A → B The rate constants of the above reaction at 200K and 300K are...

If $K$ is known at two different temperatures the activation energy can be calculated as:

At temperature 1: $ln{K}_{1} = - \dfrac{{E}_{a}}{{RT}_{1}} + lnA$

At temperature 2: $ln{K}_{2} = - \dfrac{{E}_{a}}{{RT}_{2}} + lnA$

We can subtract one of these equations from the other:

$ln{K}_{1} - ln{K}_{2} = \left(- \dfrac{{E}_{a}}{{RT}_{1}} + lnA\right) - \left(- \dfrac{{E}_{a}}{{RT}_{2}} + lnA\right)$

This equation can further simplified to:

$ln\dfrac{{K}_{1}}{{K}_{2}} = \dfrac{{E}_{a}}{R}\left(\dfrac{1}{{T}_{2}} - \dfrac{1}{{T}_{1}}\right)$

$ln\dfrac{0 . 05}{0 . 03} = \dfrac{Ea}{2 . 305 \times 8 . 314} \times \left[\dfrac{1}{200} - \dfrac{1}{300}\right]$

$⇒ 0 . 22 = \dfrac{Ea}{2 . 305 \times 8 . 314} \times \left[\dfrac{1}{600}\right]$

$⇒ {E}_{a} = 2 . 3 \times 8 . 3 \times 600 \times 0 . 22$

$⇒ {E}_{a} = 2519 . 88J ⇒ {E}_{a} = 2520J$