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Chemistry

Chemical Equilibrium

101 questions · Click an option and press Check Answer to reveal the result

29.402 atm

0.0625 atm

6.25 atm

0.00625 atm

For the reaction : PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) at 250°C the Value of Kc is 26, then the value of Kp at the same temp. will be –

0.61

0.57

0.83

0.46

The statement not applicable to an irreversible reaction is –

It goes to completion in the forward direction

On increasing the concentration of the reactants, the rate of the reaction increases

The removal of the reaction products makes the reaction faster

The addition of the reaction products does not influence the rate of reaction

When 120 g acetic acid is reacted with 138 g ethyl alcohol, the amount of alcohol consumed is 30% of its initial concentration. The number of moles of acetic acid at equilibrium are –

1.1

2.1

1.7

0.9

In the equilibrium N₂ + 3H₂ ⇌ 2NH₃ + 22 Kcal the formation of ammonia is favoured by –

Increasing the pressure

Increasing the temperature

Decreasing the pressure

Adding ammonia

For an exothermic reaction in which the number of moles of reactants are more than the number of moles of products. In order to displace the reaction in the reverse direction, what are the favourable conditions .

High pressure, low temp. and high conc. of the product

Low pressure, high temp. and high conc. of the products

Low pressure, low temp. and high conc. of the products

High pressure, high temp. and low conc. of the products

The equilibrium constant for the reaction Zn(s)+CO₂(g) ⇌ ZnO(s) + CO(g) is -

[ZnO]/[Zn]

An equilibrium mixture for the reaction 2H₂S(g) ⇌ 2H₂(g) + S₂(g) had 1 mole of H₂S, 0.2 mole of H₂ and 0.8 mole of S₂ in a 2 litre flask. The value of Kc in mol lit.⁻¹ is–

0.004

0.080

0.016

0.160

For the reaction : N₂O₃(g) ⇌ NO(g) + NO₂(g) ; total pressure = P, degree of dissociation = 50%. Then Kₚ would be –

P/3

P/2

In what manner will increase of pressure affect the equation C(s) + H₂O(g) ⇌ CO(g) + H₂(g) -

Shift in the forward direction

Shift in the reverse direction

Increase in the yield of H₂

No effect

If Kp for the reaction A(g) + 2B(g) ⇌ 3C(g) + D(g) is 0.05 atm. at 1000 K. Its Kc in terms of R will be –

20000 R

0.02 R

5 × 10⁻⁵ R

(5 × 10⁻⁵)/R

In the formation of nitric acid, N₂ and O₂ are made to combine. Thus N₂ + O₂ ⇌ 2NO − Heat Which of the following conditions will favour the formation of NO –

Low temperature

High temperature

High pressure

All are favourable

The role of catalyst in reversible reaction is-

To increase the rate of forward reaction

Decrease the rate after equilibrium

Allow equilibrium to be achieved quickly

None

If in the reaction N₂O₄(g) ⇌ 2NO₂(g), α is the part of N₂O₄ which dissociates then the number of moles at equilibrium will be -

(1−α)²

(1 + α)

In a chemical equilibrium, the equilibrium constant is found to be 2.5. If the rate constant of backward reaction is 3.2 × 10⁻², the rate constant of forward reaction is -

8.0 × 10⁻²

4.0 × 10⁻²

3.5 × 10⁻²

7.6 × 10⁻³

An unknown compound A dissociates at 500°C to give products as follows - A(g) ⇌ B(g) + C(g) + D(g) Vapour density of the equilibrium mixture is 50 when it dissociates to the extent to 10%. What will be the molecular weight of Compound A –

120

130

134

140

At 250° C, the vapour density of PCl₅ is Y (at equilibrium) and molar mass is Q (Initially). Its degree of dissociation is then equal to -

(Q − Y)/Y

(Y − Q)/Q

(Y − 2Q)/(2Q)

(Q − 2Y)/(2Y)

At 250°C and 1 atmospheric pressure, the vapour density of equilibrium mixture obtained on dissociation of PCl₅ is 57.9. What will be the degree of dissociation of PCl₅ –

