Question

A certain first-order reaction has a rate constant of 2.15×10−2 s−1 at 20 ∘C. What is the value of k at 55 ∘C if Ea = 72.0 kJ/mol ?

Answer 1

Answer:

$k_2=0.504s^{-1}$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the rate constant at 55 °C by using the temperature-variable version of the Arrhenius equation:

$ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )$

Thus, we plug in the temperatures, activation energy and universal constant of gases in consistent units to obtain:

$ln(\frac{k_2}{0.0215s^{-1}} )=-\frac{72000\frac{J}{mol}}{8.3145\frac{J}{mol*K}}(\frac{1}{55+273} -\frac{1}{20+273} ) \\\\ln(\frac{k_2}{0.0215s^{-1}} )=3.154\\\\k_2=0.0215s^{-1}exp(3.154)\\\\k_2=0.504s^{-1}$

Regards!