For ${\rm{PC}}{{\rm{l}}_5}\leftrightharpoons{\rm{PC}}{{\rm{l}}_3} + {\rm{C}}{{\rm{l}}_2}$, initial concentration of each reactant and product is 1 M. If ${K_{{\rm{eq}}}} = 0.41$ then

More ${\rm{PC}}{{\rm{l}}_5}$ will form

More ${\rm{PC}}{{\rm{l}}_3}$ will form

More ${\rm{C}}{{\rm{l}}_2}$ will form

The equilibrium constant $K$ for the reaction $2{\rm{HI}}\left( {\rm{g}} \right)\leftrightharpoons{{\rm{H}}_2}\left( {\rm{g}} \right) + {{\rm{I}}_2}\left( {\rm{g}} \right)$ at room temperature $300\;{\rm{K}}$ is ${\rm{2}}{\rm{.85}}$ and at $698\;{\rm{K}}\,\;1.84 \times {10^{ - 2}}.$ Hence the reason that ${\rm{HI}}$ exists as a stable compound at room temperature is because:

The heat of reaction at room temperature is $ - 5.31\;{\rm{kcal}}$

It decomposes so slowly that equilibrium is not readily achieved

The ${\rm{HI}}$ bond has a large covalent contribution

The $pH$ values of $0.1\;M$ solution of ${\rm{HCl}},{\rm{C}}{{\rm{H}}_3}{\rm{COOH}},{\rm{N}}{{\rm{H}}_4}{\rm{Cl}}\;{\rm{ and }}\;{\rm{C}}{{\rm{H}}_3}{\rm{COONa}}$ will have the order

${\rm{HCl}} < C{{\rm{H}}_3}{\rm{COOH}} < N{{\rm{H}}_4}{\rm{Cl}} < C{{\rm{H}}_3}{\rm{COONa}}$

${\rm{C}}{{\rm{H}}_3}{\rm{COONa}} < {\rm{N}}{{\rm{H}}_4}{\rm{Cl}} < {\rm{C}}{{\rm{H}}_3}{\rm{COOH}} < HCl$

${\rm{N}}{{\rm{H}}_4}{\rm{Cl}} < {\rm{C}}{{\rm{H}}_3}{\rm{COONa}} < {\rm{C}}{{\rm{H}}_3}{\rm{COOH}} < HCl$

For the reaction, $2{\rm{N}}{{\rm{O}}_3}\left( {\rm{g}} \right)\leftrightharpoons2{\rm{NO}}\left( {\rm{g}} \right) + {{\rm{O}}_2}\left( {\rm{g}} \right)$ $\left( {{K_c} = 1.8 \times {{10}^{ - 6}}\;{\rm{at }}184^\circ C} \right)$ $(R = 0.00831\;kJ/\left( {mol{\rm{ }}K} \right)$ When ${K_p}\;{\rm{and}}\;{K_c}$ are compared at $184^\circ C$, it is found that

Whether ${K_p}$ is greater than less than or equal to ${K_c}$ depends upon the total gas pressure

The vapour density of completely dissociated ${\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$ would be:

Slightly less than half of that of ammonium chloride

Double that of ammonium chloride

Determined by the amount of solid ammonium chloride used in the experiment

The phenomenon of interaction of anions and cations furnished by a electrolyte with the ${{\rm{H}}^ + }$ and ${\rm{O}}{{\rm{H}}^ - }$ ions of water to produce acidic nature or alkalinity is known as hydrolysis. In hydrolysis:

The pH may either increase or decrease

All the salts (except those made up with strong acid and base) undergo hydrolysis

The variation of pH depends upon the nature of salts as well as on the temperature

