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NEET • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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211 Medium
211 of 408
Understanding +4 | -1
C2H5ONa{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{ONa} acts as … in C2H5OH{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}.
View answer and explanation
Correct option: C
Strong base
Option C (Strong base) is correct because C2H5ONa{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{ONa} is the sodium salt of ethanol, and it completely dissociates in ethanol to form ethoxide ions (C2H5O−\mathrm{C}_2{\mathrm{H}_5}\mathrm{O^-}), which are strong bases.
212 Easy
212 of 408
Understanding +4 | -1
The formation of SO3\mathrm{S}{\mathrm{O}_3} takes place according to the following reaction, 2SO2+O2⇋2SO3,  ΔH=−45.2  kcal2\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3},\;\Delta H = - 45.2\;\mathrm{kcal}. The formation of SO3\mathrm{S}{\mathrm{O}_3} is favoured by
View answer and explanation
Correct option: B
Increase in pressure
Option B (Increase in pressure) is correct because, according to Le Chatelier's principle, increasing the pressure favors the side with fewer moles of gas, which is the product side in this reaction.
213 Medium
213 of 408
Understanding +4 | -1
Conjugate acid and conjugate base of HPO42−HPO_4^{2 - } respectively are
View answer and explanation
Correct option: B
H2PO4− and PO43−{H_2}PO_4^ - \,and\,PO_4^{3 - }
Option B (H2PO4− and PO43−{H_2}PO_4^ - \,and\,PO_4^{3 - }) is correct because the conjugate acid is formed by adding a proton and the conjugate base by removing a proton.
214 Medium
214 of 408
Analyzing +4 | -1
XeF6+H2O⇋XeOF4+2HFXe{F_6} + {H_2}O\underset{{}}{\overset{{}}{\leftrightharpoons}}XeO{F_4} + 2HF equilibrium constant K1{K_1} XeO4+XeF6⇋XeOF4+XeO3F2Xe{O_4} + Xe{F_6}\underset{{}}{\overset{{}}{\leftrightharpoons}}XeO{F_4} + Xe{O_3}{F_2} equilibrium constant K2{K_2} then equilibrium constant for the reaction XeO4+2HF⇋XeO3F2+H2OXe{O_4} + 2HF\underset{{}}{\overset{{}}{\leftrightharpoons}}Xe{O_3}{F_2} + {H_2}O will be
View answer and explanation
Correct option: C
K2K1\frac{{{K_2}}}{{{K_1}}}
Option C (K2K1\frac{{{K_2}}}{{{K_1}}}) is correct because the equilibrium constant for the target reaction can be obtained by dividing the second reaction's constant by the first.
215 Medium
215 of 408
Application +4 | -1
The ionization constant of HCN is 4.5×10−104.5\mathrm{ } \times \mathrm{ }{10^{ - 10}} at 298 K. Hydrolysis constant of sodium cyanide is
View answer and explanation
Correct option: C
2.22×10−−52.22\mathrm{ } \times \mathrm{ }{10^{--5}}
Option C is correct because the hydrolysis constant of a salt of a weak acid and strong base is given by Kw/Ka.
216 Easy
216 of 408
Analyzing +4 | -1
The pH\mathrm{pH} of 0.1  M\mathrm{0}\mathrm{.1}\;\mathrm{M} solution of the following salts increases in the order
View answer and explanation
Correct option: B
HCl<NH4Cl<NaCl<NaCN\mathrm{HCl} < N{\mathrm{H}_4}\mathrm{Cl} < NaCl < NaCN
Option B (HCl<NH4Cl<NaCl<NaCN\mathrm{HCl} < N{\mathrm{H}_4}\mathrm{Cl} < NaCl < NaCN) is correct because it represents the increasing order of pH from acidic to basic salts.
217 Medium
217 of 408
Understanding +4 | -1
Inert gas has been added to the following equilibrium system at constant volume SO2(g)+12O2(g)→SO3(g)S{O_2}(g) + \frac{1}{2} {O_2}(g) \to S{O_3}(g)
To which direction will the equilibrium shift?
View answer and explanation
Correct option: C
No effect
Option C (No effect) is correct because adding an inert gas at constant volume does not change the partial pressures of the reactants or products, so the equilibrium position remains unchanged.
218 Hard
218 of 408
Analyzing +4 | -1
For a concentrated solution of a weak electrolyte AxBy{A_x}{B_y} of concentration ‘CC’, the degree of dissociation 'α' is given as
View answer and explanation
Correct option: C
α=(Keq/Cx+y−1xxyy)1/(x+y)\mathrm{\alpha } = {({K_{eq}}/{C^{x + y - 1}}{x^x}{y^y})^{1/(x + y)}}
Option C is correct because it provides the accurate formula for calculating the degree of dissociation (α\alpha) for a weak electrolyte AxBy{A_x}{B_y} in a concentrated solution, considering the equilibrium constant and concentration.
219 Easy
219 of 408
Remembering +4 | -1
The Henderson Hasselbalch equation explains the relationship between
View answer and explanation
Correct option: C
pH and pKa\mathrm{pH}\text{ and }\mathrm{pKa}
Option C (pH and pKa\mathrm{pH}\text{ and }\mathrm{pKa}) is correct because the Henderson-Hasselbalch equation directly relates pH to pKa and the ratio of conjugate base and acid concentrations.
220 Easy
220 of 408
Understanding +4 | -1
What is the correct representation for the solubility product of SnS2\mathrm{Sn}{\mathrm{S}_2}?
View answer and explanation
Correct option: B
[Sn4+][S2−]2\left[ {\mathrm{S}{\mathrm{n}^{4 + }}} \right]{\left[ {{\mathrm{S}^{2 - }}} \right]^2}
Option B (Sn4+[S2−]2{\mathrm{S}{\mathrm{n}^{4 + }}} {{\left[ {{\mathrm{S}^{2 - }}} \right]}^2}) is correct because it accurately represents the solubility product expression for SnS2\mathrm{Sn}{\mathrm{S}_2} considering the correct stoichiometry and ion charges.
221 Easy
221 of 408
Understanding +4 | -1
PH{P^H} of water is 7.0 at 250C{25^0}C. If water is heated to 700C{70^0}C, then
View answer and explanation
Correct option: D
PH{P^H} will decrease but solution will be neutral
Option D (PH{P^H} will decrease but solution will be neutral) is correct because as temperature increases, the autoionization of water increases, leading to a higher concentration of both H+ and OH- ions, thus decreasing pH, but the solution remains neutral as [H+] = [OH-].
222 Easy
222 of 408
Analyzing +4 | -1
Assertion : The equilibrium constant for the reaction

