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

In a gaseous reversible reaction, ${{\rm{N}}_2} + {{\rm{O}}_2}\leftrightharpoons2{\rm{NO}} + {\rm{heat}}$ If pressure is increased then the equilibrium constant would be

Sometimes increased, sometimes decreased

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

For the reaction, ${\rm{CuS}}{{\rm{O}}_4} \cdot 5{{\rm{H}}_2}{\rm{O}}\left( s \right)\leftrightharpoons{\rm{CuS}}{{\rm{O}}_4} \cdot 3{{\rm{H}}_2}{\rm{O}}\left( s \right) + 2{{\rm{H}}_2}{\rm{O}}\left( v \right)$ which one is correct representation?

${K_p} = {\left( {{P_{{{\rm{H}}_2}{\rm{O}}}}} \right)^2}$

${K_c} = {\left[ {{{\rm{H}}_2}{\rm{O}}} \right]^2}$

${K_p} = {K_c}{\left( {RT} \right)^2}$

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

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Equilibrium practice · Mixed difficulty

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1 Hard

1 of 408

Analyzing +4 | -1

Given below are two statements: One is labelled as Assertion A and the other labelled as Reason R.
Assertion A: During the boiling of water having temporary hardness,

Mg{(HC{O_3})_2}

is converted to

MgC{O_3}

.
Reason R The solubility product of

Mg{(OH)_2}

is greater than that of

MgC{O_3}

.
In the light of the above statements, choose the most appropriate answer from the options given below

View answer and explanation

Correct option: D

A is false but R is true.

Option D (A is false but R is true) is correct because

Mg(HCO_3)_2

decomposes to

Mg(OH)_2

and

MgCO_3

upon boiling, and the solubility product of

Mg(OH)_2

is indeed greater than that of

MgCO_3

2 Hard

2 of 408

Analyzing +4 | -1

At a certain temperature,

K_{\mathrm{P}}

for the dissociation of solid

\mathrm{CaCO}_3

4.5 \times 10^{-2}

atm and for the reaction

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g}),

the

K_{\mathrm{P}}

value is

2.0 \mathrm{~atm}.

The pressure of carbon monoxide at this temperature, when solid carbon

\mathrm{CaO}

and

\mathrm{CaCO}_3

are mixed together and allowed to attain equilibrium is

View answer and explanation

Correct option: B

0.30 \mathrm{~atm}

The correct answer is B (0.30 atm) because it is derived from setting up equilibrium expressions and solving for the partial pressure of CO.

3 Easy

3 of 408

Remembering +4 | -1

Phenolphthalein shows … in acid medium.

View answer and explanation

Correct option: D

No colour

Option D (No colour) is correct because phenolphthalein is colorless in acidic solutions and turns pink in basic solutions.

4 Medium

4 of 408

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.

5 Medium

5 of 408

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.

6 Medium

6 of 408

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.

7 Easy

7 of 408

Analyzing +4 | -1

Which can act as buffer?

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because all the given options contain a weak acid/base and its salt with a strong base/acid, which are the components of a buffer solution.

8 Medium

8 of 408

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.

9 Easy

9 of 408

Remembering +4 | -1

The

pH

of gastric juice is normally:

View answer and explanation

Correct option: C

Greater than 1 and less than 3

Option C (Greater than 1 and less than 3) is correct because the normal pH of gastric juice is highly acidic, typically ranging from 1 to 3 due to the presence of hydrochloric acid (HCl).

10 Easy

10 of 408

Application +4 | -1

A buffer mixture of acetic acid and potassium acetate has

\mathrm{pH} = 5.24.

The ratio of

\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }} \right]/\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}} \right]

in this buffer is,

\left( {\mathrm{p}{K_a} = 4.74} \right):

View answer and explanation

Correct option: A

3:1

Option A (

3:1

) is correct because using the Henderson-Hasselbalch equation, the ratio of the conjugate base to the acid is calculated to be approximately 3:1.

11 Easy

11 of 408

Understanding +4 | -1

Relation between hydrolysis constant and dissociation constant are given. Which is the correct formula for

\mathrm{MgC}{\mathrm{l}_2}

View answer and explanation

Correct option: B

{K_h} = \frac{{{K_w}}}{{{K_b}}}

Option B (

{K_h} = \frac{{{K_w}}}{{{K_b}}}

) is correct because

MgCl_2

is a salt of a strong acid (HCl) and a weak base (Mg(OH)2), so its hydrolysis constant is related to the base dissociation constant.

12 Easy

12 of 408

Understanding +4 | -1

The salt that forms neutral solution in water is

View answer and explanation

Correct option: B

NaCl

Option B (

NaCl

) is correct because it is the salt of a strong acid (HCl) and a strong base (NaOH), resulting in a neutral solution when dissolved in water.

13 Medium

13 of 408

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.

