Chemical Equilibrium(12th Grade > Chemistry ) Questions and answers for exam Preparation

12th Grade > Chemistry

CHEMICAL EQUILIBRIUM MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1. Ammonia under a pressure of

15

atm at

27^{\circ} C

is heated to

347^{\circ} C

in a closed vessel in the presence of a catalyst. Under the conditions,

N H_{3}

is partially decomposed according to the equation,

2 N H_{3} ⇋ N_{2} + 3 H_{2}

.The vessel is such that the volume remains effectively constant whereas pressure increases to

50 a t m

. Calculate the percentage of

N H_{3}

actually decomposed.

65%
61.3%
62.5%
64%

Discuss Question

Answer: Option B. -> 61.3%
:
B

2 N H_{3} ⇋ N_{2} + 3 H_{2}

\begin{matrix} I n i t i a l m o l e & a & 0 & 0 \end{matrix}

Mole at equilibrium

(a - 2 x) x 3 x

Initial pressure of

N H_{3}

of a mole = 15 atm

27^{\circ} C

The pressure of 'a' mole of

N H_{3} = p a t m a t 347^{\circ} C

∴ \frac{15}{300} = \frac{p}{620}

∴

p=31 atm
At constant volume and at

347^{\circ} C

,mole

α

pressure

∴ \frac{a + 2 x}{a} = \frac{50}{31}

∴ x = \frac{19}{62}

∴ % o f N H_{3}

decomposed =

\frac{2 x}{a} \times 100

= \frac{2 \times 19 a}{62 \times a} \times 100 = 61.33 %

Question 2. Calculate

△ G^{\circ}

for conversion of oxygen to ozone

3 / 2 O_{2} (g) \to O_{3} (g) a t 298 K, i f K_{p}

for this conversion is 2.47

\times

10^{- 29}

(Use log(2.47) = 0.4)

163 KJ mol−1
2.4 X 102 KJ mol−1
1.63 KJ mol−1
2.38 X 106 KJ mol−1

Discuss Question

Answer: Option A. -> 163 KJ mol−1
:
A

△ G^{\circ} = - 2.303 R T l o g K_{p} = - 2.303 \times 8.314 \times 298 \times l o g (2.47 \times 10^{29}) \frac{j}{m o l} = 163 \frac{k j}{m o l}

Question 3. At a certain temperature and a total pressure of

10_{5}

Pa, iodine vapours contain

40 %

by volume of iodine atoms

[I_{2} (g) ⇋ 2 I (g)] . K_{p}

for the equilibrium will be

0.67
1.5
2.67 × 104
9.0 × 104

Discuss Question

Answer: Option C. -> 2.67 × 104
:
C
Partial pressure of I atoms

(p_{l}) = 40 \times 10^{5} / 100 P a = 0.4 \times 10^{5} P a

Partial pressure of

l_{2} (p_{2}) = 60 \times 10^{5} / 100 P a = 0.6 \times 10^{5} P a

K_{p} = (p_{1})^{2} / p_{2} = (0.4 \times 10^{5})^{2} / (0.6 \times 10^{5}) = 2.67 \times 10^{4}

Question 4. In a reaction

A + 2 B ⇋ 2 C; 2 m o l e s o f A, 3

mole of

B a n d 2 m o l e o f C

are placed in a 2 L flask and the equilibrium concentration of

C i s 0.5 m o l e / L .

The equilibrium constant K for the reaction is:

0.073
0.147
0.05
0.026

Discuss Question

Answer: Option C. -> 0.05
:
C

\begin{matrix} A + & 2 B ⇋ & 2 C I n i t i a l m o l e s & 2 & 3 & 2 M o l a r c o n c . & 2 / 2 = 1 m o l / L & 3 / 2 = 1.5 m o l / L & 2 / 2 = 1 m o l / L A t E q u i . C o n c & 1 + 0.25 = 1.25 m o l / L & 1.5 + 0.5 = 2 m o l / L & 0.5 m o l / L \end{matrix}

K = (0.5)^{2} / (1.25 \times (2)^{2}) = 0.05

Note that concentration of C has decreased, i.e., to attain equilibrium, reaction has shifted backwards. Decrease in conc. of C will be

0.5 m o l / L

and hence increase in concentration of B will be

0.5 m o l / L

and that of A will be

0.25 m o l / L

Question 5. The vapor density of completely dissociated

N H_{4} C l

would be

same as that of NH4Cl
double than that of NH4Cl
half than that of NH4Cl
slightly less than that of NH4Cl

Discuss Question

Answer: Option A. -> same as that of NH4Cl
:
A
On dissolution of

N H_{4} C l

into

N H_{3} + H C l

number of moles become double. Hence, volume is doubled and therefore, density is halved.

