Calculate the half life period of a first order reaction having $K=2.0~s^{-1}$

Question

Accepted Answer

The correct option is (2):

To calculate the half-life period ($t\_{1/2}$) of a first-order reaction, we use the following relationship:
$t\_{1/2} = \frac{\ln(2)}{K}$
where $K$ is the rate constant of the reaction.

We are given the rate constant $K = 2.0 \mathrm{\~s^{-1}}$.
The value of $\ln(2)$ is approximately $0.693$.

Now, substitute the values into the formula:
$t\_{1/2} = \frac{0.693}{2.0 \mathrm{\~s^{-1}}}$
$t\_{1/2} = 0.3465 \mathrm{\~s}$

Therefore, the half-life period of the first-order reaction is $0.3465$ seconds. This matches option (2). Understanding this formula is crucial for solving problems related to first-order kinetics, which are very common in chemical reactions and radioactive decay.

The incorrect option is (1):
$0.693$ second. This option would be correct if the rate constant $K$ was $1.0 \mathrm{\~s^{-1}}$, as $t\_{1/2} = 0.693/1 = 0.693 \mathrm{\~s}$. However, since $K=2.0 \mathrm{\~s^{-1}}$, this option is incorrect.

The incorrect option is (3):
$2$ second. This option incorrectly assumes that the half-life is simply the inverse of the rate constant or some other direct relationship not involving $\ln(2)$. For a first-order reaction, the half-life is inversely proportional to the rate constant, but includes the $\ln(2)$ term in the numerator.