Page No 120: - 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 4.22:

The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

Predict the rate constant at 30º and 50ºC.Draw a graph between ln k and 1/T and calculate the values of A and E_a.

Answer:

From the given data, we obtain

Slope of the line,

According to Arrhenius equation,

Again,

When  ,

Then, 

Again, when  ,

Then, at  ,

Question 4.23:

The rate constant for the decomposition of hydrocarbons is 2.418 × 10⁻⁵ s⁻¹ at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.

Answer:

k = 2.418 × 10⁻⁵ s⁻¹

T = 546 K

E_a = 179.9 kJ mol⁻¹ = 179.9 × 10³ J mol⁻¹

According to the Arrhenius equation,

= (0.3835 − 5) + 17.2082

= 12.5917

Therefore, A = antilog (12.5917)

= 3.9 × 10¹² s⁻¹ (approximately)

Question 4.24:

Consider a certain reaction A → Products with k = 2.0 × 10⁻² s⁻¹. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L⁻¹.

Answer:

k = 2.0 × 10⁻² s⁻¹

T = 100 s

[A]_o = 1.0 moL⁻¹

Since the unit of k is s⁻¹, the given reaction is a first order reaction.

Therefore, 

= 0.135 mol L⁻¹ (approximately)

Hence, the remaining concentration of A is 0.135 mol L⁻¹.

Question 4.25:

Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?

Answer:

For a first order reaction,

It is given that, t_1/2 = 3.00 hours

Therefore, 

= 0.231 h⁻¹

Then, 0.231 h⁻¹ 

Hence, the fraction of sample of sucrose that remains after 8 hours is 0.158.

Question 4.26:

The decomposition of hydrocarbon follows the equation

k = (4.5 × 10¹¹ s⁻¹) e^−28000 K/T

Calculate E_a.

Answer:

The given equation is

k = (4.5 × 10¹¹ s⁻¹) e^−28000 K/T (i)

Arrhenius equation is given by,

 (ii)

From equation (i) and (ii), we obtain

= 8.314 J K⁻¹ mol⁻¹ × 28000 K

= 232792 J mol⁻¹

= 232.792 kJ mol⁻¹

Question 4.27:

The rate constant for the first order decomposition of H₂O₂ is given by the following equation:

log k = 14.34 − 1.25 × 10⁴ K/T

Calculate E_a for this reaction and at what temperature will its half-period be 256 minutes?

Answer:

Arrhenius equation is given by,

The given equation is

From equation (i) and (ii), we obtain

= 1.25 × 10⁴ K × 2.303 × 8.314 J K⁻¹ mol⁻¹

= 239339.3 J mol⁻1 (approximately)

= 239.34 kJ mol⁻¹

Also, when t_1/2 = 256 minutes,

= 2.707 × 10⁻³ min⁻¹

= 4.51 × 10⁻⁵ s⁻¹

It is also given that, log k = 14.34 − 1.25 × 10⁴ K/T

= 668.95 K

= 669 K (approximately)

Question 4.28:

The decomposition of A into product has value of k as 4.5 × 10³ s⁻¹ at 10°C and energy of activation 60 kJ mol⁻¹. At what temperature would k be 1.5 × 10⁴ s⁻¹?

Answer:

From Arrhenius equation, we obtain

Also, k₁ = 4.5 × 10³ s⁻¹

T₁ = 273 + 10 = 283 K

k₂ = 1.5 × 10⁴ s⁻¹

Ea = 60 kJ mol⁻¹ = 6.0 × 10⁴ J mol⁻¹

Then,

= 297 K

= 24°C

Hence, k would be 1.5 × 10⁴ s⁻¹ at 24°C.

Note: There is a slight variation in this Answer and the one given in the NCERT textbook.

Question 4.29:

The time required for 10% completion of a first order reaction at 298 K is

equal to that required for its 25% completion at 308 K. If the value of A is

4 × 10¹⁰ s⁻¹. Calculate k at 318 K and Ea.

Answer:

For a first order reaction,

At 298 K, 

At 308 K, 

According to the Question,

From Arrhenius equation, we obtain

To calculate k at 318 K,

It is given that, 

Again, from Arrhenius equation, we obtain

Question 4.30:

The rate of a reaction quadruples when the temperature changes from

293 K to 313 K. Calculate the energy of activation of the reaction assuming

that it does not change with temperature.

Answer:

From Arrhenius equation, we obtain

Hence, the required energy of activation is 52.86 kJmol⁻¹.