Question

If the rate constant for a reaction triples when the temperature rises from 300. K to 310. K, what is the activation energy of the reaction

Answer 1

Answer:

$Ea=-84.9\frac{kJ}{mol}$

Explanation:

Hello,

In this case, temperature-variable version of the Arrhenius equation is applied in order to compute the activation energy as shown below:

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

Thus, we replace each variable considering k2=3*k1 (tripling):

$Ea=8.314\frac{J}{mol*K} *\frac{ln(\frac{3k_1}{k_1})}{(\frac{1}{310K} -\frac{1}{300K} )}\\\\Ea=-84.9\frac{kJ}{mol}$

Best regards.