Question

if the rate of reaction doubles for 10°c rise in temperature from 290k to 300k the activation energy of the reaction will be approximately​

Answer 1

if the rate of reaction doubles for 10°c rise in temperature from 290k to 300k the activation energy of the reaction will be aapproximately

Answer 2

Given:

T1 = 290 K

T2 = 300 K

From T1 to T2, the rate of reaction doubles.

To Find:

The activation energy of the given reaction.

Calculation:

- Let the rate constant at 290 K be k1 = k

- Since the rate of reaction doubles, hence;

The rate constant at 300 K will be k2 = 2k

- We know the formula:

log (k2 / k1) = (Ea/ 2.303 R) [1/T1 - 1/T2]

⇒ log (2k / k) = (Ea/ 2.303 × 8.314) [1/290 - 1/300]

⇒ log 2 = (Ea/ 19.147) [(300 - 290)/290 × 300]

⇒ Ea = (0.3010 × 19.147 × 290 × 300) / 10

⇒ Ea = 50140.25 J (approx)

⇒ Ea = 50.14 kJ (approx)

- So, the activation energy of the given reaction will be approximately 50.14 kJ.