Question

The reaction is run at two temperatures where temperature 1 is lower than temperature 2. Which relationship is correct for either rate constant (k1 or k2) or activation energy (Ea,1 and Ea,2)

Answer 1

Answer:

The answer is "$K_1 < K_2$"

Explanation:

In the given question, the value of the $K=Ae^{-\frac{Ea}{RT}}$ and the $T_1 <T_2$is the rate value which is the constant that is  $K_2>K_1$. As per the temperature value when its increase rate is constantly increasing. $E_a$ is activation energy it is not dependent on the temperature that why the answer is $K_1 < K_2$.