Equilibrium Test

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Equilibrium Test - 4

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Question 1
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
4 / -1

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
Question 16
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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 NO₂ and N₂O₄ gas at 25°C is 1.50 atm, what fraction of the N₂O₄ has dissociated to NO₂ ?

A
0.156

B
0.844

C
0.024

D
0.076

Solution

The correct answer is Option A.

Fraction of N₂O₄dissociated = x = 0.155 (x = mole fraction)
Question 17
4 / -1

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 N₂O₄ dissociated is

A
0.036

B
0.064

C
0.911

D
0.189
Question 18
4 / -1

Passage II
A 15 L flask at 300 K contains 64.4 g of a mixture of NO₂ and N₂O₄ in equilibrium. Given,
Q. K_c for the above equilibrium is

A
164.28

B
6.087x 10^-3

C
0.2708

D
3.693

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
4 / -1

Passage II
A 15 L flask at 300 K contains 64.4 g of a mixture of NO₂ and N₂O₄ in equilibrium. Given,
Q. Total pressure in the flask is

A
0.7427 atm

B
1.3400 atm

C
0.3714 atm

D
0.6732 atm
Question 20
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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 H₂(g)and l₂(g). After equilibrium is attained, l₂(g)was dissolved in Kl and required 100 mL of 0.1 M Na₂S₂O₃ solution.
Match the values given in Column II with the respective terms in Column I

A
a

B
b

C
c

D
d
Question 21
4 / -1

H₂O is a:

A
Proton acceptor

B
Proton donor

C
Both proton donor and proton acceptor

D
Neither proton donor, nor proton acceptor

Solution