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

12th Grade > Chemistry

CHEMICAL EQUILIBRIUM MCQs

Total Questions : 30 | Page 3 of 3 pages

Question 21. If concentration of

O H^{-}

ions in the reaction

F e (O H)_{3} (s) ⇋ F e^{3 +} + (a q) + 3 O H^{-} (a q)

is decreased by

1 / 4

times, then equilibrium concentration of

F e^{3 +}

will increase by

8 times
16 times
64 times
4 times.

Discuss Question

Answer: Option C. -> 64 times
:
C

K = [F e^{3 +}] [O H^{-}]^{3}

If concentration of

O H^{-}

ion is decreased to

\frac{1}{4}

th to keep

K

constant
Concentration of

F e^{3 +}

ions has to be increased by

64

times.

Question 22. The values of

K_{p} 1

and

K_{P} 2

for the reactions

X ⇋ Y + Z a n d A ⇋ 2 B

Are in the ratio of

9 : 1.

If the degree of dissociation of

X a n d A

be equal, then total pressure at equilibrium

(1) a n d (2)

are in the ration

3:1
1:9
36:1
1:1

Discuss Question

Answer: Option C. -> 36:1
:
C
Suppose total pressure at equilibrium for reactions 1 and 2 are

P_{1} a n d P_{2}

respectively

\begin{matrix} X ⇋ & Y + & Z I n i t i a l m o l e s & 1 & 0 & 0 A t e q u i l i b r i u m & 1 - x & x & x \end{matrix}

T o t a l m o l e s = 1 + α

Partial pressure of

X = p_{x} = \frac{(1 - x) X P 1}{1 + x}

P_{y} <= \frac{(x) X P 1}{1 + x} = P_{z}

(K_{P 1} = P_{Y} P_{Z} / P_{x} = x^{2} P_{1} / (1 - x^{2}) \approx x^{2} P_{1}

\begin{matrix} A ⇋ & 2 B h l i n e I n i t i a l m o l e s & 1 & 0 A t e q u i l i b r i u m & 1 - x & 2 x \end{matrix}

T o t a l m o l e s = 1 + α

P_{A} = \frac{(1 - x) X P 2}{1 + x} P_{B} = \frac{(2 x) X P 2}{1 + x}

K_{P 2} = P_{B}^{2} / P_{A} = 4 x^{2} P_{2} / (1 - x^{2}) \approx 4 x^{2} P_{2}

K_{P} 1 / K_{P} 2 = x^{2} P_{1} / 4 x^{2} P_{2} = 9 / 1

P_{1} / P_{2} = 36 / 1

Question 23. For a hypothetical reaction

4 A + 5 B ⇋ 4 P + 6 Q,

The equilibrium constant

K_{c}

has units

molL−1
mol−1L
(molL−1)−2
unit less

Discuss Question

Answer: Option A. -> molL−1
:
A

K_{c} = [P]^{4} [Q]^{6} / ([A]^{4} [B]^{5}) = (m o l L^{- 1})^{10} / (m o l L^{- 1})^{9} / = m o l L^{- 1}

Question 24. Equimolar concentrations of

H_{2} a n d I_{2}

are heated to equilibrium in a 2litre flask. At equilibrium, the forward and backward rate constants are found to be equal. What percentage of initial concentration of

H_{2}

has reacted at equilibrium?

33.33%
75%
50%
15%

Discuss Question

Answer: Option C. -> 50%
:
C
Solution:Correct option is c).

\begin{matrix} H_{2} (g) + & l_{2} (g) ⇋ & 2 H I (g) I n i t i a l m o l e s & 1 & 1 & 0 A t e q u i l i b r i u m & 1 - x & 1 - x & 2 x M o l a r c o n c e n t r a t i o n (m o l / L) & (1 - x) / 2 & (1 - x) / 2 & 2 x / 2 = x \end{matrix}

K = x^{2} / \frac{(1 - x) \times (1 - x)}{2 \times 2} = 4 x^{2} / (1 - x)^{2} \approx 4 x^{2}

