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QUESTION: 1

**Direction (Q. Nos. 1-13) This section contains 13 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.**

**Q. Given the reactions, **** If one mole each of A and B are take in 5 L flask at 300 K, 0.7 mole of C are formed. Molar concentration of each species at equilibrium, when one mole of each are taken initially is**

Solution:

QUESTION: 2

On taking 60.0 g CH_{3}COOH and 46.0 g CH_{3}CH_{2}OH in a 5 L flask in the presence of H_{3}0^{+} (catalyst), at 298 K 44.0 g of CH_{3}COOC_{2}H_{5} is formed at equilibrium.

If amount of CH_{3}COOH is doubled without affecting amount of CH_{3}CH_{2}OH, then CH_{3}COOC_{2}H_{5} formed is

Solution:

Molar mass of CH_{3}COOH=60gmol^{-1}

Molar mass of C_{2}H_{5}OH=46gmol^{-1}

Molar mass of CH_{3}COOC_{2}H_{5}=88gmol^{-1}

∴[CH_{3}COOH)Initial=6060×5=0.2molL^{-1}

[C_{2}H_{5}OH]Initial=4646×5=0.2molL^{-1}

[CH_{3}COOC_{2}H_{5}]eqm=4488×15=0.1molL-1

CH_{3}COOH+C_{2}H_{5}OH⇔CH_{3}COOC_{2}H_{5}+H_{2}O

Initial

0.2M 0.2M

0.2M 0.2M

At eqm.

(0.2−0.1)M (0.2−0.1)M 0.1M 0.1M

(0.2-0.1)M (0.2-0.1)M 0.1M 0.1M

∴K=[CH_{3}COOC_{2}H_{5}][H_{2}O]/[CH_{3}COOH][C_{2}H_{5}OH]

=0.1×0.1/0.1×0.1=1

In second case,

[CH_{3}COOH]Initial=0.4M

[C_{2}H_{5}OH]Initial=0.2M

If x is the amount of acid and alcohol reacted

[CH_{3}COOH]eqm.=(0.40-x)M

[C2H_{5}OH]eqm.=[0.2-x]M

[CH_{3}COOC_{2}H_{5}]eqm.=[H_{2}O]eqm.=xM

∴K=x^{2}/(0.4-x)(0.2-x)=1

x=860M

∴Moles of ethyl acetate produced =860×5=23

Mass of ethyl acetate produced =23×88=58=58.66g.

QUESTION: 3

For the reversible reaction,

In a reaction vessel, [NO]= [O_{2}]= 0.01 mol L^{-1} and [NO_{2}]= 0.1 mol L^{-1} then above reaction is

Solution:

On substituting the values of conc. of NO, O_{2} and NO_{2} in given rate equation, we get a +ve (positive) value indicating that the reaction takes place in forward direction.

QUESTION: 4

The equilibrium constant for the following reaction, is 1.6 x 10^{5} at 1024 K.

H_{2}(g) + Br_{2}(g) **⇌**2HBr(g)

HBr (g)at 10.0 bar is introduced into a sealed container at 1024 K. Thus, partial pressure of H_{2}(g)and Br_{2}(g), together is

Solution:

Squaring on both sides

⇒

⇒

⇒

=> 10 bar approximately

QUESTION: 5

At 273 K and 1 atm, 1 L of N_{2}O_{4} (g) decomposes to NO_{2}(g)a s given,

N_{2}O_{4}(g) ⇌2NO_{2}(g)

At equilibrium, original volume is 25% less than the existing volume. Percentage decomposition of N_{2}O_{4} (g) is thus,

Solution:

Let the initial volume of N_{2}O_{4} be x and initial volume of NO_{2} is 0

If the degree of dissociation is a, then the final volume of N_{2}O_{4} is x(1−a) and NO2 is 2ax.

Initial

It equilibrium

N2O4 ⟶ 2NO2

x 0

x(1−a) 2ax

Total initial volume =x+0=x

Final volume =x(1−a)+2ax=x+ax=x(1+a)

It is given that the initial volume is 25% less than the final volume

x=0.75×(1+a)

1+a=1.33

a=0.33

So %age dissociation = 33.33%

QUESTION: 6

H_{2}S (g) initially at a pressure of 10.0 atm and a temperature of 800 K, dissociates as

2H_{2}S(g) ⇌ 2H_{2}(g) + S_{2}(g)

At equilibrium, the partial pressure of S_{2} vapour is 0.020 atm . Thus, K_{p} is

Solution:

The correct answer is Option A.