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel,,br> ${\rm{PC}}{{\rm{l}}_5}\left( {\rm{g}} \right)\leftrightharpoons{\rm{PC}}{{\rm{l}}_3}\left( {\rm{g}} \right) + {\rm{C}}{{\rm{l}}_2}\left( {\rm{g}} \right)$ If total pressure at equilibrium of the reaction mixture is $p$ and degree of dissociation of ${\rm{PC}}{{\rm{l}}_5}\;{\rm{is}}\;x$, the partial pressure of ${\rm{PC}}{{\rm{l}}_3}$ will be

$\left( {\frac{{2x}}{{1 - x}}} \right)p$

$\left( {\frac{x}{{1 - x}}} \right)p$

An aqueous solution of hydrogen sulphide shows the equilibrium, ${{\rm{H}}_2}{\rm{S}}\leftrightharpoons{{\rm{H}}^ + } + {\rm{H}}{{\rm{S}}^ - }$ If dilute hydrochloric acid is added to an aqueous solution of hydrogen sulphide without any change in temperature, then:

The concentration of ${\rm{H}}{{\rm{S}}^ - }$ will decreases

The equilibrium constant will change

The concentration of ${\rm{H}}{{\rm{S}}^ - }$ will increase

The concentration of undissociated hydrogen sulphide will decrease

The correct order of increasing $\left[ {{{\rm{H}}_3}{{\rm{O}}^ + }} \right]$ in the following aqueous solutions is:

$0.01\;M\;{\rm{NaN}}{{\rm{O}}_2} < 0.01\;M\;{\rm{NaCl}} < 0.01\;M\;{{\rm{H}}_2}{\rm{S}} < 0.01\;M\;{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

$0.01\;M\;{{\rm{H}}_2}{\rm{S}} < 0.01\;M\;{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} < 0.01\;M\;{\rm{NaCl}} < 0.01\;M\;{\rm{NaN}}{{\rm{O}}_2}$

$0.01\;M\;{\rm{NaCl}} < 0.01\;M\;{\rm{NaN}}{{\rm{O}}_2} < 0.01\;M\;{{\rm{H}}_2}{\rm{S}} < 0.01\;M\;{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

$0.01\;M\;{{\rm{H}}_2}{\rm{S}} < 0.01\;M\;{\rm{NaN}}{{\rm{O}}_2} < 0.01\;M\;{\rm{NaCl}} < 0.01\;M\;{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

The equilibrium constant for the reaction,${{\rm{N}}_2}\left( {\rm{g}} \right) + {{\rm{O}}_2}\left( {\rm{g}} \right)\leftrightharpoons2{\rm{NO}}\left( {\rm{g}} \right)$ is $4 \times {10^{ - 4}}$ at 2000 K. In presence of a catalyst the equilibrium is attained ten times faster. Therefore, the equilibrium constant, in present of the catalyst, at 2000 K is:

Difficult to compute without more data

The equilibrium constant for ${N_2}(g) + {O_2}(g)\leftrightharpoons2NO\left( g \right)$ is ${K_1}$ and that for $NO(g)\leftrightharpoons\frac{1}{2}{N_2}(g) + \frac{1}{2}{O_2}(g)$ is ${K_2}$, then ${K_1}$ and ${K_2}$ related as

${K_2} = {\left( {\frac{1}{{{K_1}}}} \right)^2}$

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NEET • CHEMISTRY • Equilibrium

Equilibrium practice · Medium

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Showing 1-30 of 146 questions

1 Medium

1 of 146

Remembering +4 | -1

At the state of dynamic equilibrium, for solute

+

solvent

\leftrightharpoons

solution.

View answer and explanation

Correct option: B

Rate of dissolution = Rate of unsaturation

Option A (Rate of dissolution = Rate of crystallization) is incorrect. The correct answer is Rate of dissolution = Rate of unsaturation. At dynamic equilibrium, the rate at which the solute dissolves equals the rate at which it crystallizes out of the solution.