CaSO4⋅5H2O(s)=CaSO4⋅3H2O(s)+2H2O(g) is KC=[CaSO4⋅3H2O][H2O]2[CaSO4⋅5H2O]\begin{aligned} & {CaSO}_4 \cdot 5 {H}_2 {O}({s})={CaSO}_4 \cdot 3 {H}_2 {O}({s})+2 {H}_2 {O}({g}) \\ & \text { is } {K}_{{C}}=\frac{\left[{CaSO}_4 \cdot 3 {H}_2 {O}\right]\left[{H}_2 {O}\right]^2}{\left[{CaSO}_4 \cdot 5 {H}_2 {O}\right]} \end{aligned}

Reason : Equilibrium constant is the ratio of the product of molar concentration of the substances produced to the product of the molar concentrations of reactants with each concentrations term raised to the power equal to the respective stoichiometric constant.
View answer and explanation
Correct option: D
If assertion is false but reason is true
Option D (If assertion is false but reason is true) is correct because the assertion incorrectly formulates the equilibrium constant for a heterogeneous reaction, while the reason correctly states the general definition of the equilibrium constant.
223 Easy
223 of 408
Analyzing +4 | -1
Which of the following is not a buffer solution?
View answer and explanation
Correct option: C
50  mL  0.1  M  CH3COOH  +  100  mL  0.1  M  NaOH50\;\mathrm{mL}\;0.1\;\mathrm{M}\;\mathrm{ C}{\mathrm{H}_3}\mathrm{COOH}\; + \;100\;\mathrm{mL}\;0.1\;\mathrm{M}\;\mathrm{ NaOH}
Option C (50  mL  0.1  M  CH3COOH  +  100  mL  0.1  M  NaOH50\;\mathrm{mL}\;0.1\;\mathrm{M}\;\mathrm{ C}{\mathrm{H}_3}\mathrm{COOH}\; + \;100\;\mathrm{mL}\;0.1\;\mathrm{M}\;\mathrm{ NaOH}) is not a buffer solution because the strong base (NaOH) is in excess, completely neutralizing the weak acid (CH3COOH) and resulting in a basic solution, not a buffer.
224 Hard
224 of 408
Understanding +4 | -1
(A): Benzene is amphiprotic solvent.