2H_{2}S(g) ⇌ 2H_{2}(g) + S_{2(g)}

Pressure

at t=0 Pi − −

at eqm Pi−P 2P P

as P=0.02 thus Pi−P=10−0.02

Pi=10 2P=0.04

Kp = 3.23×10^{−7} atm.

QUESTION: 7

A sample of N_{2}O_{4}(g)with a pressure of 1.00 atm is placed in a flask. When equilibrium is reached, 20% of N_{2}O_{4}(g)has been converted to NO_{2}(g)

If the original pressure is made 10% of the earlier pressure, then per cent dissociation will be

Solution:

Correct answer is A.

QUESTION: 8

Following equilibrium is set up at 1000 K and 1 bar in a 5 L flask,

At equilibrium, NO_{2} is 50% o f the total volume. Thus, equilibrium constant K_{c} is

Solution:

The correct answer is Option A.

N2O4 ⇌ 2NO2

Initial 1 0

Equilibrium 1−x 2x

Total moles = 1 - x + 2x

NO2 is 50% of the total volume when equilibrium is set up.

Thus, the volume fraction (at equilibrium) of NO2 = 50/100 = 0.5 = ½

So, 2x / (1+x) = ½

=> x = ⅓

For 1 litre;

Kc = [NO2] / [N2O4]

= [4*(1/9)] / [⅔]

= 0.66;

For 5 litres;

Kc = 0.66 / 5

= 0.133

Thus, option A is correct.

QUESTION: 9

Following equilibrium is set up at 298 K in a 1 L flask.

If one starts with 2 moles of A and 1 mole of B, it is found that moles of B and D are equal.Thus K_{c} is

Solution:

For the equilibrium reaction:

A+2B ⇌ 2C+D

volume of flask = 1L

Initial moles of A = 2 mol

initial concentration of A=[A]i = 2 M

initial mole of B = 1 mol

[B]i = 1 M

[A]_{eq }= 2-x, [B]_{eq} = 1-2x, [C]_{eq} = x, [D]_{eq} = 3x

Given [D]_{eq} = 1 * 1L

= 1 M

Thus x = 1M

[A]_{eq} = 1, [B]_{eq} = -1, [C]_{eq} = 1, [D] = 3

Kc = {([D]eq)^{3} * ([C]eq)}/{[A]eq * ([B]eq)^{2}}

= Kc = {(3)^{3*}1}/{1*(-1)^{2}}

= 27/1

= 27

QUESTION: 10

At 700 K and 350 bar, a 1 : 3 mixture of N_{2}(g) and H_{2}(g) reacts to form an equilibrium mixture containing X (NH_{3})= 0.50. Assuming ideal behaviour K_{p} for the equilibrium reaction,

Solution:

The correct answer is option A

2.03x 10^{-4}

The given equation is :-

N2(g)+3H2(g) ⇌ 2NH_{3}(g)

Initial moles : 1 3 0

At eqm ; (1−x) (3−3x) (2x)

(let)

Total moles of equation

=1 − x + 3 − 3x + 2x = (4−2x)

Now, X(NH_{3}) =

⇒ 2x = 2 − x

⇒ 3x = 2 ⇒ x = 0.66 =

32

Now, at equation, moles of N_{2}= 1/3, moles of NH_{3} = 4/3

moles of H2 =3 − 2 = 1

QUESTION: 11

Equilibrium constant for the reaction,

is 1.8 x 109. Hence, equilibrium constant for

Solution:

The correct answer is Option A.

NH_{4}OH + H+ ⇌ NH_{4}+ + H_{2}O

∴

Again,

NH_{3} + H_{2}O → NH_{4}OH ⇌ NH

4++ OH-

= [OH

-] [H+] (∵H2O is in excess)

=K_{w}

=1×10^{-14}

∴K = K×1×10^{-14}

=1.8×10^{9}×10^{-14}

=1.8×10^{-5}

QUESTION: 12

A gaseous phase reaction taking place in 1L flask at 400 K is given,

Starting with 1 mole N_{2} and 3 moles H_{2}, equilibrium mixture required 250 mL of 1M H_{2}SO_{4} . Thus, K_{c} is

Solution:

QUESTION: 13

For the following equilibrium starting with 2 moles SO_{2} and 1 mole O_{2} in 1 L flask,

Equilibrium mixture required 0.4 mole in acidic medium. Hence, K_{c} is

Solution:

0.4m of KMnO_{4} = 1 mole of SO_{2}

= 1 mole of SO_{3 }

2SO_{2} + O_{2} ⇌ 2SO_{3}

2 1 0

1 0.5 1

K = [SO_{3}]^{2}/[SO_{2}]^{2} [O_{2}]

= 12/(1^{2}*0.5)

= 2

QUESTION: 14

**Direction (Q. Nos. 14 and 15) This sectionis based on statement I and Statement II. Select the correct answer from the code given below**.

