Equilibrium Test 5

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Equilibrium Test 5

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Question 1
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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

A
a

B
b

C
c

D
d
Question 2
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On taking 60.0 g CH₃COOH and 46.0 g CH₃CH₂OH in a 5 L flask in the presence of H₃0⁺ (catalyst), at 298 K 44.0 g of CH₃COOC₂H₅ is formed at equilibrium.
If amount of CH₃COOH is doubled without affecting amount of CH₃CH₂OH, then CH₃COOC₂H₅ formed is

A
20.33 g

B
22.0 g

C
44.0 g

D
58.66 g

Solution

Molar mass of CH₃COOH=60gmol^-1
Molar mass of C₂H₅OH=46gmol^-1
Molar mass of CH₃COOC₂H₅=88gmol^-1
∴[CH₃COOH)Initial=6060×5=0.2molL^-1
[C₂H₅OH]Initial=4646×5=0.2molL^-1
[CH₃COOC₂H₅]eqm=4488×15=0.1molL-1
CH₃COOH+C₂H₅OH⇔CH₃COOC₂H₅+H₂O
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=[CH₃COOC₂H₅][H₂O]/[CH₃COOH][C₂H₅OH]
=0.1×0.1/0.1×0.1=1
In second case,
[CH₃COOH]Initial=0.4M
[C₂H₅OH]Initial=0.2M
If x is the amount of acid and alcohol reacted
[CH₃COOH]eqm.=(0.40-x)M
[C2H₅OH]eqm.=[0.2-x]M
[CH₃COOC₂H₅]eqm.=[H₂O]eqm.=xM
∴K=x²/(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
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For the reversible reaction,

In a reaction vessel, [NO]= [O₂]= 0.01 mol L^-1 and [NO₂]= 0.1 mol L^-1 then above reaction is

A
shifted in forward direction

B
shifted in backward direction

C
in equilibrium

D
not predictable in the absence of required data

Solution

On substituting the values of conc. of NO, O₂ and NO₂ in given rate equation, we get a +ve (positive) value indicating that the reaction takes place in forward direction.
Question 4
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The equilibrium constant for the following reaction, is 1.6 x 10⁵ at 1024 K.
H₂(g) + Br₂(g) ⇌2HBr(g)
HBr (g)at 10.0 bar is introduced into a sealed container at 1024 K. Thus, partial pressure of H₂(g)and Br₂(g), together is

A
10 bar

B
0.05 bar

C
0.025 bar

D
0.10 bar

Solution

Squaring on both sides

⇒
⇒
⇒
=> 10 bar approximately
Question 5
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At 273 K and 1 atm, 1 L of N₂O₄ (g) decomposes to NO₂(g)a s given,
N₂O₄(g) ⇌2NO₂(g)
At equilibrium, original volume is 25% less than the existing volume. Percentage decomposition of N₂O₄ (g) is thus,

A
25%

B
50%

C
66.66%

D
33.33%

Solution

Let the initial volume of N₂O₄ be x and initial volume of NO₂ is 0
If the degree of dissociation is a, then the final volume of N₂O₄ 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
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H₂S (g) initially at a pressure of 10.0 atm and a temperature of 800 K, dissociates as
2H₂S(g) ⇌ 2H₂(g) + S₂(g)
At equilibrium, the partial pressure of S₂ vapour is 0.020 atm . Thus, K_p is

A
3.23x 10^-7

B
6.45x 10^-7

C
1.55x 10⁶

D
6.20x 10⁷

Solution

The correct answer is Option A.

2H₂S(g) ⇌ 2H₂(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
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A sample of N₂O₄(g)with a pressure of 1.00 atm is placed in a flask. When equilibrium is reached, 20% of N₂O₄(g)has been converted to NO₂(g)
If the original pressure is made 10% of the earlier pressure, then per cent dissociation will be

A
20%

B
42%

C
54%

D
62%

Solution

Correct answer is A.
Question 8
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Following equilibrium is set up at 1000 K and 1 bar in a 5 L flask,

At equilibrium, NO₂ is 50% o f the total volume. Thus, equilibrium constant K_c is

A
0.133

B
0.266

C
0.200

D
0.400

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
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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

A
9.0

B
15.0

C
3.0

D
0.0667

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
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At 700 K and 350 bar, a 1 : 3 mixture of N₂(g) and H₂(g) reacts to form an equilibrium mixture containing X (NH₃)= 0.50. Assuming ideal behaviour K_p for the equilibrium reaction,

A
2.03x 10^-4

B
3.55x 10^-3

C
1.02 x 10^-4

D
3.1 x 10^-4

Solution

The correct answer is option A
2.03x 10^-4
The given equation is :-
N2(g)+3H2(g) ⇌ 2NH₃(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(NH₃) =
⇒ 2x = 2 − x
⇒ 3x = 2 ⇒ x = 0.66 =
32
Now, at equation, moles of N₂= 1/3, moles of NH₃ = 4/3
moles of H2 =3 − 2 = 1
Question 11
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Equilibrium constant for the reaction,

is 1.8 x 109. Hence, equilibrium constant for

A
1.8x 10^-5

B
1.8x 10^-7

C
1.8x 10^-9

D
5.55x 10^-10

Solution

The correct answer is Option A.
NH₄OH + H+ ⇌ NH₄+ + H₂O
∴
Again,
NH₃ + H₂O → NH₄OH ⇌ NH
4++ OH-

= [OH
-] [H+] (∵H2O is in excess)
=K_w
=1×10^-14
∴K = K×1×10^-14
=1.8×10⁹×10^-14
=1.8×10^-5
Question 12
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A gaseous phase reaction taking place in 1L flask at 400 K is given,

Starting with 1 mole N₂ and 3 moles H₂, equilibrium mixture required 250 mL of 1M H₂SO₄ . Thus, K_c is

A
0.006

B
0.08

C
0.029

D
2.05
Question 13
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For the following equilibrium starting with 2 moles SO₂ and 1 mole O₂ in 1 L flask,

Equilibrium mixture required 0.4 mole in acidic medium. Hence, K_c is

A
2.0

B
0.4

C
1.6

D
2.6

Solution

0.4m of KMnO₄ = 1 mole of SO₂
= 1 mole of SO₃
2SO₂ + O₂ ⇌ 2SO₃
2 1 0
1 0.5 1
K = [SO₃]²/[SO₂]² [O₂]
= 12/(1²*0.5)
= 2
Question 14
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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 N₂, 3 moles H₂ and 2 moles NH₃ in 1L flask is in equilibrium.
Statement II
Reaction quotient of the given quantities is less than the equilibrium constant.

A
Both Statement I and Statement II are correct and Statement II is the corect explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is incorrect but Statement II is correct
Question 15
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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

A
Both Statement I and Statement II are correct and Statement II is the corect explanation of Statement I

B
Both Statement I and Statement II are correct and Statement II is not the correct explanation of Statement I

C
Statement I is correct but Statement II is incorrect

D
Statement II is incorrect but Statement II is correct