Question

At 600.0 k the rate constant is 6.1× 10–8 s–1. what is the value of the rate constant at 790.0 k?

Answer 1

Missing details. Complete text is:
"The following reaction has an activation energy of 262 kJ/mol.

C4H8(g) --> 2C2h4(g)

At 600.0 K the rate constant is 6.1× 10–8 s–1. What is the value of the rate constant at 790.0 K?"

To solve the exercise, we can use Arrhenius equation:

`ln ( (K_2)/(K_1) ) = (Ea)/(R) ( (1)/(T_1)- (1)/(T_2))`

where K are the reaction rates, Ea is the activation energy, R=8.314 J/mol*K and T are the temperatures. Using
`T_1=600 K` and
`T_2=790 K`, and Ea=262 kJ/mol = 262000 J/mol, on the right side of the equation we have

`(Ea)/(R) ( (1)/(T_1)- (1)/(T_2))=12.64`

And so

`ln ( (K_2)/(K_1))=12.64`

And using
`K_1=6.1\cdot 10^(-8) s^(-1)`, we find K2:

`K_2 = K_1 e^(12.64)=0.0188 s^(-1)`