14 Hard

14 of 408

Analyzing +4 | -1

What is the correct relationship between the

\mathrm{pHs}

of isomolar solutions of sodium oxide

\left(\mathrm{pH}_1\right)

, sodium sulphide

\left(\mathrm{pH}_2\right)

, sodium selenide

\left(\mathrm{pH}_3\right)

and sodium telluride

\left(\mathrm{pH}_4\right)

View answer and explanation

Correct option: A

\mathrm{pH}_1 > \mathrm{pH}_2 > \mathrm{pH}_3 > \mathrm{pH}_4

Option A (

\mathrm{pH}_1 > \mathrm{pH}_2 > \mathrm{pH}_3 > \mathrm{pH}_4

) is correct because the basic strength of the oxides/chalcogenides decreases down the group.

15 Easy

15 of 408

Understanding +4 | -1

In a gaseous reversible reaction,

{\mathrm{N}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{NO} + \mathrm{heat}

If pressure is increased then the equilibrium constant would be

View answer and explanation

Correct option: A

Unchanged

Option A (Unchanged) is correct because increasing the pressure does not affect the equilibrium constant, only the equilibrium position, for a reaction with an equal number of moles of gas on both sides.

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16 Medium

16 of 408

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.

17 Easy

17 of 408

Understanding +4 | -1

In a reversible reaction two substance are in equilibrium. If the concentration each one is doubled, the equilibrium constant will be

View answer and explanation

Correct option: C

Constant

Option C (Constant) is correct because the equilibrium constant is only affected by temperature and is independent of the initial concentrations of reactants or products.

18 Medium

18 of 408

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.

19 Easy

19 of 408

Understanding +4 | -1

For the reaction,

\mathrm{CuS}{\mathrm{O}_4} \cdot 5{\mathrm{H}_2}\mathrm{O}\left( s \right)\leftrightharpoons\mathrm{CuS}{\mathrm{O}_4} \cdot 3{\mathrm{H}_2}\mathrm{O}\left( s \right) + 2{\mathrm{H}_2}\mathrm{O}\left( v \right)

which one is correct representation?

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because the equilibrium constant only depends on the partial pressure of water vapor, and the relationship between Kp and Kc is correctly represented.

20 Easy

20 of 408

Understanding +4 | -1

In which of the following reactions, the value of

{K_p}

will be equal to

{K_c}

View answer and explanation

Correct option: C

{\mathrm{H}_2} + {\mathrm{I}_2}\leftrightharpoons2\mathrm{HI}

Option C (

{\mathrm{H}_2} + {\mathrm{I}_2}\leftrightharpoons2\mathrm{HI}

) is correct because the change in the number of moles of gas (delta n) is zero, making Kp equal to Kc.

21 Easy

21 of 408

Application +4 | -1

Consider the following reaction equilibrium

{\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)

Initially,

1\;mole

{\mathrm{N}_2}

and

3\;mole

{\mathrm{H}_2}

are take in a

2\;L

flask. At equilibrium state if, the number of moles of

{\mathrm{N}_2}

is 0.6, what is the total number of moles of all gases present in the flask?

View answer and explanation

Correct option: C

3.2

Option C (3.2) is correct because it represents the total number of moles of all gases present at equilibrium, calculated based on the change in moles of nitrogen.

22 Medium

22 of 408

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).

23 Medium

23 of 408

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}

24 Hard

24 of 408

Application +4 | -1

A vessel at 1000 K contains

\mathrm{C}{\mathrm{O}_2}

with a pressure of 0.5 atm. Some of the

\mathrm{C}{\mathrm{O}_2}

is converted into

\mathrm{CO}

on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of

{K_p}

View answer and explanation

Correct option: A

1.8 atm

Option A (1.8 atm) is correct because the equilibrium constant is calculated based on the partial pressures of the reactants and products at equilibrium.

25 Medium

25 of 408

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.

26 Easy

26 of 408

Understanding +4 | -1

For the reversible reactions.

CaC{O_{3\left( S \right)}}Ca{O_{\left( s \right)}} + C{O_{2\left( g \right)}}

, Which of the following does not affect the equilibrium state ?

View answer and explanation

Correct option: C

Addition of CaO(s)

Option C (Addition of CaO(s)) is correct because adding a solid reactant or product does not affect the equilibrium state, as the activity of a pure solid is considered to be 1 and does not appear in the equilibrium constant expression.

27 Easy

27 of 408

Application +4 | -1

A reaction is,

A + B\; \to C + D

. Initially we start with equal concentrations of

A

and

B

. At equilibrium, we find the moles of

C

are two times of

A

. What is the equilibrium constant of the reaction?

View answer and explanation

Correct option: B

Option B (4) is correct because the equilibrium constant is calculated based on the equilibrium concentrations, where [C] = 2[A].

28 Medium

28 of 408

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.

29 Easy

29 of 408

Understanding +4 | -1

Which one is hard base?

View answer and explanation

Correct option: D

{\mathrm{F}^ - }

Option D (

{\mathrm{F}^ - }

) is the correct answer because fluoride ion is small, highly charged, and not easily polarizable, making it a hard base.

30 Medium

30 of 408

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.

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Key Takeaways

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