Step 1: Determine the order of reaction.
The problem involves percentages of completion and time, which suggests a first-order reaction. The formula for the time required to achieve a certain completion in a first-order reaction is:
\[
t = \frac{2.303}{k} \log \frac{[A]_0}{[A]},
\]
where:
- \( t \) is the time,
- \( k \) is the rate constant,
- \( [A]_0 \) is the initial concentration,
- \( [A] \) is the concentration at time \( t \).
Step 2: Calculate the rate constant \( k \).
For 50\% completion, \( [A]_0/[A] = 2 \). Substituting \( t = 15 \, \text{minutes} \):
\[
15 = \frac{2.303}{k} \log 2.
\]
\[
k = \frac{2.303 \log 2}{15}.
\]
Using \( \log 2 = 0.3010 \):
\[
k = \frac{2.303 \times 0.3010}{15} = 0.04627 \, \text{min}^{-1}.
\]
Step 3: Calculate the time for 90\% completion.
For 90\% completion, \( [A]_0/[A] = 10 \). Substituting into the formula:
\[
t = \frac{2.303}{k} \log 10.
\]
Using \( \log 10 = 1 \):
\[
t = \frac{2.303}{0.04627} \times 1.
\]
\[
t = 49.44 \, \text{minutes}.
\]
Step 4: Final Answer.
The time required to complete 90\% of the reaction is:
\[
\boxed{49.44 \, \text{minutes}}.
\]