Question

For a first order reaction, the time required for completion of \(75\%\) of the reaction is \(40\) minutes. The half-life of the reaction is:

• A. \(10\) min
• B. \(20\) min
• C. \(40\) min
• D. \(80\) min

Hint:

For first order reactions: \[ 75\% \text{ completion}=2t_{1/2} \] because after two half-lives only \(25\%\) reactant remains.

Answer 1

The Correct Option is B

Concept: For first order reactions: \[ t = \frac{2.303}{k}\log\frac{a}{a-x} \] and: \[ t_{1/2} = \frac{0.693}{k} \]

Step 1: Interpret \(75\%\) completion.
If \(75\%\) reaction is completed, then: \[ 25\% \] reactant remains. Hence: \[ \frac{a}{a-x}=4 \]

Step 2: Apply integrated rate equation.
\[ 40=\frac{2.303}{k}\log4 \] Since: \[ \log4=0.6021 \] \[ 40=\frac{2.303\times0.6021}{k} \] \[ 40=\frac{1.386}{k} \] \[ k=\frac{1.386}{40} \]

Step 3: Calculate half-life.
\[ t_{1/2}=\frac{0.693}{k} \] Substituting: \[ t_{1/2} = \frac{0.693}{1.386/40} \] \[ t_{1/2}=20\text{ min} \] Hence: \[ \boxed{20\text{ min}} \]