Skip to content
Question of 117

Q.The rate constants of a reaction at 500 K and 700 K are 0.025 sec^-1 and 0.075 sec^-1 respectively. Calculate the energy of activation of the reaction. (R = 8.314 JK^-1 and log 3 = 0.477)

Odisha ChseOdisha CHSE +2 Science Board Exam 2019Subjective· 3mImportance★★★★★
0% · 0/117 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Ea ≈ 15.98 kJ/mol (about 1.6 × 10^4 J/mol).

Arrhenius two-temperature form:

log(k2/k1) = (Ea / 2.303R) × (1/T1 − 1/T2)

Given k1 = 0.025 s^-1 at T1 = 500 K and k2 = 0.075 s^-1 at T2 = 700 K.

k2/k1 = 0.075/0.025 = 3, so log(k2/k1) = log 3 = 0.477.

1/T1 − 1/T2 = 1/500 − 1/700 = (700 − 500)/(500 × 700) = 200/350000 = 5.714 × 10^-4 K^-1.

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.