A sample of pure P C l 5 was introduced into an evacuated vessel at 473 K . After equilibrium was attained, concentration of P C l 5 was found to b e 0.5 × 10 - 1 m o l L - 1 . If value of K c is 8.3 × 10 - 3 , what are the concentrations of P C l 3 and C l 2 at equilibrium? P C l 5 g ⇌ P C l 3 g + C l 2 ( g )

Given : Temperature = 473 K After equilibrium concentration of P C l 5 = 0.5 × 10 - 1 m o l L - 1 K c = 8.3 × 10 - 3 , Let the concentrations of both P C l 3 a n d C l 2 at equilibrium be x m o l L - 1 . The given reaction is: P C l 5 g ⇌ PCl 3 ( g ) + Cl 2 ( g ) At equilibrium 0.5 × 10 - 1 m o l L - 1 x m o l L - 1 x m o l L - 1 Given the value of the equilibrium constant, K c is 8.3 × 10 - 3 . Now, we can write the expression for equilibrium as: P C l 3 1 C l 2 1 P C l 5 1 = K c ⇒ x × x 0.5 × 10 - 1 = 8.3 × 10 - 3 ⇒ x 2 = 4.15 × 10 - 4 ⇒ x = 2 . 04 × 10 - 2 = 0.0204 ≈ 0.02 Therefore, at equilibrium, P C l 3 = C l 2 = x = 0.02 m o l L - 1

One of the reaction that takes place in producing steel from iron ore is the reduction of iron II oxide by carbon monoxide to give iron metal and C O 2 . F e O s + C O g → F e s + C O 2 g ; K p = 0.265 atm at 1050 K What are the equilibrium partial pressures of C O and C O 2 at 1050 K if the initial partial pressures are: P C O = 1.4 atm and P CO 2 = 0 .80 atm?

Given: K p = 0.265 atm Temperature = 1050 K Initial partial pressures of CO ( P C O ) = 1.4 atm Initial partial pressures of C O 2 ( P C O 2 ) = 0.80 atm For the given reaction, FeO s Initially , + CO g 1 .4 atm ⇌ Fe s + CO 2 ( g ) 0 .80 atm Q p = p C O 2 P C O = 0.80 1.4 = 0.571 It is given that K p = 0.265 . Since Q p > K p , the reaction will proceed in the backward direction. Therefore, we can say that the pressure of C O will increase while the pressure of C O 2 will decrease. Now, let the increase in pressure of C O = decrease in pressure of C O 2 be p . Then, we can write, FeO ( s ) Initially , + CO ( g ) 1 .4 atm ↔ Fe s + CO 2 ( g ) 0 .80 atm K p = p C O 2 P C O ⇒ 0.265 = 0.80 - p 1.4 + p ⇒ 0.371 + 0.265 p = 0.80 - p ⇒ 1.265 p = 0.429 ⇒ p = 0.339 atm Therefore, equilibrium partial pressure of C O 2 ( P C O 2 ) = 0.80 - 0.339 = 0.461 atm And, equilibrium partial pressure of C O ( P C O ) = 1.4 + 0.339 = 1.739 atm

Bromine monochloride, B r C l decomposes into bromine and chlorine and reaches the equilibrium: 2 B r C l g ⇌ B r 2 g + C l 2 ( g ) for which K c = 32 at 500 K . If initially pure B r C l is present at a concentration of 3.3 × 10 - 3 m o l L - 1 , what is its molar concentration in the mixture at equilibrium?

Let the amount of bromine and chlorine formed at equilibrium be x . The given reaction is: 2 B r C l g ⇌ Br 2 ( g ) + Cl 2 ( g ) Initial conc. 3.3 × 10 - 3 0 0 At equilibrium 3.3 × 10 - 3 - 2 x x x Now, we can write, K c = B r 2 C l 2 B r C l 2 ⇒ 32 = x × x 3.3 × 10 - 3 - 2 x 2 By taking root both sides ⇒ x 3.3 × 10 - 3 - 2 x = 5.66 ⇒ x = 18.678 × 10 - 3 - 11.32 x ⇒ 12.32 x = 18.678 × 10 - 3 ⇒ x = 1 . 5 × 10 - 3 Therefore, at equilibrium, B r C l = 3.3 × 10 - 3 - 2 x = 3.3 × 10 - 3 - 2 × 1.5 × 10 - 3 = 3.3 × 10 - 3 - 3.0 × 10 - 3 = 0.3 × 10 - 3 = 3.0 × 10 - 4 m o l L - 1

Calculate a) ∆ G ∘ and b) the equilibrium constant for the formation of N O 2 from N O and O 2 at 298 K NO g + 1 2 O 2 g ⇌ NO 2 ( g ) where ∆ 1 G o N O 2 = 52.0 k J / m o l ∆ G o N O = 87.0 kJ / mol ∆ G o O 2 = 0 k J / m o l

For the given reaction, NO g + 1 2 O 2 g ⇌ NO 2 ( g ) ∆ 1 G o N O 2 = 52.0 k J / m o l ∆ G o N O = 87.0 k J / m o l , ∆ G o O 2 = 0 k J / m o l Formula : Δ G o = Δ G o Products - ΔG o Reactants Δ G o = 52.0 - 87.0 + 0 = - 35 . 0 k J m o l - 1 Formula : Δ G o = - R T l n K c Δ G o = - 2.303 R T l o g K c Put all values in above formula log K c = - 35.0 × 10 - 3 - 2.303 × 8.314 × 298 log K c = 6.134 &there4; K c = antilog 6.134 K c = 1.36 × 10 6 Hence, the equilibrium constant for the given reaction K c is 1.36 × 10 6 .