2 Medium

2 of 146

Application +4 | -1

For

\mathrm{PC}{\mathrm{l}_5}\leftrightharpoons\mathrm{PC}{\mathrm{l}_3} + \mathrm{C}{\mathrm{l}_2}

, initial concentration of each reactant and product is 1 M. If

{K_{\mathrm{eq}}} = 0.41

then

View answer and explanation

Correct option: C

\mathrm{PC}{\mathrm{l}_5}

will form

Option C (More PCl5 will form) is correct because the reaction quotient Q is greater than Keq, indicating the reaction will shift to the left to form more PCl5.

3 Medium

3 of 146

Understanding +4 | -1

Select the

\mathrm{p}{K_a}

value of the strongest acid from the following

View answer and explanation

Correct option: A

1.0

Option A (

1.0

) is the correct answer because a lower

\mathrm{p}{K_a}

value indicates a stronger acid.

4 Medium

4 of 146

Application +4 | -1

The

\mathrm{p}{K_b}

value of

\mathrm{N}{\mathrm{H}_3}

is 5. Calculate the pH of the buffer solution,

\mathrm{1}\;L

of which contains

0.01\mathrm{ }M\;\mathrm{ }

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}\,\mathrm{ and}\,0.10\,\mathrm{M }\,\mathrm{N}{\mathrm{H}_4}\mathrm{OH}

View answer and explanation

Correct option: D

Option D (10) is correct because it is calculated using the Henderson-Hasselbalch equation for a basic buffer.

5 Medium

5 of 146

Application +4 | -1

The vapour density of

{\mathrm{N}_2}{\mathrm{O}_4}

at a certain temperature is 30. What is the percentage dissociation of

{\mathrm{N}_2}{\mathrm{O}_4}

at this temperature?

View answer and explanation

Correct option: C

53.3\%

Option C (

53.3\%

) is correct because it is calculated using the relationship between vapour density, molecular weight, and degree of dissociation.

6 Medium

6 of 146

Analyzing +4 | -1

The equilibrium constant

K

for the reaction

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

300\;\mathrm{K}

\mathrm{2}\mathrm{.85}

and at

698\;\mathrm{K}\,\;1.84 \times {10^{ - 2}}.

Hence the reason that

\mathrm{HI}

exists as a stable compound at room temperature is because:

View answer and explanation

Correct option: C

The heat of reaction at room temperature is

- 5.31\;\mathrm{kcal}

Option A is correct because the slow decomposition rate of HI at room temperature prevents the attainment of equilibrium, making it appear stable.

7 Medium

7 of 146

Remembering +4 | -1

Nucleophiles are:

View answer and explanation

Correct option: B

Lewis bases

Option B (Lewis bases) is correct because nucleophiles are electron-rich species that donate electron pairs, which is the definition of a Lewis base.

8 Medium

8 of 146

Analyzing +4 | -1

The

pH

values of

0.1\;M

solution of

\mathrm{HCl},\mathrm{C}{\mathrm{H}_3}\mathrm{COOH},\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}\;\mathrm{ and }\;\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}

will have the order

View answer and explanation

Correct option: A

\mathrm{HCl} < C{\mathrm{H}_3}\mathrm{COOH} < N{\mathrm{H}_4}\mathrm{Cl} < C{\mathrm{H}_3}\mathrm{COONa}

Option A (

\mathrm{HCl} < C{\mathrm{H}_3}\mathrm{COOH} < N{\mathrm{H}_4}\mathrm{Cl} < C{\mathrm{H}_3}\mathrm{COONa}

) is correct because it correctly orders the pH values from most acidic (lowest pH) to most basic (highest pH).

9 Medium

9 of 146

Application +4 | -1

For the reaction,

2\mathrm{N}{\mathrm{O}_3}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{NO}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)

\left( {{K_c} = 1.8 \times {{10}^{ - 6}}\;\mathrm{at }184^\circ C} \right)

(R = 0.00831\;kJ/\left( {mol\mathrm{ }K} \right)

When

{K_p}\;\mathrm{and}\;{K_c}

are compared at

184^\circ C

, it is found that

View answer and explanation

Correct option: D

{K_p}

is greater than

{K_c}

Option D (

{K_p}

is greater than

{K_c}

) is correct because for this reaction, the change in the number of moles of gas is positive, leading to

{K_p} > {K_c}

10 Medium

10 of 146

Understanding +4 | -1

The vapour density of completely dissociated

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

would be:

View answer and explanation

Correct option: B

Half of that of ammonium chloride

Option B is correct because upon complete dissociation, one mole of NH4Cl yields two moles of products, halving the vapor density.