(R): Amphiprotic solvents are those which neither donate nor accept protons
View answer and explanation
Correct option: D
Both (A) and (R) is false
Option D is correct because Benzene is not amphiprotic and amphiprotic solvents both donate and accept protons.
225 Medium
225 of 408
Application +4 | -1
For reaction A(g)+B(g)AB(g)A(g) + B(g)AB(g), We start with 2 moles of A and B each. At equilibrium 0.8 moles of AB is formed. Then how much of A changes to AB.
View answer and explanation
Correct option: B
40%40\%
Option B (40%40\%) is correct because 0.8 moles of AB formed from 2 moles of A represents a 40% conversion.
226 Medium
226 of 408
Analyzing +4 | -1
What happens to pH\mathrm{pH} of a solution when NH4Cl\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} crystal is added to a dilute solution of NH4OH\mathrm{N}{\mathrm{H}_4}\mathrm{OH}?
View answer and explanation
Correct option: A
Decreases
Option A (Decreases) is correct because adding NH4Cl\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} to NH4OH\mathrm{N}{\mathrm{H}_4}\mathrm{OH} introduces a common ion (NH4+\mathrm{N}{\mathrm{H}_4}^+), suppressing the ionization of NH4OH\mathrm{N}{\mathrm{H}_4}\mathrm{OH} and decreasing the pH.
227 Medium
227 of 408
Application +4 | -1
Liquid ammonia ionises to a slight extent. At −50∘C,- 50^\circ C,&nbsp; its self ionisation constant, KNH3=[NH4+][NH2−]=10−30.{K_{\mathrm{N}{\mathrm{H}_3}}} = \left[ {\mathrm{NH}_4^ + } \right]\left[ {\mathrm{NH}_2^ - } \right] = {10^{ - 30}}.&nbsp; How many amide ions, are present per cm3\mathrm{c}{\mathrm{m}^3} of pure liquid ammonia? (Assume N=6.0×1023N = 6.0 \times {10^{23}})
View answer and explanation
Correct option: B
6×105 ions6 \times {10^5}\,\mathrm{ions}
Option B (6×1056 \times 10^5 ions) is correct because it represents the calculated number of amide ions per cm³ based on the self-ionization constant.
228 Hard
228 of 408
Analyzing +4 | -1
Statement (S): Metal oxides are usually either acidic or amphoteric but Mn2O7\mathrm{Mn}_2 \mathrm{O}_7 is acidic.

Explanation (E): On dissolution in water, Mn2O7\mathrm{Mn}_2 \mathrm{O}_7 forms per manganic acid.
View answer and explanation
Correct option: C
Both S and E are correct and E is correct explanation for S
Option C is correct. Mn2O7 is acidic because it dissolves in water to form permanganic acid (HMnO4). The high oxidation state of Mn (+7) contributes to its acidic nature.
229 Medium
229 of 408
Analyzing +4 | -1
In which case pH{p^H} will not change on dilution
View answer and explanation
Correct option: D
In all cases
Option D (In all cases) is correct because in all the given cases, the pH will not change significantly upon dilution. For a buffer, the ratio of salt to acid remains constant. For CH3COONH4, it's a salt of a weak acid and weak base. For NaH2PO4, it's an amphoteric salt.
230 Easy
230 of 408
Application +4 | -1
The pHpH of a solution is 5.05.0. To this solution sufficient acid is added to decrease the pH  to  2.0pH\;to\;2.0. The increase in hydrogen ion concentration is:
View answer and explanation
Correct option: A
1000  times1000\;times
Option A (1000 times) is correct because a decrease in pH from 5.0 to 2.0 represents a 1000-fold increase in hydrogen ion concentration.
231 Easy
231 of 408
Analyzing +4 | -1
10 mL10\,\mathrm{mL} of a solution contains 0.1 M NH4Cl+0.01 M NH4OH.0.1\,M\,\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} + 0.01\,M\,\mathrm{N}{\mathrm{H}_4}\mathrm{OH}\mathrm{.} Which addition would not change the pH\mathrm{pH} of the solution?
View answer and explanation
Correct option: A
Adding 1 mL1\,\mathrm{mL} water
Option A (Adding 1 mL1\,{\mathrm{mL}} water) is correct because adding water to a buffer solution does not change the ratio of the weak acid/base and its conjugate, thus maintaining the pH.
232 Hard
232 of 408
Understanding +4 | -1
Assertion :- Boric acid behaves as a weak monobasic acid.

