NTA Abhyas JEE Main2020ChemistryChemical KineticsPractice
What is the activation energy for the decomposition of N 2 O 5 as, N 2 O 5 ⇌ 2 N O 2 + 1 2 O 2 If the values of rate constant are 3.45 × 10 - 5 at 27 ℃ and = 6.9 × 10 - 3 a t 67 ℃ ?
Options
- A112.3 kJ mol − 1
- B200 .5   kJ   mol − 1
- C149 .5   kJ   mol − 1
- D11 .25   kJ   mol − 1
Correct answer
A. 112.3 kJ mol − 1
Step-by-step solution
To be solved with the help of formula, log k 2 k 1 = E a 2 .303 R T 2 − T 1 T 1 T 2 T 1 = 273 + 27 = 300 K T 2 = 273 + 67 = 340 K log 6.9 × 10 − 3 3.45 × 10 − 5 = E a 2.303 × 8.31 340 − 300 340 × 300 log 200 = E a 19.1379 × 40 102000 2.3010 = E a 19.14 × 4 10200 E a = 19.14 × 10200 × 2.3010 4 = 112304 .907 J mol − 1 = 112 .3 kJ mol − 1