K = K_{f} / K_{b} = 1

4 x^{2} = 1 x = \frac{1}{2}, i . e ., 50 %

Question 25. For the reaction

2 H I (g) ⇋ H_{2} (g) + I_{2} (g),

the degree of dissociation α of

H I (g)

is related to the equilibrium constant

(K_{p})

by the expression

1+2√Kp2
√1+2Kp√2
√2Kp√2+2Kp
2√Kp1+2√Kp

Discuss Question

Answer: Option D. -> 2√Kp1+2√Kp
:
D

\begin{matrix} 2 H I (g) ⇋ & H_{2} (g) + & l_{2} (g) A t e q u i l i b r i u m & 1 - α & α / 2 & α / 2 \end{matrix}

Total moles at equilibrium =1

P H I = (1 - α) \times P, P H_{2} = α / 2 \times P P l_{2} = α / 2 \times P

K_{p} = P H_{2} \times P l_{2} / (P H I)^{2}

Substituting above values we get,

K_{p} = a^{2} / [4 \times (1 - α^{2})]

solving,

α / 1 - α = 2 (K_{p})^{\frac{1}{2}}

α = \frac{2 \sqrt{K p}}{1 + 2 \sqrt{K p}}

Question 26. The equilibrium constant for the reaction

2 S O_{2} + O_{2} ⇋ 2 S O_{3} i s 900 a t m^{- 1} a t 800 K .

A mixture containing

S O_{3} a n d O_{2}

having initial pressures of

1 a t m a n d 2 a t m

is heated to equilibrate. The partial pressure of

O_{2}

at equilibrium will be

0.97atm
0.012atm
2.01atm
0.024atm

Discuss Question

Answer: Option C. -> 2.01atm
:
C
Correct option is c). Considering the backward reaction we have,

\begin{matrix} 2 S O_{3} ⇋ & 2 S O_{2} + & O_{2} & K_{P} = 1 / 900 a t m I n i t i a l p r e s s u r e & 1 a t m & 0 & 2 a t m P r e s s u r e a t e q u i l i b r i u m & 1 - x & x & 2 + x / 2 \end{matrix}

Let partial pressure of

S O_{2} = p_{1}

and partial pressure of

O_{2} = p_{2}

and partial pressure of

S O_{3} = p_{3}

K <_{p} = (p_{1})^{2} \times (p_{2}) / (p_{3})^{2} = x^{2} (2 + x / 2) / (1 - x)^{2} = 1 / 900

A s K_{p}

for this reaction is very small,

x << 1.

Taking

2 + x / 2

as x and

(1 - x)

as x and solving for x, we get

X = 0.0236

Hence, at equilibrium, partial pressure of

S O_{3} = 1 - X = 0.9764 a t m,

Partial pressure of

S O_{2} = x = 0.0236 a t m

And Partial pressure of

O_{2} = 2 + x / 2 = 2.0118 a t m

Question 27. If

K_{p}

for a reaction

A (g) + 2 B (g) ⇋ 3 C (g) + D (g)

is 0.05atm at 1000K. Its

K_{c}

in terms of R will be

2000 R
0.02 R
10−6 R
5× 10−5/R

Discuss Question

Answer: Option D. -> 5× 10−5/R
:
D

△ n_{g} = 4 - 3 = 1, K_{P} = K_{C} (R T)^{△ n}

0.05 = K_{c} (1000 R), K_{c} = 5 \times 10^{- 5} / R

Question 28. The equilibrium

P_{4} (s) + 6 C l_{2} (g) ⇋ 4 P C l_{3} (g)

Is attained by mixing equal moles of

P_{4} a n d C l_{2}

inana evacuated vessel. Then at equilibrium

[Cl2]>[PCl3]
[Cl2]>[P4]
[P4]>[Cl2]
[PCl3]>[P4]

Discuss Question

Answer: Option C. -> [P4]>[Cl2]
:
C
Solution:Correct option is c).
$\begin{matrix} P_{4} (s) + & 6 C l_{2} (g) ⇋ & 4 P C l_{3} (g) I n i t i a l m o l e s & 1 & 1 & 0 A t e q u i l i b r i u m & 1 - x & 1 - 6 x & 4 x \end{matrix}$
$A s (1 - x) > (1 - 6 x), H e n c e [P_{4}] > [C l_{2}] .$