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:
$\ln( \frac{K_2}{K_1} ) = \frac{Ea}{R} ( \frac{1}{T_1}- \frac{1}{T_2} )$
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
$\frac{Ea}{R}( \frac{1}{T_1}- \frac{1}{T_2} )=12.38$
And so
$\ln( \frac{K_2}{K_1})=12.38$
And using $K_1=6.1\cdot 10^{-8} s^{-1}$ , we find K2:
$K_2=K_1 e^{12.38}=0.0145 s^{-1}$