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NEET • CHEMISTRY • Equilibrium

Equilibrium practice · Mixed difficulty

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271 Hard
271 of 408
Analyzing +4 | -1
aA⇌bB+cC,ΔH=−x KCala A \rightleftharpoons b B+c C, \Delta H=-x \text { KCal}

If high pressure and low temperature are the favourable condition for the formation of the product in above reaction, Hence:
View answer and explanation
Correct option: A
a>b+ca>b+c
Option A (a>b+ca>b+c) is correct because high pressure favors the side with fewer moles of gas, and low temperature favors exothermic reactions (negative ΔH\Delta H).
272 Hard
272 of 408
Application +4 | -1
1 mole of 'A\mathrm{A}', 1.5 mole of 'B\mathrm{B}' and 2 mole of 'C\mathrm{C}' are taken in a vessel of volume one litre. At equilibrium concentration of C\mathrm{C} is 0.5 mole/L0.5 \mathrm{~mole} / \mathrm{L}. Equilibrium constant for the reaction A(g)+B(g)⇌C(g)\mathrm{A}{(\mathrm{g})}+\mathrm{B}{(\mathrm{g})} \rightleftharpoons \mathrm{C}{(\mathrm{g})} is
View answer and explanation
Correct option: B
0.0660.066
Option B (0.066) is correct because it represents the equilibrium constant calculated based on the given changes in concentration.
273 Easy
273 of 408
Understanding +4 | -1
Conjugate acid and bases for H2PO4−\mathrm{H}_2 \mathrm{PO}_4^{-} ion respectively are
View answer and explanation
Correct option: D
H3PO4\mathrm{H}_3 \mathrm{PO}_4 and HPO42−\mathrm{HPO}_4^{2-}
Option D (H3PO4\mathrm{H}_3 \mathrm{PO}_4 and HPO42−\mathrm{HPO}_4^{2-}) is correct because H3PO4\mathrm{H}_3 \mathrm{PO}_4 is formed by adding a proton to H2PO4−\mathrm{H}_2 \mathrm{PO}_4^{-} and HPO42−\mathrm{HPO}_4^{2-} is formed by removing a proton from H2PO4−\mathrm{H}_2 \mathrm{PO}_4^{-}.
274 Hard
274 of 408
Understanding +4 | -1
Fe3+(aq)+YellowSCN−(aq)⇌[Fe(SCN)]2+Red\mathop {{{F}}{{{e}}^{3 + }}({{aq}}) + }\limits_{{\text{Yellow}}} {{SC}}{{{N}}^ - }({{aq}}) \rightleftharpoons \mathop {{{[{{Fe}}({{SCN}})]}^{2 + }}}\limits_{{\text{Red}}} Addition of the following increases the intensity of red colour.
View answer and explanation
Correct option: A
KSCN{KSCN}
Option A is correct because adding KSCN increases the concentration of SCN-, shifting the equilibrium to the right and increasing the intensity of the red color.
275 Hard
275 of 408
Application +4 | -1
Ammonium carbamate dissociates as
NH2COONH4(s)⇋2NH3(g)+CO2(g)\mathrm{NH}_2 \mathrm{COONH}_{4}(\mathrm{s}) \leftrightharpoons 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})
In a closed vessel containing ammonium carbamate in equilibrium, ammonia is added such that the partial pressure of NH3\mathrm{NH}_3 now equals to the original total pressure. The ratio of total pressure now to the original pressure is
View answer and explanation
Correct option: B
3127\frac{31}{27}
Option B (31/27) is correct. The ratio of total pressure is calculated based on the equilibrium constant and the change in partial pressures due to the addition of ammonia.
276 Medium
276 of 408
Understanding +4 | -1
When pressure is applied to the equilibrium system ice r water. Which of the following phenomenon will happen?
View answer and explanation
Correct option: C
More water will be formed
Option C (More water will be formed) is correct because increasing pressure favors the side with the smaller volume, which is water in this case.
277 Medium
277 of 408
Understanding +4 | -1
Which one of the following salts on being dissolved in water gives pH  >  7  at  25∘C\mathrm{pH}\; > \;7\;\mathrm{at}\;25^\circ C
View answer and explanation
Correct option: A
KCN\mathrm{KCN}
