Two reaction, `(I)A rarr` Products and `(II) B rarr` Products, follow first order kinetics. The rate of reaction `(I)` is doubled when the temperature is raised form `300 K` to `310K`. The half life for this reaction at `310K` is `30 min`. At the same temperature `B` decomposes twice as fast as `A`. If the energy of activation for reaction `(II)` is twice that of reaction `(I)`, (a) calculate the rate of constant of reaction `(II)` at `300 K`.

Explanation

To solve the problem step by step, we will follow the information provided in the question and the video transcript.

### Step 1: Determine the rate constant \( K_2 \) at 310 K for reaction (I)
Given that the rate of reaction (I) is doubled when the temperature is raised from 300 K to 310 K, we can express this relationship as:
\[ \frac{K_2}{K_1} = 2 \]
where \( K_1 \) is the rate constant at 300 K and \( K_2 \) is the rate constant at 310 K.
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