11 Medium

11 of 146

Analyzing +4 | -1

Consider the following salts, which one(s) when dissolved in water will produce a basic solution?

\begin{array}{cccccccccccccccccccc} {{\text{I}}{\text{.}}\,\,\,RbCl{O_4}}&{{\text{II}}{\text{.}}\,\,\,NaN{O_2}}&{{\text{III}}{\text{.}}\,\,\,N{H_4}Cl}&{{\text{IV}}{\text{.}}\,\,\,NaCl} \end{array}

View answer and explanation

Correct option: B

Only II

Option B (Only II) is correct because

NaN{O_2}

is a salt of a weak acid (

HN{O_2}

) and a strong base (NaOH), which hydrolyzes in water to produce a basic solution.

12 Medium

12 of 146

Understanding +4 | -1

The interfering radicals interfere in the test of usual inorganic analysis after II group analysis due to:

View answer and explanation

Correct option: C

Their insoluble nature in alkaline medium

Option C (Their insoluble nature in alkaline medium) is correct because interfering radicals precipitate in alkaline medium, hindering the analysis of subsequent groups.

13 Medium

13 of 146

Understanding +4 | -1

The phenomenon of interaction of anions and cations furnished by a electrolyte with the

{\mathrm{H}^ + }

and

\mathrm{O}{\mathrm{H}^ - }

ions of water to produce acidic nature or alkalinity is known as hydrolysis. In hydrolysis:

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because all the given statements accurately describe the phenomenon of hydrolysis.

14 Medium

14 of 146

Understanding +4 | -1

When hydrogen molecules decomposed into it’s atoms which conditions gives maximum yield of H atom?

View answer and explanation

Correct option: A

High temperature and low pressure

Option A (High temperature and low pressure) is correct because the decomposition of hydrogen molecules into atoms is an endothermic process, and increasing the temperature and decreasing the pressure will favor the formation of H atoms.

15 Medium

15 of 146

Application +4 | -1

Buffer capacity of a buffer solution is x, the volume of 1M added to 100 ml of this solution if the change of

{P^H}

by 1 is

View answer and explanation

Correct option: C

100x\,ml

Option C (100x ml) is correct because buffer capacity is defined as the moles of acid or base needed to change the pH by 1, and this relates directly to the volume of 1M acid added.

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16 Medium

16 of 146

Remembering +4 | -1

The principal buffer present in human blood

View answer and explanation

Correct option: D

\mathrm{H}_2 \mathrm{CO}_3+\mathrm{HCO}_3^{-}

Option D (

\mathrm{H}_2 \mathrm{CO}_3+\mathrm{HCO}_3^{-}

) is correct because the carbonic acid/bicarbonate buffer system is the primary buffer system in human blood, maintaining blood pH within a narrow range.

17 Medium

17 of 146

Application +4 | -1

Degree of dissociation of

\mathrm{N}{\mathrm{H}_4}\mathrm{OH}

in water is

1.8 \times {10^{ - 5}}

, then hydrolysis constant of

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

View answer and explanation

Correct option: D

5.55 \times {10^{ - 10}}

Option D is correct because the hydrolysis constant of NH4Cl can be calculated using the formula Kh = Kw/Kb.