Question 6. Assertion (A): Water boiling in an open vessel at a fixed temperature is an example of physical equilibrium.
Reason (R): Temperature does not change during the process of boiling. Which of the following is correct?

Both A & R are correct and R is correct explanation of A.
Both A & R are correct but R is not a correct explanation of A.
A is true but R is false
A is false but R is true.

Discuss Question

Answer: Option D. -> A is false but R is true.
:
D
Equilibrium is possible only in a closed vessel. Hence A is false.
Boiling occurs at a constant temperature at a given pressure. Hence R is true.

Question 7. Which of the following is incorrect about the following reaction –
$2 S O_{2} (g) + O_{2} (g) ⇋ 2 S O_{3} (g) △ H = - 45 k C a l$

High pressure will be favorable for production of SO3 .
High temperature will favor forward reaction.
Decreasing SO3 concentration will reduce the rate of backward reaction.
Doubling the volume of container while keeping temperature same, will shift the equilibrium in reverse direction.

Discuss Question

Answer: Option B. -> High temperature will favor forward reaction.
:
B
a) Increasing pressure favors the reaction in the direction in which no. of moles decreases. As number of moles is decreasing in forward direction, hence increasing pressure will favor the production of

S O_{3}

b) Increasing temperature favors endothermic reaction. Here forward reaction is exothermic; hence high temperature will not favor forward reaction.
c)

k [S O_{3}]

Hence it will decrease on reducing conc. of

S O_{3}

d) Increasing the volume reduces the pressure. Hence it will favor backward reaction in this case as number of moles is increasing in backward direction.

Question 8. For the gas phase reaction

C_{2} H_{4} + H_{2} ⇌ C_{2} H_{6},

Δ H

= -32.7kcal carried out in a vessel, the equilibrium concentration of

C_{2} H_{4}

can be increased by -

increasing the temperature
Increasing concentration of H2
Decreasing temperature
Increasing pressure

Discuss Question

Answer: Option A. -> increasing the temperature
:
A
Since backward reaction is endothermic, increase in temperature favors backward reaction

Question 9. An amount of solid

N H_{4} H S

is placed in a flask already containing ammonia gas at a certain temperature and

0.50

atm pressure. Ammonium hydrogen sulphide decomposes to yield

N H_{3}

and

H_{2} S

gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to

0.84

atm. The equilibrium constant for

N H_{4} H S

decomposition at this temperature is

0.30
0.18
0.17
0.11

Discuss Question

Answer: Option D. -> 0.11
:
D

N H_{4} H S (s) ⇋ N H_{3} (g) + H_{2} S (g)

\begin{matrix} I n i t i a l l y, & a & 0.5 a t m & 0 a t e q u i l i b r i u m & . . . . . . & 0.5 + x & x \end{matrix}

Total Pressure at eqilibrium

= 0.5 + x + x = 0.84

\Rightarrow x = 0.17

K_{p} = P_{N H_{3}} (g) \times P_{H_{2} S} (g) = (0.5 + x) x = (0.5 + 0.17) \times 0.17 = 0.1139 a t m^{2}

Question 10. For the reaction

P C l_{5} (g) ⇋ P C l_{3} (g) + C l_{2} (g),

the forward reaction at constant temperature is favored by –

Introducing an inert gas at constant volume
Introducing chlorine gas at constant volume
Introducing an inert gas at constant pressure
Decreasing the volume of container

Discuss Question

Answer: Option C. -> Introducing an inert gas at constant pressure
:
C
a) Introducing an inert gas at constant volume does not change the partial pressure of gases. Hence it does not affect the equilibrium.
b) Introducing chlorine gas at constant volume will favor the backward reaction.
c) Introducing inert gas at constant pressure means, one has to increase the volume of the container given temperature is also not changing.

\begin{matrix} P V    m u s t i n c r e a s e = n    i n c r e a s i n g R T \end{matrix}

Increasing container volume will favor the forward reaction. Henceoption (c) is correct.
d) Decreasing the container volume will favor backward reaction.