1.00

0.90

0.80

0.65

During thermal dissociation of gas, the vapour density -

Remains same

Will be increased

Will be decreased

Some times increases some times decreases

According to Law of Mass action, the rate of reaction is directly proportional to -

molarities of the reactants

normalities of the reactants

molalities of the reactants

mole fractions of the reactants

(a + b) > (c + d)

(a + b) − (c + d) = 0

(c + d) > (a + b)

(a + c) = (b + d)

0.02

0.2

50.0

50/RT

The equilibrium constant for the reaction 2X(g) + Y(g) ⇌ 2Z(g) is 2.25. What would be the concentration of Y at equilibrium with 2.0 moles of X and 3.0 moles of Z in one litre vessel at equilibrium –

1.0 moles

2.25 moles

2.0 moles

4.0 moles

K₁ and K₂ are the rate constants of forward and backward reactions. The equilibrium constant K of the reaction is -

K₁ × K₂

K₁ − K₂

K₁/K₂

(K₁ + K₂)/(K₁ − K₂)

At 500K, the equilibrium constant for the reaction cis-C₂H₂Cl₂ ⇌ trans-C₂H₂Cl₂ is 0.6. At the same temperature, the equilibrium constant for the reaction trans-C₂H₂Cl₂ ⇌ cis-C₂H₂Cl₂ will be –

0.66

1.67

0.76

2.6

The Kc for the reaction A + B ⇌ C + D is 9. If one mole of each of A and B are initially mixed and there is no change in volume, the number of moles of C formed is –

0.50

0.75

0.90

1.5

At any moment before a reversible reaction attains equilibrium it is found that –

The rate of the forward reaction is increasing and that of backward reaction is decreasing

The rate of the forward reaction is decreasing and that of backward reaction is increasing

The rate of both forward and backward reactions is increasing

The rate of both forward and backward reactions is decreasing

2SO₂ + O₂ ⇌ 2SO₃ + 47 Kcal. If large excess of oxygen is added to the above system in equilibrium this will -

Shift the equilibrium to the right

Shift the equilibrium to the left

Cause no change in the equilibrium

Produce only enormous amount of heat

The ratio of Kp / Kc for the reaction : CO(g) + 1/2 O₂(g) ⇌ CO₂(g) is –

1.0

1/√RT

(RT)¹ᐟ²

The relation between Kp and Kc for the reaction A + B ⇌ C + 2D is –

Kp = Kc(RT)⁻¹

KpKc⁻¹ = RT

KcKp⁻¹ = RT

Kp = Kc(RT)³

In the reaction C₂H₄(g) + H₂(g) ⇌ C₂H₆(g), the equilibrium constant can be expressed in units of -

litre⁻¹ mol⁻¹

mol² litre⁻²

litre mol⁻¹

mol litre⁻¹

For the reaction A(g) + B(g) ⇌ C(g) + D(g), the degree of dissociation α would be –

√K/(√K + 1)

√K + 1

√K ± 1

√K − 1

The reversible reaction [Cu(NH₃)₄]²⁺ + SO₃²⁻ ⇌ [Cu(NH₃)₃SO₃] + NH₃ is at equilibrium. What would not happen if ammonia is added –

[SO₃²⁻] would increase

[Cu(NH₃)₃SO₃] would increase

The value of equilibrium constant would not change

[Cu(NH₃)₄]²⁺ would increase

At a total equilibrium pressure of 1.0 atmosphere, the degree of dissociation of phosgene is 0.2 COCl₂(g) ⇌ CO(g) + Cl₂(g) Now the same equilibrium is established at the same temperature in presence of N₂ gas at a partial pressure of 0.4 atm. in a total pressure of 1.0 atm. The new degree of dissociation, α, is –

α > 0.2

α = 0.2

α = 0

The reaction in which the yield of the products can not be increased by the application of high pressure is –

PCl₃(g) + Cl₂(g) ⇌ PCl₅(g)

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

N₂(g) + O₂(g) ⇌ 2NO(g)

