Question

At 600.0 k the rate constant is 6.1× 10–8 s–1. what is the value of the rate constant at 785.0 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 785.0 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 T1=600 K and T2=785 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) )=12.38`
And so

`\ln( (K_2)/(K_1))=12.38`
And using
`K_1=6.1\cdot 10^(-8) s^(-1)` , we find K2:

`K_2=K_1 e^(12.38)=0.0145 s^(-1)`