Question

The time required for completion of \(93.75\%\) of a first order reaction is \(x\) minutes. The half life of it (in minutes) is

• A. \(x/8\)
• B. \(x/2\)
• C. \(x/4\)
• D. \(x/3\)

Hint:

For first order reactions, the fraction remaining after \(n\) half-lives is: \[ \left(\frac12\right)^n \] Use this relation directly to connect percentage completion with half-life.

Answer 1

The Correct Option is C

Step 1: Determine the fraction of reactant remaining.
It is given that \(93.75\%\) of the reaction is completed.
Therefore, percentage of reactant left unreacted is:
\[ 100-93.75=6.25\% \] Convert this percentage into fraction form:
\[ 6.25\%=\frac{6.25}{100} \] \[ =0.0625 \] \[ = \frac{1}{16} \] Thus, after time \(x\), only:
\[ \frac{1}{16} \] of the original reactant remains.

Step 2: Use the property of first order reactions.
For a first order reaction:
after one half-life, amount remaining becomes:
\[ \frac{1}{2} \] after two half-lives:
\[ \left(\frac12\right)^2=\frac14 \] after three half-lives:
\[ \left(\frac12\right)^3=\frac18 \] after four half-lives:
\[ \left(\frac12\right)^4=\frac1{16} \] Thus, \(\frac1{16}\) of reactant remains after \(4\) half-lives.

Step 3: Relate the given time with half-life.
Given that the time required to reach \(\frac1{16}\) of the initial concentration is \(x\) minutes.
Therefore:
\[ x = 4t_{1/2} \] where \(t_{1/2}\) is the half-life.

Step 4: Calculate the half-life.
Rearranging:
\[ t_{1/2}=\frac{x}{4} \]

Step 5: Match with the given options.
The correct option is:
\[ (3)\;\frac{x}{4} \]

Step 6: Final conclusion.
Hence, the half-life of the reaction is:
\[ \boxed{\frac{x}{4}} \]