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

In A + B + Heat ⇌ C, then reaction is favoured at –

High temperature and high pressure

High temperature and low pressure

Low temperature and high pressure

Low pressure and low temperature

When the equilibrium : 2NH₃ ⇌ N₂ + 3H₂ has been established, NH₃ is found to be 20% dissociated. The ratio of total number of moles at equilibrium to the moles of NH₃ at equilibrium is –

3/2

2/3

3/1

1/3

For the equilibrium reaction : 2HCl(g) ⇌ H₂(g) + Cl₂(g) the equilibrium constant is 1.0 × 10⁻⁵ what is the concentration of HCl if the equilibrium concentrations of H₂ and Cl₂ are 1.2 × 10⁻³ M and 1.2 × 10⁻⁴ M respectively –

12 × 10⁻² M

12 × 10⁻⁴ M

12 × 10⁻³ M

12 × 10⁴ M

In the melting of ice, which one of the conditions will be more favourable –

High temperature and high pressure

Low temperature and low pressure

Low temperature and high pressure

High temperature and Low pressure

For the reaction C(s) + CO₂(g) ⇌ 2CO(g) the partial pressure of CO₂ and CO are 2.0 and 4.0 atm respectively at equilibrium. The Kₚ for the reaction is –

0.5

4.0

8.0

32.0

The reaction in which the yield of the products cannot be increased by the application of high pressure is –

PCl₃(g) + Cl₂(g) ⇌ PCl₅(g)

N₂(g) + O₂(g) ⇌ 2NO(g)

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

In Bodenstein experiment, one mol H₂ and two mol I₂ are taken initially in a one litre flask. If at equilibrium 0.5 mol H₂ are present, the equilibrium concentrations of I₂ and HI in mol l⁻¹ are –

0.5 , 1.0

1.0 , 0.5

1.5 , 1.0

1.5 , 0.5

8 × 10⁻³

6.25 × 10⁴

6.25 × 10⁸

8 × 10⁴

The equilibrium constant of the reaction A + B ⇌ C + D is 10. If rate constant of forward reaction is 203, the rate constant of backward reaction is –

20.3

10.3

2.03

203

Which of the following will shift the reaction PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) to the left side-

Addition of PCl₅

Increase in pressure

Decrease in temperature

Catalyst

The value of Kp for the reaction 2H₂S(g) ⇌ 2H₂(g) + S₂(g) is 1.2 × 10⁻² at 1065°C. The value of Kc for this reaction is–

1.2 × 10⁻²

> 1.2 × 10⁻²

0.02

0.03

0.04

0.05

At a given temperature the following reaction is allowed to reach equilibrium in a vessel of volume V₁ litre. The degree of dissociation is α₁. If by keeping the temperature fixed the volume of the reaction vessel is doubled (assuming the degrees of dissociation to be small) the new degree of dissociation shall be – PCl₅ ⇌ PCl₃ + Cl₂

2α₁

√(α₁/2)

√(2α₁)

√2 α₁

When H₂ is added to an equilibrium mixture 2HI(g) ⇌ H₂(g) + I₂(g), at constant temperature, the-

Value of Kₚ decreases

Value of Kₚ increases

The degree of dissociation of HI decreases

Degree of dissociation of HI increases

In the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the forward reaction is exothermic and the backward reaction is endothermic. In order to produce more heat it is necessary –

To add ammonia

To add N₂ and H₂

Increasing the concentration of N₂, H₂ and NH₃ equally

None of the above

Increasing concentration of the reactant

Presence of catalyst

Method of writing balanced equation (or stoichiometry of reaction)

Time taken by the chemical reaction

√2

√3

√5

Which among the following conditions, increases the yield of the product in the equilibrium, 3A(g) + B(g) ⇌ 4C(g) + heat

Increase in pressure

Increase in volume

Increase in temperature

Decrease in temperature

At 444° C, the equilibrium constant K for the reaction 2AB(g) ⇌ A₂(g) + B₂(g) is 1/64. The degree of dissociation of AB will be -

10%

20%

30%

50%

In closed vessel PCl₅ undergoes thermal dissociation as – PCl₅ ⇌ PCl₃ + Cl₂, the mole fraction of PCl₃ at equilibrium is 0.4. The total pressure is 2.0 atmosphere. The partial pressure of Cl₂ at equilibrium is –

