Question

The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K, 600.0 K, the rate constant, k , k, is 6.1 × 10 − 8 s − 1 . 6.1×10−8 s−1. What is the value of the rate constant at 775.0 K?

Answer 1

Answer:
`4.3* 10^(-13)s^(-1)`

Step-by-step explanation:

According to the Arrhenius equation,

`K=A* e^{(-Ea)/(RT)}`

or,

`\log ((K_2)/(K_1))=(Ea)/(2.303* R)[(1)/(T_1)-(1)/(T_2)]`

where,

`K_1` = rate constant at
`600.0K` =
`6.1* 10^(-8)s^(-1)`

`K_2` = rate constant at
`775.0` =
`?`

`Ea` = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

`T_1` = initial temperature =
`600.0K`

`T_2` = final temperature =
`775.0K`

Now put all the given values in this formula, we get

`\log ((6.1* 10^(-8))/(K_2))=(262000)/(2.303* 8.314J/mole.K)[(1)/(600.0K)-(1)/(775.0K)]`

`\log ((6.1* 10^(-8)s^)/(K_2))=5.150`

`((6.1* 10^(-8))/(K_2))=141253.8`

Therefore, the value of the rate constant at 775.0 K is
`4.3* 10^(-13)s^(-1)`