Question

The rate constants for the first-order decomposition of a compound are 5.22× 10–4 s–1 at 43°C and 2.91 × 10–3 s–1 at 62°C. What is the value of the activation energy for this reaction? (R = 8.31 J/(mol · K)) a. 79.5 kJ/mol b. 34.5 kJ/mol c. 0.751 kJ/mol d. 0.87104 kJ/mol e. 2 kJ/mol

Answer 1

Activation energy of the reaction is 79.5 kJ/mol

Step-by-step explanation:

According to Arrhenius equation for a reaction-

`k=Ae^{((-E_(a))/(RT))}`

where k is the rate constant, A is the Arrhenius constant,
`E_(a)` is the activation energy and T is temperature in kelvin

For the given two different set of condition, we can write-

at
`43^(0)\textrm{C}`,
`5.22* 10^(-4)=Ae^{((-E_(a))/(8.31* 316))}`............(1)

at
`62^(0)\textrm{C}`,
`2.91* 10^(-3)=Ae^{((-E_(a))/(8.31* 335))}`............(2)

`Eq-(1)/ Eq-(2)` gives-

`(5.22* 10^(-4))/(2.91* 10^(-3))=e^{(E_(a))/(8.31)((1)/(335)-(1)/(316))}`

Solving this equation we get
`E_(a)=79.5 kJ/mol`

So activation energy of the reaction is 79.5 kJ/mol