2.5

0.8

0.5

None

H₂ + I₂ ⇌ 2HI

N₂ + 3H₂ ⇌ 2NH₃

N₂ + O₂ ⇌ 2NO

CO + H₂O ⇌ CO₂ + H₂

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + Heat –

The value of Δng for the reaction : 2Hg(l) + Cl₂(g) ⇌ Hg₂Cl₂(s) is –

−1

−2

What is wrong about equilibrium state –

ΔG(equi) = 0

The reaction ceases at equilibrium

Equilibrium constant is independent of initial concentrations of reactants

Catalyst has no effect on equilibrium state

Consider the reaction, CaCO₃(s) ⇌ CaO(s) + CO₂(g); in closed container at equilibrium. What would be the effect of addition of CaCO₃ on the equilibrium concentration of CO₂ -

Increases

Decreases

Remains unaffected

Data is not sufficient to predict it

At a total pressure of 2 atmospheres, the number of moles of N₂, H₂ and NH₃ at equilibrium in a gaseous mixture are 1, 3 and 2. The partial pressure of NH₃ in atmospheres is –

1.0

0.5

0.33

0.67

(RT)⁻³ , (RT)² , (RT)⁰

(RT)⁻³ , (RT)⁻² , (RT)⁻¹

(RT)⁻³ , (RT)² , (RT)

None of the above

The equilibrium constants for the reaction X₂ ⇌ 2X at 300K and 600K are 10⁻⁸ and 10⁻³ respectively. The reaction is –

Exothermic

Endothermic

Thermo neutral

Slow

0.625

0.0625

6.25

0.00625

Adding inert gas to system N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at equilibrium at constant volume will be –

N₂ and H₂ are formed in abundance

N₂, H₂ and NH₃ will have the same molar concentration

The production of ammonia increases

No change in the equilibrium

Dissociation of phosphorus pentachloride is favoured by –

High temperature and high pressure

High temperature and low pressure

Low temperature and low pressure

Low temperature and high pressure

Which of the following reaction will be favoured at low pressure -

H₂(g) + I₂(g) ⇌ 2HI(g)

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

N₂(g) + O₂(g) ⇌ 2NO(g)

In a 10 litre box 2.5 mole hydroiodic acid is taken. After equilibrium 2HI ⇌ H₂ + I₂ the concentration of HI is found to be 0.1 mol l⁻¹. The concentration of [H₂] at equilibrium in mol l⁻¹, is –

2.4

0.15

1.5

7.5 × 10⁻²

In the reaction A(g) + B(g) ⇌ C(g), the backward reaction is favoured by -

Increase in pressure

Decrease in pressure

Neither increase nor decrease in pressure

Data unpredictable

In the study of the reaction – Cl₂ + PCl₃ ⇌ PCl₅ partial pressures of Cl₂, PCl₃ and PCl₅ at equilibrium are 0.1, 0.1 and 0.2 atm respectively at 250° C. At the same temperature, in another experiment on the same reaction, at equilibrium the partial pressures of PCl₃ and Cl₂ are half those in the first experiment. The partial pressure of the PCl₅ at equilibrium in the second experiment is –

One-fourth of the first

Half of the first

One-eighth of the first

One-third of the first

The oxidation of SO₂ to SO₃ by oxygen is an exothermic reaction. The yield of SO₃ will be maximum if –

Temperature is increased and pressure is kept constant

Temperature is reduced and pressure is increased

Both temperature and pressure are increased

Both temperature and pressure are decreased

N₂(g) + O₂(g) ⇌ 2NO(g)

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

2NH₃(g) ⇌ N₂(g) + 3H₂(g)

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

In an equilibrium reaction for which ΔG° = 0, the equilibrium constant K should be equal to –

According to Le Chatelier principle, an increase in the temperature of the following reaction will N₂ + O₂ ⇌ 2NO − 43200 cal

Increase the yield of NO

Decrease the yield of NO

Not effect on the yield of NO

Not help the reaction to proceed

N₂O₄ dissociates as N₂O₄(g) ⇌ 2NO₂(g) at 273 K and 2 atm pressure. The equilibrium mixture has a vapour density of 41. What will be the degree of dissociation -