MEDIUM

11th CBSE

IMPORTANT

Earn 100

Ethyl acetate is formed by the reaction between ethanol and acetic acid, and the equilibrium is represented as:
${C H}_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ {C H}_{3} {C O O C}_{2} H_{5} (l) + H_{2} O (l)$
At $293 K$ , if one starts with $1.00 m o l$ of acetic acid and $0.18 m o l$ of ethanol, there is $0.171 m o l$ of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.

Important Questions on Equilibrium

MEDIUM

11th CBSE

IMPORTANT

Textbook for Class XI Chemistry Part I>Chapter 7 - Equilibrium>EXERCISES>Q 7.18

Ethyl acetate is formed by the reaction between ethanol and acetic acid, and the equilibrium is represented as:

{C H}_{3} C O O H (l) + C_{2} H_{5} O H (l) ⇌ {C H}_{3} {C O O C}_{2} H_{5} (l) + H_{2} O (l)

Starting with

0.5 m o l

of ethanol and

1.0 m o l

of acetic acid and maintaining it at

293 K

0.214 m o l

of ethyl acetate is found after some time. Has equilibrium been reached?

MEDIUM

11th CBSE

IMPORTANT

Textbook for Class XI Chemistry Part I>Chapter 7 - Equilibrium>EXERCISES>Q 7.19

A sample of pure

{P C l}_{5}

was introduced into an evacuated vessel at

473 K .

After equilibrium was attained, concentration of

{P C l}_{5}

was found to be

0.5 \times 10^{- 1} m o l L^{- 1}

. If value of

K_{c}

8.3 \times 10^{- 3}

, what are the concentrations of

{P C l}_{3}

and

{C l}_{2}

at equilibrium?

{P C l}_{5} (g) ⇌ {P C l}_{3} (g) + {C l}_{2} (g)

MEDIUM

11th CBSE

IMPORTANT

Textbook for Class XI Chemistry Part I>Chapter 7 - Equilibrium>EXERCISES>Q 7.2

One of the reaction that takes place in producing steel from iron ore is the reduction of iron

(II)

oxide by carbon monoxide to give iron metal and

{C O}_{2} .

F e O (s) + C O (g) \to F e (s) + C O_{2} (g); K p = 0.265

atm at

1050 K

What are the equilibrium partial pressures of

C O

and

{C O}_{2}

1050 K

if the initial partial pressures are:

P_{C O} = 1.4

atm and

P_{CO 2} = 0 .80

atm?

MEDIUM

11th CBSE

IMPORTANT

Textbook for Class XI Chemistry Part I>Chapter 7 - Equilibrium>EXERCISES>Q 7.21

Equilibrium constant,

K_{c}

for the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {N H}_{3} (g)

500 K

0.061

At a particular time, the analysis shows that composition of the reaction mixture is

3.0 m o l L^{- 1} N_{2}, 2.0 m o l L^{- 1} H_{2}

and

0.5 m o l L^{- 1} N H_{3} .

Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium?

HARD

11th CBSE

IMPORTANT

Textbook for Class XI Chemistry Part I>Chapter 7 - Equilibrium>EXERCISES>Q 7.22

Bromine monochloride,

B r C l

decomposes into bromine and chlorine and reaches the equilibrium:

2 B r C l (g) ⇌ {B r}_{2} (g) + {C l}_{2} (g)

for which

K_{c} = 32

500 K

. If initially pure

B r C l

is present at a concentration of

3.3 \times 10^{- 3} m o l L^{- 1},

what is its molar concentration in the mixture at equilibrium?

HARD

11th CBSE

IMPORTANT

Textbook for Class XI Chemistry Part I>Chapter 7 - Equilibrium>EXERCISES>Q 7.23

At $1127 K$ and $1$ atm pressure, a gaseous mixture of $CO$ and ${CO}_{2}$ in equilibrium with solid carbon has $90.55 %$ $CO$ by mass

$C (s) + {CO}_{2} (g) \to 2 CO (g)$

Calculate $K_{c}$ for this reaction at the above temperature.

EASY

11th CBSE

IMPORTANT

Textbook for Class XI Chemistry Part I>Chapter 7 - Equilibrium>EXERCISES>Q 7.24

Calculate a)

∆ G^{\circ}

and b) the equilibrium constant for the formation of

{N O}_{2}

from

N O

and

O_{2}

298 K

NO (g) + \frac{1}{2} O_{2} (g) ⇌ {NO}_{2} (g)

where

∆_{1} G^{o} (N O_{2}) = 52.0 k J / m o l

∆ G^{o} (N O) = 87.0 kJ / mol

∆ G^{o} (O_{2}) = 0 k J / m o l

EASY

11th CBSE

IMPORTANT

Textbook for Class XI Chemistry Part I>Chapter 7 - Equilibrium>EXERCISES>Q 7.25

Does the number of moles of reaction product increase, decrease or remain same when the following equilibria is subjected to a decrease in pressure by increasing the volume?

${PCl}_{5} (g) ⇌ {PCl}_{3} (g) + {Cl}_{2} (g)$

Important Questions on Equilibrium

Learn and Prepare for any exam you want !