$$[A] \xrightarrow{} [B]$$
Reactant $$\to$$ Product
If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is $$x \times 10^{-6} s^{-1}$$. The value of $$x$$ is ____

We need to find the rate constant for a first-order reaction where the concentration of A becomes half in 70 minutes. The concentration of A becomes half of its initial concentration in 70 minutes, which means the half-life is given by $$t_{1/2} = 70 \text{ minutes} = 70 \times 60 = 4200 \text{ seconds}$$.

For a first-order reaction the rate constant is calculated using $$k = \frac{0.693}{t_{1/2}}$$. Substituting the value of the half-life gives $$k = \frac{0.693}{4200}$$, which evaluates to $$k = 1.65 \times 10^{-4} \text{ s}^{-1}$$. This can also be expressed as $$k = 165 \times 10^{-6} \text{ s}^{-1}$$. Therefore, $$x = 165$$, and the answer is 165.