14.2%

16.2%

12.2%

None

The unit of equilibrium constant for the reaction H₂ + I₂ ⇌ 2HI is –

Mole⁻¹ litre

Mole⁻² litre

Mole litre

None

(a) Kp > Kc ; (b) Kp = Kc

A chemical reaction A ⇌ B is said to be in equilibrium when -

Complete conversion of A to B has taken place

Conversion of A to B is only 50% complete

Only 10% conversion of A to B has taken place

The rate of transformation of A to B is just equal to rate of transformation of B to A in the system

On adding inert gas to the equilibrium PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) at constant pressure. The degree of dissociation will remain –

Unchanged

Decreased

Increased

None of these

The correct expression for equilibrium constant, Kc, for the reaction : H₂(g) + 1/2 O₂(g) ⇌ H₂O(g) is –

[H₂O]/([H₂][O₂]¹ᐟ²)

The reaction A(g) + B(g) ⇌ C(g) + D(g) proceeds to right hand side upto 99.9%. The equilibrium constant K for the reaction will be -

10⁴

10⁵

10⁶

10⁸

The equilibrium constant for the reaction N₂(g) + O₂(g) ⇌ 2NO(g) is 4 × 10⁻⁴ at 2000 K. In the presence of catalyst, the equilibrium is attained 10 times faster. The equilibrium constant in the presence of catalyst, at 2000 K is–

40 × 10⁻⁴

4 × 10⁻⁴

4 × 10⁻²

Difficult to compute without more data

The value of equilibrium constant for a reaction depends on-

Temperature

Pressure

Volume

Catalytic agent

In the reaction, 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) + X cals, most favourable condition of temperature and pressure for greater yield of SO₃ are -

Low temperature and low pressure

High temperature and low pressure

High temperature and high pressure

Low temperature and high pressure

A(l) + B(l) ⇌ C(l) + D(l) One mole of A and one mole of B are mixed in a volume of one litre. If 0.9 mole per litre of C is found at equilibrium, the equilibrium constant Kc is –

10⁻¹

1.5

0.9 × 10⁶

7.5 × 10²

1.95 × 10⁻³

1.95 × 10³

Which of the following statements is false in case of equilibrium state –

There is no apparent change in properties with time

It is dynamic in nature

It can be attained from either side of the reaction

It can be attained from the side of the reactants only

When a reversible reaction has reached the state of equilibrium -

The forward reaction stops

The backward reaction stops

The whole reaction stops

The forward and backward reaction proceed with same speed

For the reaction at equilibrium – CO₂(g) + H₂(g) ⇌ CO(g) + H₂O(g) [P – total pressure and x = degree of dissociation]

PCO₂ and PH₂ is equal to [(1−x)/2]P

PCO and PH₂ is equal to [(1−x)/2]P

PH₂ and PH₂O is equal to [x/2]P

PCO₂ and PH₂O is equal to [x/2]P

Which of the following conditions will be favourable for the formation of HX according to the gaseous equilibrium : H₂X₂ + heat ⇌ 2HX –

High temperature and high pressure

High temperature and low pressure

Low temperature and low pressure

High pressure and low temperature

Which of the following equilibrium is not affected by pressure -

N₂(g) + O₂(g) ⇌ 2NO(g)

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

2O₃(g) ⇌ 3O₂(g)

2NO₂(g) ⇌ N₂O₄(g)

Chemical equilibrium is dynamic in nature because –

The equilibrium in maintained quickly

Conc. of reactants and products become same at equilibrium

Conc. of reactants and products are constant but different

Both forward and backward reactions occur at equilibrium with same speed

100

For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), the forward reaction at constant temperature is favoured by -

Increasing the volume of container

Introducing an inert gas at constant pressure

Introducing PCl₅ at constant volume

All of these

101

For dissociation of PCl₅,mole fraction of Cl₂ at equilibrium is 0.3. The total pressure is 1.0 atmosphere. The partial pressure of PCl₃ at equilibrium is –

1.3

0.3

0.7

0.4