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:
[tex]ln ( \frac{K_2}{K_1} ) = \frac{Ea}{R} ( \frac{1}{T_1}- \frac{1}{T_2}) [/tex]
where K are the reaction rates, Ea is the activation energy, R=8.314 J/mol*K and T are the temperatures. Using [tex]T_1=600 K[/tex] and [tex]T_2=790 K[/tex], and Ea=262 kJ/mol = 262000 J/mol, on the right side of the equation we have
[tex] \frac{Ea}{R} ( \frac{1}{T_1}- \frac{1}{T_2})=12.64[/tex]
And so
[tex]ln ( \frac{K_2}{K_1})=12.64
[/tex]
And using [tex]K_1=6.1\cdot 10^{-8} s^{-1}[/tex], we find K2:
[tex]K_2 = K_1 e^{12.64}=0.0188 s^{-1}[/tex]
