Nitric oxide reacts with Br 2 and gives nitrosyl bromide as per reaction given below: 2 N O ( g ) + B r 2 ( g ) ⇌ 2 N O B r ( g ) When 0.087 m o l of N O and 0.0437 m o l of B r 2 are mixed in a closed container at a constant temperature, 0 . 0518 mol of N O B r is obtained at equilibrium. Calculate equilibrium amount of N O and B r 2 .

Mole of N O B r is obtained at equilibrium= 0 . 0518 The given reaction is: 2 N O ( g ) 2 m o l + B r 2 ( g ) 1 m o l ⇋ 2 N O B r ( g ) 2 m o l According to stoichiometry, 2 moles of N O B r is formed from 2 moles of NO and 1 mole of B r 2 . Therefore, 0.0518 mole of N O B r is formed from 0.0518 2 mole of Br 2 or 0.0259 mole of Br 2 The amount of N O and Br 2 present initially is as follows: [ N O ] = 0.087 mol [ B r 2 ] = 0.0437 mol Therefore, the amount of NO present at equilibrium = 0.087 – 0.0518 = 0.0352 mole And, the amount of B r 2 present at equilibrium = 0.0437 – 0.0259 = 0.0178 mol .

At 1400 K , K c = 2 . 5 × 10 - 3 for the reaction CH 4 ( g ) + 2 H 2 S ( g ) ⇌ CS 2 ( g ) + 4 H 2 ( g ) . A 10 . 0 L reaction vessel at 1400 K contains 2 . 00 mole of CH 4 , 3 . 0 mol of CS 2 , 3 . 0 mole of H 2 4 . 0 mole and of H 2 S . Then in which direction reaction will proceed to reach equilibrium.

At 1400 K , K c = 2 . 5 × 10 - 3 for the reaction CH 4 ( g ) + 2 H 2 S ⇌ CS 2 ( g ) + 4 H 2 ( g ) . A 10 . 0 L reaction vessel at 1400 K contains 2 . 0 mol of CH 4 , 3 . 0 mol of CS 2 , 3 . 0 mol of H 2 and 4 . 0 mol of H 2 S . The reaction will proceed from right to left to reach equilibrium. The reaction quotient Q = [ CS 2 ] [ H 2 ] 2 [ CH 4 ] [ H 2 S ] 2 Q = 3 . 0 × ( 3 . 0 ) 4 2 . 0 × ( 4 . 0 ) 4 = 0 . 0759 But the equilibrium constant K C = 2 . 5 × 10 - 3 Hence, the reaction quotient Q is much greater than the equilibrium constant K Q < K Hence, the concentration of products is greater than the equilibrium concentration and the concentration of reactants is lesser than the equilibrium concentration. The system will shift left to reach equilibrium till Q becomes equal to K .

The exothermic formation of ClF 3 is represented by the following equation: Cl 2 g + 3 F 2 g ⇌ 2 ClF 3 g ; ΔH = - 329 kJ Which of the following will increase the quantity of ClF 3 in the equilibrium mixture of Cl 2 , F 2 and ClF 3 ?

Increasing the volume of the container.

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IMPORTANT

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Discuss the effect of temperature on the equilibrium constant. How does it change for $(a)$ exothermic reaction $(b)$ endothermic reaction $(c)$ having zero heat of reaction?

Important Questions on Equilibrium

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 21

A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium, the partial pressure of HI(g) is 0.04 atm. What is K_p for the given equilibrium ?

MEDIUM

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 22

The following reaction has attained equilibrium:

$CO (g) + 2 H_{2} (g) ⇌ {CH}_{3} OH (g), ΔH ° = - 92.0 kJ {mol}^{- 1}$

What will happen if:

(i) volume of the reaction vessel is suddenly reduced to half?

(ii) the partial pressure of hydrogen is suddenly doubled?

(iii) an inert gas is added to the system at constant volume.

HARD

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 23

What is $K_{c}$ for the following equilibrium when the equilibrium concentration of each substance is,

$[{SO}_{2}] = 0.60 M, [O_{2}] = 0.82 M, [{SO}_{3}] = 1.90 M 2 {SO}_{2 (g)} + O_{2 (g)} ⇌ 2 {SO}_{3 (g)} ||$

HARD

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IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 24

Nitric oxide reacts with

{Br}_{2}

and gives nitrosyl bromide as per reaction given below:

2 N O (g) + B r_{2} (g) ⇌ 2 N O B r (g)

When

0.087 m o l

N O

and

0.0437 m o l

{B r}_{2}

are mixed in a closed container at a constant temperature,

0.0518 mol

N O B r

is obtained at equilibrium. Calculate equilibrium amount of

N O

and

{B r}_{2}

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 25

1400 K, K_{c} = 2.5 \times 10^{- 3}

for the reaction

{CH}_{4} (g) + 2 H_{2} S (g) ⇌ {CS}_{2} (g) + 4 H_{2} (g)

. A

10.0 L

reaction vessel at

1400 K

contains

2.00 mole

{CH}_{4}, 3.0 mol

{CS}_{2}, 3.0 mole

H_{2}

4.0 mole

and of

H_{2} S

. Then in which direction reaction will proceed to reach equilibrium.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 26

473 K

, equilibrium constant

K_{c}

for decomposition of phosphorus pentachloride,

{P C l}_{5}

8.3 \times 10^{- 3} .

If decomposition is depicted as,

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

∆_{r} H^{e} = 124.0 k J {m o l}^{- 1}

(a) write an expression for

K_{c}

for the reaction.

(2 m a r k)

(b) what is the value of

K_{c}

for the reverse reaction at the same temperature?

(1 m a r k)

{P C l}_{5}

is added

(1 m a r k)

(ii) pressure is increased

(1 m a r k)

(iii) the temperature is increased?

(1 m a r k)

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 27

At a certain temperature and total pressure of

10^{5} Pa

, iodine vapour contains

40 %

by volume of

I

atoms

I_{2} (g) ⇌ 2 I (g)

Calculate

K_{p}

for the equilibrium.

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Chemistry>Chapter 5 - Equilibrium>Exercise-1>Q 28

Consider the following reaction $N_{2} O_{4} (g) ⇌ 2 {NO}_{2} (g) ΔH = 58.6 kJ$

What will be the effect of the following changes on the concentration of $N_{2} O_{4}$ at equilibrium ?

(i) Increasing the pressure
(ii) Increasing the temperature
(iii) Increasing the volume
(iv) Adding more ${NO}_{2} (g)$ to the system without changing temperature and pressure
(v) Adding catalyst.

Discuss the effect of temperature on the equilibrium constant. How does it change for a exothermic reaction b endothermic reaction c having zero heat of reaction?

Important Questions on Equilibrium

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Discuss the effect of temperature on the equilibrium constant. How does it change for $(a)$ exothermic reaction $(b)$ endothermic reaction $(c)$ having zero heat of reaction?