Option A (KCN) is correct because KCN is a salt of a strong base (KOH) and a weak acid (HCN), so it will undergo hydrolysis to produce a basic solution with pH > 7.
278 Hard
278 of 408
Understanding +4 | -1
Dissociation constant of two acids HA\mathrm{HA} and HB\mathrm{HB} are respectively 4×10−104 \times 10^{-10} and 1.8×10−5.1.8 \times 10^{-5}. The pH\mathrm{pH} value of which acid will be higher for a given molarity of their aqueous solution?
View answer and explanation
Correct option: A
HA\text{HA}
Option A (HA) is correct because the acid with the smaller dissociation constant (HA) will have a lower concentration of H+ ions and therefore a higher pH for the same molarity.
279 Hard
279 of 408
Application +4 | -1
Ag++NH3  ⇋[Ag(NH3)]+;    K1=3.5×10−3\mathrm{A}{\mathrm{g}^ + } + \mathrm{N}{\mathrm{H}_3}\;\leftrightharpoons{\left[ {\mathrm{Ag}\left( {\mathrm{N}{\mathrm{H}_3}} \right)} \right]^ + };\;\;{K_1} = 3.5 \times {10^{ - 3}}
[Ag(NH3)]++NH3  ⇋[Ag(NH3)2]+;{\left[ {\mathrm{Ag}\left( {\mathrm{N}{\mathrm{H}_3}} \right)} \right]^ + } + \mathrm{N}{\mathrm{H}_3}\;\leftrightharpoons{\left[ {\mathrm{Ag}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}} \right]^ + };
K2=1.7×10−3{K_2} = 1.7 \times {10^{ - 3}}
Then the formation constant of[Ag(NH3)2]+{\left[ {\mathrm{Ag}{{\left( {\mathrm{N}{\mathrm{H}_3}} \right)}_2}} \right]^ + } is
View answer and explanation
Correct option: A
6.08×10−66.08 \times {10^{ - 6}}
Option A (6.08×10−66.08 \times {10^{ - 6}}) is correct because the overall formation constant is the product of the individual stepwise formation constants.
280 Easy
280 of 408
Analyzing +4 | -1
Which one of the following statements is not true?
View answer and explanation
Correct option: C
The pH of 1×10−8  M  HCl1 \times {10^{ - 8}}\;M\;\mathrm{HCl} is 8
Option C is correct because the pH of a very dilute strong acid like 1×10−8M  HCl1 \times 10^{-8} M \;HCl cannot be 8. It must be less than 7, considering the contribution of water's autoionization.
281 Easy
281 of 408
Understanding +4 | -1
For which reaction Kp{K_p} is less than Kc{K_c}?
View answer and explanation
Correct option: C
2SO2+O2⇋2SO32\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3}
Option C (2SO2+O2⇋2SO32\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3}) is correct because in this reaction, Δn < 0, which makes Kp < Kc.
282 Medium
282 of 408
Remembering +4 | -1
van′t  Hoff′s\mathrm{va}{\mathrm{n}^\mathrm{'}}\mathrm{t }\;\mathrm{Hoff's} equation giving the effect of temperature on chemical equilibrium is represented as:
View answer and explanation
Correct option: C
d  In  KpdT=ΔHRT2\frac{{d\;\mathrm{In}\;{K_p}}}{{dT}} = \frac{{\Delta H}}{{R{T^2}}}
Option C (d  In  KpdT=ΔHRT2\frac{{d\;\mathrm{In}\;{K_p}}}{{dT}} = \frac{{\Delta H}}{{R{T^2}}}) is correct because this is the correct form of the van't Hoff equation, relating the change in the equilibrium constant with temperature to the enthalpy change.
283 Medium
283 of 408
Analyzing +4 | -1
Final pressure is higher than initial pressure of a container filled with an ideal gas at constant temperature. What will be the value of equilibrium constant?
View answer and explanation
Correct option: D
K<1.0K < 1.0
Option D (K<1.0K < 1.0) is correct because if the final pressure is higher than the initial pressure at constant temperature, it means the reaction favors the side with more moles of gas, indicating that the equilibrium constant is less than 1.
284 Hard
284 of 408
Application +4 | -1
For the decomposition of PCl5( g)\mathrm{PCl}_5(\mathrm{~g}) in a closed vessel, the degree of dissociation is ' x\mathrm{x} ' at a total pressure ' P\mathrm{P} '.