18 Medium

18 of 146

Application +4 | -1

Phosphorus pentachloride dissociates as follows, in a closed reaction vessel,,br>

\mathrm{PC}{\mathrm{l}_5}\left( \mathrm{g} \right)\leftrightharpoons\mathrm{PC}{\mathrm{l}_3}\left( \mathrm{g} \right) + \mathrm{C}{\mathrm{l}_2}\left( \mathrm{g} \right)

If total pressure at equilibrium of the reaction mixture is

p

and degree of dissociation of

\mathrm{PC}{\mathrm{l}_5}\;\mathrm{is}\;x

, the partial pressure of

\mathrm{PC}{\mathrm{l}_3}

will be

View answer and explanation

Correct option: A

\left( {\frac{x}{{x + 1}}} \right)p

Option A (

\left( {\frac{x}{{x + 1}}} \right)p

) is correct because it represents the partial pressure of PCl3 at equilibrium, derived from the degree of dissociation and total pressure.

19 Medium

19 of 146

Application +4 | -1

{K_c}

for the reaction :

{\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]^ - }\leftrightharpoons\mathrm{A}{\mathrm{g}^ + } + 2\mathrm{C}{\mathrm{N}^ - },

the equilibrium constant at

25^\circ C\mathrm{ }\;is\mathrm{ }\;4.0\mathrm{ } \times \mathrm{ }{10^{ - 19}},

then the silver ion concentration in a solution which was originally

\mathrm{0}\mathrm{.1}\;\mathrm{molar}\;\mathrm{in}\;\mathrm{KCN}

and

\mathrm{0}\mathrm{.03}\;\mathrm{molar}\;\mathrm{in}\;\mathrm{AgN}{\mathrm{O}_3}

is :

View answer and explanation

Correct option: B

7.5 \times {10^{ - 18}}

Option B (

7.5 \times {10^{ - 18}}

) is correct because it calculates the silver ion concentration by considering the formation of the complex ion and using the equilibrium constant.

20 Medium

20 of 146

Analyzing +4 | -1

The reactions,

\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})

and

\mathrm{COCl}_2(\mathrm{~g})=\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_2(\mathrm{~g})

are simultaneously in equilibrium in an equilibrium box at constant volume. A few moles of

\mathrm{CO}(\mathrm{g})

are later introduced into the vessel. After some time, the new equilibrium concentration of

View answer and explanation

Correct option: C

\mathrm{PCl}_5

will become less

Option C (PCl5 will become less) is correct because adding CO shifts the second equilibrium to the left, reducing Cl2, which then shifts the first equilibrium to the left, increasing PCl5 consumption and thus decreasing its concentration.

21 Medium

21 of 146

Analyzing +4 | -1

The reaction that proceeds in the forward direction is :

View answer and explanation

Correct option: B

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} + \mathrm{NaOH} \to {\mathrm{H}_2}\mathrm{O} + \mathrm{N}{\mathrm{H}_3} + \mathrm{NaCl}

Option B (

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} + \mathrm{NaOH} \to {\mathrm{H}_2}\mathrm{O} + \mathrm{N}{\mathrm{H}_3} + \mathrm{NaCl}

) is correct because it represents a strong base reacting with a weak acid salt to form a weaker base and a salt, which is a favorable reaction.

22 Medium

22 of 146

Analyzing +4 | -1

From separate solutions of four sodium salts

\mathrm{Na}W,\mathrm{Na}X,\mathrm{Na}Y\;and\;\mathrm{Na}Z

had

\mathrm{pH}\;\;7.0,9.0,10.0\;and\;11.0

respectively. When each solution was

0.1\;M

, the strongest acid is:

View answer and explanation

Correct option: A

HW

Option A (

HW

) is correct because the sodium salt with the lowest pH (pH 7.0) corresponds to the strongest acid.

23 Medium

23 of 146

Application +4 | -1

The

pH

of a

{10^{ - 9}}

M solution of

\mathrm{HCl}

in water is

View answer and explanation

Correct option: D

Between 6 and 7

Option D (Between 6 and 7) is correct because the pH of a very dilute acid solution cannot be greater than 7.

