The rate constant for the first order de | Class 12 Chemistry Chapter Chemical Kinetics, Chemical Kinetics NCERT Solutions

Question:

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

log k = 14.34 - 1.25 x 104 K/T

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

Answer:

Arrhenius equation is given by,

k= Ae -Ea/RT  

⇒In k = In A - Ea/RT

⇒In k = Log A - Ea/RT

⇒ Log k = Log A - Ea/2.303RT         (i)

The given equation is

Log k = 14.34 - 1.25 104 K/T             (ii)

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

Ea/2.303RT  = 1.25 104 K/T  

⇒ E=1.25 × 104K × 2.303 × R

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

= 239339.3 J mol - 1 (approximately)

= 239.34 kJ mol - 1

 

Also, when t1/2= 256 minutes,

k = 0.693 / t1/2

= 0.693 / 256

= 2.707 × 10 - 3 min - 1

= 4.51 × 10 - 5s - 1

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

= 668.95 K

= 669 K (approximately)


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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 27: The rate constant for the first order decomposition of H2O2 is given by the following equation: l....