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

The answer is "
`K_1 < K_2`"

Step-by-step explanation:

In the given question, the value of the
`K=Ae^{-(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`.