1. Write the reaction and relate moles formed to moles reacted

Given: $A \to 2B$

Initially, there is $1$ mole of $A$.

If $x$ moles of $A$ react, then since $1$ mole of $A$ produces $2$ moles of $B$, $\text{moles of } B = 2x$

We are told that after $100$ min, $0.2$ moles of $B$ are formed. So, $2x = 0.2$ $x = 0.1$

Thus, moles of $A$ left after $100$ min are $1 - 0.1 = 0.9$

2. Use first-order kinetics

For a first-order reaction, $k = \frac{1}{t} \ln \frac{[A]_0}{[A]_t}$

Here,

• $t = 100$ min
• $[A]_0 = 1$
• $[A]_t = 0.9$

So, $k = \frac{1}{100} \ln \frac{1}{0.9} = \frac{1}{100} \ln \frac{10}{9}$

Now, $\ln \frac{10}{9} = \ln 10 - \ln 9$

Also, $\ln 9 = \ln(3^2) = 2\ln 3$

Using standard approximation, $\ln \frac{10}{9} \approx 0.105$

Hence, $k \approx \frac{0.105}{100} = 0.00105\ \text{min}^{-1}$

3. Calculate half-life

For a first-order reaction, $t_{1/2} = \frac{\ln 2}{k}$

Using $\ln 2 = 0.69$, $t_{1/2} = \frac{0.69}{0.00105}$

$t_{1/2} \approx 657.14\ \text{min}$

Rounded to nearest integer, $\boxed{657}$

4. Comparison with stored correct answer

Stored correct answer: 600to700

Our derived integer answer is $657$, which lies in the range $600$ to $700$. So it agrees with the stored answer range.