water lowers the vapor pressure according to Raoult's Law.
58 Medium
58 of 146
Application +4 | -1
The equilibrium constant for the reaction between CH4(g)C{H_4}\left( g \right) and H2S(g){H_2}S\left( g \right) to form CS2(g)C{S_2}\left( g \right) and H2(g){H_2}\left( g \right) , at 1173 K is 3.6. For the following composition of the reaction mixture, [CH4]=1.0 M, [H2S]=1.20M   [CS2]=2.0M,  [H2]=1.8M\left[ {C{H_4}} \right] = 1.0\,M,\,\left[ {{H_2}S} \right] = 1.20M\,\,\,\left[ {C{S_2}} \right] = 2.0M,\,\,\left[ {{H_2}} \right] = 1.8M decide which of the following option is correct? [CH4(g)+2H2S(g)CS2(g)+4H2(g)]\left[ {C{H_4}\left( g \right) + 2{H_2}S\left( g \right)C{S_2}\left( g \right) + 4{H_2}\left( g \right)} \right]
View answer and explanation
Correct option: C
Reaction will shift to form more H2S{H_2}S
Option C is correct because the reaction quotient Q is greater than Kc, indicating that the reaction will shift to the left, forming more H2S.
59 Medium
59 of 146
Understanding +4 | -1
The buffering action of an acidic buffer is maximum when its pHpH is equal to
View answer and explanation
Correct option: D
pKa\mathrm{p}{K_a}
Option D (pKa\mathrm{p}{K_a}) is correct because the buffering action is maximum when the pHpH is equal to the pKapK_a of the weak acid.
60 Medium
60 of 146
Understanding +4 | -1
For which of the following reactions,Kp=Kc{K_p} = {K_c} ?
View answer and explanation
Correct option: B
N2+O2⇋2NO{\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO}
Option B (N2+O2⇋2NO{\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO}) is correct because in this reaction, the number of moles of gaseous reactants equals the number of moles of gaseous products, making Δn=0\Delta n = 0.
Revision notes and key takeaways

Review Equilibrium in your textbook or class notes, then use this set to check your understanding.

Showing 31-60 of 146 questions
0 /146
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.