1. Identify the order of reaction

Given rate law: $r = k[A]$ This is a first-order reaction.

2. Use first-order integrated rate law

For a first-order reaction: $[A]_t = [A]_0 e^{-kt}$ Or, $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]_t}$

3. Use the information for 50% decomposition

If $50\%$ of $A$ is decomposed, then: $[A]_t = \frac{[A]_0}{2}$ Given time = 120 min.

So, $120 = \frac{2.303}{k} \log 2$ Since for first-order reaction, half-life is: $t_{1/2} = \frac{0.693}{k}$ Thus, $120 = \frac{0.693}{k}$ So, $k = \frac{0.693}{120}$

4. Find time for 90% decomposition

If $90\%$ is decomposed, then $10\%$ remains: $[A]_t = 0.1[A]_0$

Using first-order formula: $t = \frac{2.303}{k} \log \frac{[A]_0}{0.1[A]_0}$ $t = \frac{2.303}{k} \log 10$ Since $\log 10 = 1$, $t = \frac{2.303}{k}$

Substitute $k = \frac{0.693}{120}$: $t = \frac{2.303}{0.693/120}$ $t = 120 \times \frac{2.303}{0.693}$ $t \approx 120 \times 3.323$ $t \approx 398.8 \text{ min}$

5. Final integer answer

$t \approx 399 \text{ minutes}$

6. Comparison with stored answer

Derived answer = $399$

Stored correct answer = $399$

They agree.