Free Objective Test 02 Practice Test - 11th and 12th

Question 1

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 \times 10^{- 5} / R

SOLUTION

Solution : 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 2

For a hypothetical reaction
$4 A + 5 B ⇋ 4 P + 6 Q,$
The equilibrium constant $K_{c}$ has units

m o l L^{- 1}

m o l^{- 1} L

(m o l L^{- 1})^{- 2}

D. unit less

SOLUTION

Solution : 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 3

Consider the following reactions in which all the reactants and products are in gaseous state.
$2 P Q ⇋ P_{2} + Q_{2}; K_{1} = 2.5 \times 10^{5}$
$P Q + \frac{1}{2} R_{2} ⇋ P Q R, K_{2} = 5 \times 10^{- 3}$

The value of K₃ for the equilibrium
$1 / 2 P_{2} + 1 / 2 Q_{2} + 1 / 2 R_{2} ⇋ P Q R$

2.5 \times 10^{- 3}

2.5 \times 10^{3}

1.0 \times 10^{- 5}

5 \times 10^{3}

SOLUTION

Solution : C

Correct option is c).
For $2 P Q ⇋ P_{2} + Q_{2}, K_{1} = [P_{2}] [Q_{2}] / [P Q]^{2}$
For $P Q + \frac{1}{2} R_{2} ⇋ P Q R, K_{2} = \frac{[P Q R]}{([P Q] [R]^{\frac{1}{2}})}$
For the required equilibrium
$1 / 2 P_{2} + 1 / 2 Q_{2} + 1 / 2 R_{2} ⇋ P Q R$
$K = [P Q R] / ([P_{2}]^{\frac{1}{2}} [Q_{2}]^{\frac{1}{2}} [R_{2}]^{\frac{1}{2}} = \frac{K 2}{\sqrt{K T}} = 5 \times 10^{- 3} / (2.5 \times 10^{5})^{\frac{1}{2}} = 5 \times 10^{- 3} / 5 \times 10^{2} = 10^{- 5}$

Question 4

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 \times 10^{4}

9.0 \times 10^{4}

SOLUTION

Solution : 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 5

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:

A. 0.073

B. 0.147

C. 0.05

D. 0.026

SOLUTION

Solution : 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 6

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

[C l_{2}] > [P C l_{3}]

[C l_{2}] > [P_{4}]

[P_{4}] > [C l_{2}]

[P C l_{3}] > [P_{4}]

SOLUTION

Solution : C

Solution:Correct option is c).

$\begin{matrix} P_{4} (s) + & 6 C l_{2} (g) ⇋ & 4 P C l_{3} (g) I n i t i a l m o l e s & 1 & 1 & 0 A t e q u i l i b r i u m & 1 - x & 1 - 6 x & 4 x \end{matrix}$

$A s (1 - x) > (1 - 6 x), H e n c e [P_{4}] > [C l_{2}] .$

Question 7

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 %

SOLUTION

Solution : 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 8

At $30^{\circ} C, K_{p}$ for the dissociation reaction:
$S O_{2} C l_{2} (g) ⇋ S O_{2} (g) + C l_{2} (g) i s 2.9 x 10^{- 2}$ ~atm. If the total pressure is 1 atm, the degree of dissociation of $S O_{2} C l_{2} i s (a s s u m e 1 - α^{2} = 1)$

85 %

12 %

17 %

35 %

SOLUTION

Solution : C

Correct option is c).
$\begin{matrix} S O_{2} C l_{2} (g) ⇋ & S O_{2} (g) + & C l_{2} (g) 1 & 0 & 0 1 - α & α & α \end{matrix}$
Total moles = $1 + α$
$P s o_{2} C l_{2} = 1 - α / 1 + α, P s o_{2} = α / 1 + α, P c l_{2} = α / 1 + α$
$K_{p} = (α / 1 + α)^{2} / (1 - α / 1 + α) = α^{2} / 1 - α^{2} \approx α^{2}$
$α = \sqrt{K p} = (2.9 \times 10^{- 2} = 0.17, i . e ., 17 %$

Question 9

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

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

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

\frac{\sqrt{2 K p}}{\sqrt{2 + 2 K p}}

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

SOLUTION

Solution : 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 10

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

0.97 a t m

0.012 a t m

2.01 a t m

0.024 a t m

SOLUTION

Solution : 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 11

The vapor density of completely dissociated $N H_{4} C l$ would be

A. same as that of

N H_{4} C l

B. double than that of

N H_{4} C l

C. half than that of

N H_{4} C l

D. slightly less than that of

N H_{4} C l

SOLUTION

Solution : 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 12

Pure ammonia is placed in a vessel at a temperature where its dissociation is appreciable. At equilibrium,

K_{p}

does not change significantly with pressure

α

does not change with pressure

C. concentration of

N H_{3}

does not change with pressure

D. concentration of

H_{2}

is less than that of

N_{2}

SOLUTION

Solution : A

$K_{p}$ is constant and does not change with pressure.

Question 13

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

A. 8 times

B. 16 times

C. 64 times

D. 4 times.

SOLUTION

Solution : 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 14

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

A. 3:1

B. 1:9

C. 36:1

D. 1:1

SOLUTION

Solution : 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 15

1 mole of $N_{2} O_{4} (g)$ at 300K is kept in a closed container under 1 atm pressure. It is heated to 600K when 20 $%$ by mass of $N_{2} O_{4} (g)$ decomposes to $N O_{2} (g)$ . The resultant pressure is

A. 1.2 atm

B. 2.4 atm

C. 2.0 atm

D. 1.0 atm

SOLUTION

Solution : B

Correct option is b).
$\begin{matrix} N_{2} O_{4} ⇋ & 2 N O_{2} I n i t i a l m o l e s & 1 & 0 \end{matrix}$

Total moles=0.8+0.4=1.2

Applying, $P V = n R T$ , at constant volume,

$\frac{P 1}{n 1 T 1} = \frac{P}{n 2 T 2}$

$\frac{1}{1 X 300} = \frac{P_{2}}{1.2 X 600}$

$P_{2} = 2.4 a t m .$

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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