Question

C4H8(g)⟶2C2H4(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, is 6.1×10−8 s−1.6.1×10−8 s−1. What is the value of the rate constant at 725.0 K?

Answer 1

Rate constant at 725 K is
`5.2* 10^(-4)s^(-1)`

Step-by-step explanation:

According to Arrhenius equation for a reaction-

`ln((k_(2))/(k_(1)))=(E_(a))/(R)((1)/(T_(1))-(1)/(T_(2)))`

where
`k_(2)` and
`k_(1)` are rate constants of reaction at
`T_(2)` and
`T_(1)` temperatures (in kelvin) respectively.

`E_(a)` is activation energy of reaction.

Here
`T_(1)`= 600 K ,
`k_(1)`=
`6.1* 10^(-8)s^(-1)`

`T_(2)`= 725 K,
`E_(a)`= 262 kJ/mol and R = 8.314 J/(mol.K)

So plugin all the values in the above equation-

`ln((k_(2))/(6.1* 10^(-8)s^(-1)))=(262* 10^(3)J/mol)/(8.314J/(mol.K))* ((1)/(600K)-(1)/(725K))`

So,
`k_(2)` =
`5.2* 10^(-4)s^(-1)`

Hence rate constant at 725 K is
`5.2* 10^(-4)s^(-1)`