PCl5( g)⇌PCl3( g)+Cl2( g);Kp\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g}) ; \mathrm{Kp}

Which among the following relations is correct?
View answer and explanation
Correct option: A
x=KpKp+Px=\sqrt{\frac{K_p}{K_p+P}}
Option A (x=KpKp+Px=\sqrt{\frac{K_p}{K_p+P}}) is correct because it represents the correct relationship between the degree of dissociation 'x', Kp, and total pressure 'P' for the given reaction.
285 Easy
285 of 408
Understanding +4 | -1
Which of the following can act both as Bronsted acid and Bronsted base?
View answer and explanation
Correct option: B
HCO3−HCO_3^ -
Option B (HCO3−HCO_3^ -) is correct because it can both donate a proton to become CO32−CO_3^{2-} (acting as an acid) and accept a proton to become H2CO3H_2CO_3 (acting as a base).
286 Medium
286 of 408
Understanding +4 | -1
The pH of pure water at 500C{50^0}C is
View answer and explanation
Correct option: C
Less than 7
Option C (Less than 7) is correct because the autoionization of water increases with temperature, leading to a higher concentration of both H+ and OH- ions, but neutrality is still maintained, resulting in a pH less than 7.
287 Easy
287 of 408
Application +4 | -1
The pHpH of a 0.0001N0.0001\mathrm{ }N solution of KOHKOH will be
View answer and explanation
Correct option: C
10
Option C (10) is correct because a 0.0001 N solution of KOH is a 10-4 M solution of a strong base, so the pOH is 4 and the pH is 14 - 4 = 10.
288 Medium
288 of 408
Remembering +4 | -1
A solution that has reserve acidity (or) alkalinity is called
View answer and explanation
Correct option: C
Buffer solution
Option C (Buffer solution) is correct because a buffer solution is specifically designed to resist changes in pH by having reserve acidity or alkalinity.
289 Hard
289 of 408
Analyzing +4 | -1
MY and NY3\mathrm{N}{\mathrm{Y}_\mathrm{3}}, two nearly insoluble salts, have the same Ksp{\mathrm{K}_{\mathrm{sp}}} values of 6.2×10−136.2 \times \mathrm{1}{\mathrm{0}^{ - 13}} at room temperature. Which statement would be true in regard to MY and NY3\mathrm{N}{\mathrm{Y}_\mathrm{3}}?
View answer and explanation
Correct option: B
The molar solubility of MY in water is less than that of NY3\mathrm{N}{\mathrm{Y}_\mathrm{3}}.
Option B (The molar solubility of MY in water is less than that of NY3\mathrm{N}{\mathrm{Y}_\mathrm{3}}) is correct because, for the same KspK_{sp}, salts with more ions have higher molar solubility.
290 Easy
290 of 408
Application +4 | -1
40% of a mixture of 0.2 mole of N2{\mathrm{N}_2} and 0.6 mole of H2{\mathrm{H}_2} react to give NH3\mathrm{N}{\mathrm{H}_3} according to the equation,N2(g)+3H2(g)⇋2NH3(g){\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right) at constant temperature and pressure. Then the ratio of the final volume to the initial volume of gases is :
View answer and explanation
Correct option: A
4:54\mathrm{ }:\mathrm{ }5
Option A (4:54\mathrm{ }:\mathrm{ }5) is correct because the ratio of final to initial volume is calculated based on the stoichiometry of the reaction and the percentage conversion.
291 Hard
291 of 408
Analyzing +4 | -1
Consider the following reversible chemical reactions
A2(g)+B2(g)⇋K12AB(g)..... (i)6AB(g)⇋K23A2(g)+3B2(g) ..... (ii)\begin{array}{l} {A_2}(g) + {B_2}(g)\underset{{}}{\overset{{{K_1}}}{\leftrightharpoons}}2AB(g).....\,\left( \mathrm{i} \right)\\ 6AB(g)\underset{{}}{\overset{{{K_2}}}{\leftrightharpoons}}3{A_2}(g) + 3{B_2}(g)\,.....\,\left( {\mathrm{ii}} \right) \end{array}
The relation between K1{K_1} and K2{K_2} is
View answer and explanation
Correct option: C
K2=K1−3{K_2} = K_1^{ - 3}
The correct answer is C (K2=K1−3{K_2} = K_1^{ - 3}) because it correctly relates the equilibrium constants of the forward and reverse reactions with appropriate powers.
292 Easy
292 of 408
Remembering +4 | -1
The pH of a solution is 4. The hydrogen ion concentration of the solution in mol/L is
View answer and explanation
Correct option: B
10−4{10^{ - 4}}
Option B (10−4{10^{ - 4}}) is correct because pH is defined as the negative logarithm of the hydrogen ion concentration, so if pH = 4, then [H+] = 10-4 M.
293 Easy
293 of 408
Understanding +4 | -1
The aqueous solution of disodium hydrogen phosphate is:
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Correct option: C
Basic
Option C (Basic) is correct because disodium hydrogen phosphate (Na2HPO4Na_2HPO_4) is a salt of a weak acid and a strong base, resulting in a basic aqueous solution.
294 Easy
294 of 408
Analyzing +4 | -1
If K1{K_1} and K2{K_2} are equilibrium constants for reactions (I) and (II) respectively for,

