Chemical Equilibrium(11th And 12th > Chemistry ) Questions and answers for exam Preparation

11th And 12th > Chemistry

CHEMICAL EQUILIBRIUM MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

Chemical equilibrium is dynamic in nature because

Equilibrium is maintained rapidly
The concentration of reactants and products become same at equilibrium
The concentration of reactants and products are constant but different
Both forward and backward reactions occur at all times with same speed

Discuss Question

Answer: Option D. -> Both forward and backward reactions occur at all times with same speed
:
D

Chemical equilibrium is dynamic in nature because, reactants change into products and products change into reactants even after equilibrium is achieved. But rate of forward and backward reactions are same.

Question 2.

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 3.

Hl was heated in a closed tube at $440^{\circ} c$ till equilibrium is obtained. At this temperature $22 %$ of Hl was dissociated. The equilibrium constant for this dissociation will be

0.282
0.0796
0.0199
1.99

Discuss Question

Answer: Option C. -> 0.0199
:
C

$2 H I ⇋ H_{2} + l_{2}$
$\begin{matrix} I n i t i a l n o o f m o l e s - & 2 m o l e s & 0 & 0 a t e q u i l i b r i u m & 2 m o l e s - 0.44 & 0.22 & 0.22 \end{matrix}$
$K_{c} = \frac{[H_{2}] [l_{2}]}{[H I]^{2}} = \frac{0.22 \times 0.22}{{1.56}^{2}} = 0.0199$

Question 4.

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 5.

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 6.

The reaction quotient (Q) for a reaction is given as $Q = \frac{[N H_{3}]^{2}}{[N_{2}] [H_{2}]^{3}}$ . The reaction will proceed from left to right if ........

$Q = 0$
$Q > k_{c}$
$Q < k_{c}$
$Q = k_{c}$

Discuss Question

Answer: Option C. ->

Q < k_{c}

:
C

Reaction moves in forward direction if $Q < K_{c}$
Reaction moves in backward direction if $Q > K_{c}$
Reaction remains at equilibrium if $Q = K_{c}$

Question 7.

At a temperature of $1000 K$ , equilibrium constant $K_{c}$ for the following reaction is $100 m o l^{2} L^{- 2}$ .
$N_{2} (g) + 3 H_{2} (g) ⇋ 2 N H_{3} (g)$
What will be the value of $K_{p}$ for the reaction below?
$N H_{3} (g) ⇋ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g)$

0.1 atm
8.21 atm
831.4 atm
8.314 atm

Discuss Question

Answer: Option B. -> 8.21 atm
:
B

$N H_{3} (g) ⇋ \frac{1}{2} N_{2} (g) + \frac{3}{2} H_{2} (g) K_{c} = K_{c}^{'}$
$K_{c}^{'} = (\frac{1}{100})^{\frac{1}{2}} = 0.1$
$K_{p} = K_{c}^{'} \times (R T)^{△ n} = 0.1 \times (0.0821 \times 1000)^{1} = 8.21 a t m$
Notes: Use $R = 0.0821 a t m l m o l^{- 1} K^{- 1}$ as unit of concentration has been taken as $m o l / l$ while calculating $K_{c} .$

Question 8.

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)

$163 K J m o l^{- 1}$
$2.4 X 10^{2} K J m o l^{- 1}$
$1.63 K J m o l^{- 1}$
$2.38 X 10^{6} K J m o l^{- 1}$

Discuss Question

Answer: Option A. ->

163 K J m o l^{- 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 9.

At equilibrium, reaction $A ⇋ B$ is at the stage of half completion. Which one of the followings is true?

$△ G^{\circ} = 0$
$△ G^{\circ} > 0$
$△ G^{\circ} < 0$
$△ G^{\circ} = - R T l n 2$

Discuss Question

Answer: Option A. -> $△ G^{\circ} = 0$
:
A

At the stage of half completion of the reaction
$[A] = [B]$
$K_{e q} = \frac{[B]}{[A]} = 1$
$△ G^{\circ} = - R T l n k_{e q} = 0$

Question 10.

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 $S O_{3}$ .
High temperature will favor forward reaction.
Decreasing $S O_{3}$ 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.