Question

In a first order reaction the initial concentration of reactant decreases to 50% in 100 minutes. how long will it take for 90 % of the reactant to react ?

• A. 180 minutes
• B. 882.82 minutes
• C. 560.25 minutes
• D. 332.32 minutes

Hint:

Find k from the half-life, then use t equals 2.303 over k times log of initial over remaining concentration.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a first order reaction, \(k=\dfrac{0.693}{t_{1/2}}\) and \(t=\dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}\).

Step 2: Find k
The concentration falls to 50 percent in 100 min, so \(t_{1/2}=100\text{ min}\) and \(k=\dfrac{0.693}{100}=6.93\times10^{-3}\text{ min}^{-1}\).

Step 3: Calculation
For 90 percent reaction, \([A]=0.1[A]_0\), so \(\log\dfrac{[A]_0}{[A]}=\log10=1\).
\(t=\dfrac{2.303}{6.93\times10^{-3}}\times1=332.3\text{ min}\).

Final Answer:
90 percent of the reactant reacts in about 332.32 minutes. \[ \boxed{\text{(D) }332.32\text{ minutes}} \]