**Q. Statement I **

If K_{c} = 4 for the following equilibrium,

Then mixture of 1 mole N_{2}, 3 moles H_{2} and 2 moles NH_{3} in 1L flask is in equilibrium.

**Statement II**

Reaction quotient of the given quantities is less than the equilibrium constant.

Solution:

QUESTION: 15

**Statement I **

1 mole A(g) and 1 mole B(g)give 0.5 mole of C(g)and 0.5 mole D(g) at equilibrium.

On taking 2 moles each of A(g)and B(g), percentage dissociation A(g)and B(g) is also doubled.

**Statement II**

Equilibrium constant, K_{c} = 1

Solution:

QUESTION: 16

**Direction (Q. Nos. 16-19) This section contains a paragraph, each describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given options (a),(b),(c),(d)**

Passage I

The equilibrium reaction has been thoroughly studied K_{p} = 0.148 at 298 K

If the total pressure in a flask containing NO_{2} and N_{2}O_{4} gas at 25°C is 1.50 atm, what fraction of the N_{2}O_{4} has dissociated to NO_{2} ?

Solution:

The correct answer is Option A.

Fraction of N_{2}O_{4 }dissociated = x = 0.155 (x = mole fraction)

QUESTION: 17

Passage I

The equilibrium reaction has been thoroughly studied K_{p} = 0.148 at 298 K

**Q. If the volume of the container is increased so that the total equilibrium pressure falls to 1.00 atm, then fraction of N _{2}O_{4} dissociated is**

Solution:

QUESTION: 18

Passage II

A 15 L flask at 300 K contains 64.4 g of a mixture of NO_{2} and N_{2}O_{4} in equilibrium. Given,

**Q. K _{c} for the above equilibrium is **

Solution:

Kp = Kc(RT)^{∆}n

Kp = 6.67 ,

∆n = moles of products - moles of reactants = 1-2 = -1

R = 0.0821 L atm mol-¹K-¹

T = 300K

Substitute these values in the formula,

=> Kc = 6.67×0.0821×300

Kc = 164.28.

QUESTION: 19

Passage II

A 15 L flask at 300 K contains 64.4 g of a mixture of NO_{2} and N_{2}O_{4} in equilibrium. Given,

**Q. Total pressure in the flask is **

Solution:

QUESTION: 20

**Direction (Q. Nos. 20) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.**

Q. In the following equilibrium, is set up at 298 K in a 5 L flask. Experiment is carried by taking 0.1 mole each of H_{2}(g)and l_{2}(g). After equilibrium is attained, l_{2}(g)was dissolved in Kl and required 100 mL of 0.1 M Na_{2}S_{2}O_{3} solution.

Match the values given in Column II with the respective terms in Column I

Solution:

*Answer can only contain numeric values

QUESTION: 21

**Direction (Q. Nos. 21) This section contains 2 questions. when worked out will result in an integer from 0 to 9 (both inclusive)**

**Q. For the equilibrium in gaseous phase in 2 L flask we start with 2 moles of SO _{2} and 1 mole of O_{2} at 3 atm, **

When equilibrium is attained, pressure changes to 2.5 atm. Hence, equilibrium constant K_{c} is

Solution:

The correct answer is 4

2SO_{2}(g) + O_{2}(g) ⇋ 2SO3

Initial moles 2 1

At equilibrium 2 - 2x 1 - x 2x

Net moles at equilibrium = 2 - 2x + 1 - x + 2x

=(3 - x)moles

Initial:

moles = 3,

Pressure = 3 atm,

Volume = 2L,

PV = nRT

3 x 2 = 3RT -------- 1

At equilibrium

Moles = 3 - x,

Pressure = 2.5 atm

Volume = 2L

P‘V = n’RT ---------- 2

Divide eqn 2 by 1

⇒2.5 = 3 - x

⇒x = 0.5

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