N2+O2⇋2NO     …(i){\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO}\,\,\,\,\, \ldots \left( \mathrm{i} \right)

12N2+12O2⇋NO    …(ii)\frac{1}{2}{\mathrm{N}_2} + \frac{1}{2}{\mathrm{O}_2}\leftrightharpoons\mathrm{NO}\,\,\,\, \ldots \left( {\mathrm{ii}} \right)

Then:
View answer and explanation
Correct option: B
K2=K1{K_2} = \sqrt {{K_1}}
Option B (K2=K1{K_2} = \sqrt {{K_1}}) is correct because reaction (II) is half of reaction (I), so its equilibrium constant is the square root of the first.
295 Hard
295 of 408
Analyzing +4 | -1
A 50 ml50 \mathrm{~ml} of mixture of He,CO\mathrm{He}, \mathrm{CO} and C2H6\mathrm{C}_2 \mathrm{H}_6 was exploded in 120 ml120 \mathrm{~ml} of oxygen. On cooling, the gases occupied 142 ml142 \mathrm{~ml}. After treatment with KOH\mathrm{KOH} the volume reduced by 26 ml26 \mathrm{~ml} and again on treatment with alkaline pyrogallol the volume further reduced, then wrong one is ___
View answer and explanation
Correct option: D
V(CO\mathrm{V (CO} gas)=8 ml)=8 \mathrm{~ml}
Option D (V(CO gas) = 8 ml) is wrong. The correct volume of CO gas is 16 ml, calculated from the oxygen consumed and the stoichiometry of the combustion reaction.
296 Medium
296 of 408
Understanding +4 | -1
In a vessel containing SO3, SO2,\mathrm{S}{\mathrm{O}_3},\,\mathrm{S}{\mathrm{O}_2}, and O2{\mathrm{O}_2} at equilibrium, some helium gas in introduced so that, the total pressure increase, while temperature and volume remain constant. According to Le-Chatelier’s principle the dissociation of SO3:\mathrm{S}{\mathrm{O}_3}:
View answer and explanation
Correct option: C
Remains unaltered
Option C (Remains unaltered) is correct because the addition of an inert gas at constant volume does not affect the partial pressures of the reactants and products, thus the equilibrium remains unchanged.
297 Hard
297 of 408
Understanding +4 | -1
Solution of aniline hydrochloride is XX due to hydrolys of YY. XX and YY are
View answer and explanation
Correct option: B
acidic, C6H5NH3+\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3^{+}
Option B (acidic, C6H5NH3+\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3^{+}) is correct because aniline hydrochloride is a salt of a weak base (aniline) and a strong acid (HCl), and the anilinium ion undergoes hydrolysis to produce an acidic solution.
298 Medium
298 of 408
Understanding +4 | -1
Which of the following is correct for the reaction?
N2(g)+3H2(g)⇋2NH3(g){\mathrm{N}_2}\left( \mathrm{g} \right) + 3{\mathrm{H}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{N}{\mathrm{H}_3}\left( \mathrm{g} \right)
View answer and explanation
Correct option: B
Kp<Kc{K_p} < {K_c}
Option B is correct because for the given reaction, Δn < 0, which implies Kp < Kc.
299 Medium
299 of 408
Application +4 | -1
5 moles of SO2  \mathrm{S}{\mathrm{O}_2}\; and 5 moles of O2  {\mathrm{O}_2}\; are allowed to react to forms SO3\mathrm{S}{\mathrm{O}_3} in a closed vessel. At the equilibrium stage 60% of SO2  \mathrm{S}{\mathrm{O}_2}\; is used up. The total number of moles of   SO2  ,O2  and  SO3\;\mathrm{S}{\mathrm{O}_2}\;,{\mathrm{O}_2}\;and\;\mathrm{S}{\mathrm{O}_3} in the vessel now is
View answer and explanation
Correct option: A
8.58.5
Option A (6.08×10−66.08 \times {10^{ - 6}}) is correct because the overall formation constant is the product of the individual stepwise formation constants.
300 Easy
300 of 408
Analyzing +4 | -1
Assertion : For the reaction,

2NH3( g)⇌N2( g)+3H2( g)2 {NH}_3({~g}) \rightleftharpoons {N}_2({~g})+3 {H}_2({~g}), the unit of KpK_p will be atm.

Reason : Unit of KpK_p is (atm)Δn(a t m)^{\Delta n}.
View answer and explanation
Correct option: D
If assertion is false but reason is true
Option D is correct because the assertion is false as the unit of Kp is atm2, and the reason is true as it correctly states the general formula for the unit of Kp.
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