24 Medium

24 of 146

Understanding +4 | -1

In an equilibrium reaction

{H_2}(g) + {I_2}(g)2HI(g);\Delta H = - 3000

cal. Which factor favours dissociation of HI

View answer and explanation

Correct option: C

High Temperature

Option C (High Temperature) is correct because the reverse reaction (dissociation of HI) is endothermic, and increasing the temperature favors the endothermic reaction.

25 Medium

25 of 146

Analyzing +4 | -1

An aqueous solution of hydrogen sulphide shows the equilibrium,

{\mathrm{H}_2}\mathrm{S}\leftrightharpoons{\mathrm{H}^ + } + \mathrm{H}{\mathrm{S}^ - }

If dilute hydrochloric acid is added to an aqueous solution of hydrogen sulphide without any change in temperature, then:

View answer and explanation

Correct option: D

The concentration of

\mathrm{H}{\mathrm{S}^ - }

will decreases

Option D is correct because adding HCl increases the concentration of H+, which shifts the equilibrium to the left, decreasing the concentration of HS- according to Le Chatelier's principle.

26 Medium

26 of 146

Analyzing +4 | -1

The correct order of increasing

\left[ {{\mathrm{H}_3}{\mathrm{O}^ + }} \right]

in the following aqueous solutions is:

View answer and explanation

Correct option: C

0.01\;M\;\mathrm{NaN}{\mathrm{O}_2} < 0.01\;M\;\mathrm{NaCl} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

Option C (

0.01\;M\;\mathrm{NaN}{\mathrm{O}_2} < 0.01\;M\;\mathrm{NaCl} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S} < 0.01\;M\;{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

) is correct because it correctly orders the solutions from least to greatest acidity based on their chemical properties.

27 Medium

27 of 146

Analyzing +4 | -1

Which of the following has highest pH?

View answer and explanation

Correct option: D

\frac{\mathrm{M}}{4}\mathrm{Ca}{\left( {\mathrm{OH}} \right)_2}

Option D (

\frac{\mathrm{M}}{4}\mathrm{Ca}{\left( {\mathrm{OH}} \right)_2}

) is correct because it provides the highest hydroxide ion concentration due to being a strong base and having two hydroxide ions per formula unit.

28 Medium

28 of 146

Understanding +4 | -1

For the reactions,

A + B + Q\;\leftrightharpoons\;C + D

  if the temperature is increased then concentration of the products will

View answer and explanation

Correct option: A

Increase

Option A is correct because increasing the temperature favors the endothermic reaction, which in this case, is the forward reaction, thus increasing the concentration of products.

29 Medium

29 of 146

Understanding +4 | -1

The equilibrium constant for the reaction,

{\mathrm{N}_2}\left( \mathrm{g} \right) + {\mathrm{O}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{NO}\left( \mathrm{g} \right)

4 \times {10^{ - 4}}

at 2000 K. In presence of a catalyst the equilibrium is attained ten times faster. Therefore, the equilibrium constant, in present of the catalyst, at 2000 K is:

View answer and explanation

Correct option: B

4 \times {10^{ - 4}}

Option B (

4 \times {10^{ - 4}}

) is correct because a catalyst only affects the rate of reaction, not the equilibrium constant.

30 Medium

30 of 146

Analyzing +4 | -1

The equilibrium constant for

{N_2}(g) + {O_2}(g)\leftrightharpoons2NO\left( g \right)

{K_1}

and that for

NO(g)\leftrightharpoons\frac{1}{2}{N_2}(g) + \frac{1}{2}{O_2}(g)

{K_2}

, then

{K_1}

and

{K_2}

related as

View answer and explanation

Correct option: A

{K_1} = {\left( {\frac{1}{{{K_2}}}} \right)^2}

The correct answer is A (

{K_1} = {(\frac{1}{{{K_2}}})^2}

) because it correctly relates the equilibrium constants of the forward and reverse reactions.

Revision notes and key takeaways

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