Reason :- Boric acid contains hydrogen bonds in its structure.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because while both statements are true, hydrogen bonding doesn't explain the weak monobasic nature.
233 Medium
233 of 408
Analyzing +4 | -1
Assuming complete dissociation which of the following aqueous solutions will have the same pH value?

(i)100 mL of 0.01 M HCl
(ii)100 mL of 0.01 M H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}
(iii)50 mL of 0.01 M HCl
(iv)Mixture of 50 mL of 0.02 M H2SO4{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} and 50 mL of 0.02 M NaOH\mathrm{NaOH}
View answer and explanation
Correct option: D
(i), (iv)
Option D ((i), (iv)) is correct because both solutions have the same final concentration of H+H^+ ions after considering neutralization in (iv).
234 Hard
234 of 408
Application +4 | -1
The dissociation constant for acetic acid and HCN\mathrm{HCN} at 25∘C25^\circ C are 1.5×10−31.5 \times {10^{ - 3}} and 4.5×10−104.5 \times {10^{ - 10}} respectively. the equilibrium constant for the equilibrium,

CN−+CH3COOH  ⇋  HCN+CH3COO−\mathrm{C}{\mathrm{N}^ - } + \mathrm{C}{\mathrm{H}_3}\mathrm{COOH}\;\leftrightharpoons\;\mathrm{HCN} + \mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }
View answer and explanation
Correct option: A
3.0×1043.0 \times {10^4}
Option A (3.0×1043.0 \times {10^4}) is correct because it is calculated by dividing the dissociation constant of acetic acid by the dissociation constant of HCN.
235 Easy
235 of 408
Application +4 | -1
The percentage of pyridine (C5H5N)({C_5}{H_5}N) that forms pyridinium ion (C5H5N+H)({C_5}{H_5}{N^ + }H) in a 0.10 M aqueous pyridine solution (Kb for C5H5N=1.7×10−9)({K_b}\,\mathrm{for}\,{C_5}{H_5}N = 1.7 \times {10^{ - 9}}) is
View answer and explanation
Correct option: C
0.00013
Option C (0.00013) is correct because it represents the calculated percentage of pyridine that forms pyridinium ion based on the given Kb value and concentration.
236 Medium
236 of 408
Understanding +4 | -1
Change in pH of a solution from 1 to 2, indicates
View answer and explanation
Correct option: B
Decrease in H+{H^ + } concentration by 10 times
Option B (Decrease in H+{H^ + } concentration by 10 times) is correct because pH is a logarithmic scale, and an increase of 1 pH unit represents a 10-fold decrease in H+ concentration.
237 Medium
237 of 408
Understanding +4 | -1
Which among the following is an electron deficient compound?
View answer and explanation
Correct option: C
BF3\mathrm{B}{\mathrm{F}_3}
Option C (BF3\mathrm{BF_3}) is correct because Boron in BF3 has only 6 electrons in its valence shell, making it electron deficient.
238 Easy
238 of 408
Analyzing +4 | -1
The equilibrium constant, KK for 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 room temperature is 2.85 and that at 698  K698\;K, it is 1.4×10−21.4 \times {10^{ - 2}}. This implies that
View answer and explanation
Correct option: D
HI\mathrm{HI} is less stable than H2{\mathrm{H}_2} and I2{\mathrm{I}_2}
Option D is correct because the equilibrium constant decreases with increasing temperature, indicating that the reverse reaction (formation of HI) is exothermic, meaning HI is less stable than H2 and I2.
239 Easy
239 of 408
Understanding +4 | -1
Which of the following molecules acts as a Lewis acid?
View answer and explanation
Correct option: B
(CH3)3B{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{B}
Option B ((CH3)3B( {\mathrm{C}{\mathrm{H}_3}} )_3\mathrm{B}) is correct because boron in trimethylborane has an incomplete octet and can accept a pair of electrons, thus acting as a Lewis acid.
240 Medium
240 of 408
Understanding +4 | -1
A solution of sodium borate has a pH\mathrm{pH} of approximately:
View answer and explanation
Correct option: A
>7> 7
Option A (>7> 7) is correct because sodium borate is the salt of a weak acid and a strong base, resulting in a basic solution with a pHpH greater than 7.
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