Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 7.7:

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Answer:

For a pure substance (both solids and liquids),

Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.

Question 7.8:

Reaction between N₂ and O₂ takes place as follows:

If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which Kc = 2.0 × 10^–37, determine the composition of equilibrium mixture.

Answer:

Let the concentration of N₂O at equilibrium be x.

The given reaction is:

Therefore, at equilibrium, in the 10 L vessel:

The value of equilibrium constant i.e.,  = 2.0 × 10^–37 is very small. Therefore, the amount of N₂ and O₂ reacted is also very small. Thus, x can be neglected from the expressions of molar concentrations of N₂ and O₂.

Then,

Now,

Question 7.9:

Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:

When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br₂.

Answer:

The given reaction is:

Now, 2 mol of NOBr are formed from 2 mol of NO. Therefore, 0.0518 mol of NOBr are formed from 0.0518 mol of NO.

Again, 2 mol of NOBr are formed from 1 mol of Br.

Therefore, 0.0518 mol of NOBr are formed from  mol of Br, or

0.0259 mol of NO.

The amount of NO and Br present initially is as follows:

[NO] = 0.087 mol [Br₂] = 0.0437 mol

Therefore, the amount of NO present at equilibrium is:

[NO] = 0.087 – 0.0518

= 0.0352 mol

And, the amount of Br present at equilibrium is:

[Br₂] = 0.0437 – 0.0259

= 0.0178 mol

Question 7.10:

At 450 K, K_p= 2.0 × 10¹⁰/bar for the given reaction at equilibrium.

What is Kc at this temperature?

Answer:

For the given reaction,

Δn = 2 – 3 = – 1

T = 450 K

R = 0.0831 bar L bar K^–1 mol^–1

 = 2.0 × 10¹⁰ bar ^–1

We know that,

Question 7.11:

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?

Answer:

The initial concentration of HI is 0.2 atm. At equilibrium, it has a partial pressure of 0.04 atm. Therefore, a decrease in the pressure of HI is 0.2 – 0.04 = 0.16. The given reaction is:

Therefore,

Hence, the value of K_p for the given equilibrium is 4.0.

Question 7.12:

A mixture of 1.57 mol of N₂, 1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction 

Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer:

The given reaction is:

Now, reaction quotient Q_c is:

Since , the reaction mixture is not at equilibrium.

Again,  . Hence, the reaction will proceed in the reverse direction.

Question 7.13:

The equilibrium constant expression for a gas reaction is,

Write the balanced chemical equation corresponding to this expression.

Answer:

The balanced chemical equation corresponding to the given expression can be written as:

Question 7.14:

One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to

725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,

Calculate the equilibrium constant for the reaction.

Answer:

The given reaction is:

Therefore, the equilibrium constant for the reaction,

Question 7.15:

At 700 K, equilibrium constant for the reaction

is 54.8. If 0.5 molL^–1 of HI₍g₎ is present at equilibrium at 700 K, what are the concentration of H₂₍g₎ and I₂₍g₎ assuming that we initially started with HI₍g₎ and allowed it to reach equilibrium at 700 K?

Answer:

It is given that equilibrium constant for the reaction

 is 54.8.

Therefore, at equilibrium, the equilibrium constant  for the reaction

will be  .

Let the concentrations of hydrogen and iodine at equilibrium be x molL^–1

.

Hence, at equilibrium,

Question 7.16:

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

2 ICl(g) ⇌  I₂(g) + Cl₂(g) ; K_C = 0.14

Answer:

The given reaction is:

2 ICl(g)    ⇌   I₂(g)  +  Cl₂(g) Initial conc.                0.78 M           0           0 At equilibrium         (0.78 – 2x) M     x M        x M

Hence, at equilibrium,

[ICl]=[I2] = 0.167 M[ICl] = (0.78 -2×0.167)M        =0.446 M