Page No 118: - Chapter 4 Chemical Kinetics Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 4.6:

A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

(i) doubled (ii) reduced to half?

Answer:

Letthe concentration of the reactant be [A] = a

Rate of reaction, R = k [A]²

= ka²

(i)If the concentration of the reactant is doubled, i.e. [A] = 2a, then the rate of the reaction would be

= 4ka²

= 4 R

Therefore, the rate of the reaction would increase by 4 times.

(ii) If the concentration of the reactant is reduced to half, i.e.  , then the rate of the reaction would be

Therefore, the rate of the reaction would be reduced to

Question 4.7:

What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?

Answer:

The rate constant is nearly doubled with a rise in temperature by 10° for a chemical reaction.

The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation,

where, k is the rate constant,

A is the Arrhenius factor or the frequency factor,

R is the gas constant,

T is the temperature, and

Ea is the energy of activation for the reaction

Question 4.8:

In a pseudo first order hydrolysis of ester in water, the following results were obtained:

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.

(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.

Answer:

(i) Average rate of reaction between the time interval, 30 to 60 seconds, 

= 4.67 × 10⁻³ mol L⁻¹ s⁻¹

(ii) For a pseudo first order reaction,

For t = 30 s, 

= 1.911 × 10⁻² s⁻¹

For t = 60 s, 

= 1.957 × 10⁻² s⁻¹

For t = 90 s, 

= 2.075 × 10⁻² s⁻¹

Then, average rate constant, 

Question 4.9:

A reaction is first order in A and second order in B.

(i) Write the differential rate equation.

(ii) How is the rate affected on increasing the concentration of B three times?

(iii) How is the rate affected when the concentrations of both A and B are doubled?

Answer:

(i) The differential rate equation will be

(ii) If the concentration of B is increased three times, then

Therefore, the rate of reaction will increase 9 times.

(iii) When the concentrations of both A and B are doubled,

Therefore, the rate of reaction will increase 8 times.

Question 4.10:

In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below:

Answer:What is the order of the reaction with respect to A and B?

Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore,

Dividing equation (i) by (ii), we obtain

Dividing equation (iii) by (ii), we obtain

= 1.496

= 1.5 (approximately)

Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.

Question 4.11:

The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

Answer:Determine the rate law and the rate constant for the reaction.

Let the order of the reaction with respect to A be x and with respect to B be y.

Therefore, rate of the reaction is given by,

According to the Question,

Dividing equation (iv) by (i), we obtain

Dividing equation (iii) by (ii), we obtain

Therefore, the rate law is

Rate = k [A] [B]²

From experiment I, we obtain

= 6.0 L² mol⁻² min⁻¹

From experiment II, we obtain

= 6.0 L² mol⁻² min⁻¹

From experiment III, we obtain

= 6.0 L² mol⁻² min⁻¹

From experiment IV, we obtain

= 6.0 L² mol⁻² min⁻¹

Therefore, rate constant, k = 6.0 L² mol⁻² min⁻¹

Question 4.12:

The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

The given reaction is of the first order with respect to A and of zero order with respect to B.Answer:

Therefore, the rate of the reaction is given by,

Rate = k [A]¹ [B]⁰

⇒ Rate = k [A]

From experiment I, we obtain

2.0 × 10⁻² mol L⁻¹ min⁻¹ = k (0.1 mol L⁻¹)

⇒ k = 0.2 min⁻¹

From experiment II, we obtain

4.0 × 10⁻² mol L⁻¹ min⁻¹ = 0.2 min⁻¹ [A]

⇒ [A] = 0.2 mol L⁻¹

From experiment III, we obtain

Rate = 0.2 min⁻¹ × 0.4 mol L⁻¹

= 0.08 mol L⁻¹ min⁻¹

From experiment IV, we obtain

2.0 × 10⁻² mol L⁻¹ min⁻¹ = 0.2 min⁻¹ [A]

⇒ [A] = 0.1 mol L⁻¹