The reaction is:

$$ A(g) \longrightarrow B(g) $$

This is a first-order reaction.

Initial concentration of reactant A, $[A]_0 = 0.2 \mathrm{~mol} \mathrm{~L}^{-1}$.

Time elapsed, $t = 10 \mathrm{~minutes}$.

Concentration of product B after 10 minutes, $[B]_t = 0.18 \mathrm{~mol} \mathrm{~L}^{-1}$.

Our goal is to find the rate constant, $k$.

First, we need to find the concentration of reactant A that is left after 10 minutes, which we call $[A]_t$.

Look at the stoichiometry of the reaction: $A \longrightarrow B$. For every one mole of A that reacts, one mole of B is formed.

So, the concentration of A that has reacted is equal to the concentration of B that has been formed.

Concentration of A reacted = Concentration of B formed = $0.18 \mathrm{~mol} \mathrm{~L}^{-1}$.

Now, we can calculate the concentration of A remaining after 10 minutes:

$$ [A]_t = \text{Initial concentration of A} - \text{Concentration of A reacted} $$

$$ [A]_t = [A]_0 - 0.18 $$

$$ [A]_t = 0.2 - 0.18 = 0.02 \mathrm{~mol} \mathrm{~L}^{-1} $$

So, after 10 minutes, the concentration of A is $0.02 \mathrm{~mol} \mathrm{~L}^{-1}$.

Now, we will use the integrated rate law for a first-order reaction. You might remember this from your NCERT textbook:

$$ k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t} $$

where:

Let's substitute the values we have into this equation:

$$ k = \frac{2.303}{10 \mathrm{~min}} \log \frac{0.2}{0.02} $$

Now, let's simplify the expression inside the logarithm:

$$ \frac{0.2}{0.02} = 10 $$

So, the equation becomes:

$$ k = \frac{2.303}{10 \mathrm{~min}} \log(10) $$

We know that the value of $\log(10)$ (to the base 10) is 1.

$$ k = \frac{2.303}{10 \mathrm{~min}} \times 1 $$

$$ k = 0.2303 \mathrm{~min}^{-1} $$

Therefore, the rate constant for the reaction is $0.2303 \mathrm{~min}^{-1}$.

Looking at the options provided, this matches Option A.

So, the correct answer is Option A.

Chemical Kinetics

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