Question

The time needed for completion of 80% is \( y \) times the half-life period of a first order reaction. What is the value of \( y \)?

• A. 0.648
• B. 3.46
• C. 2.32
• D. 0.322

Hint:

For first-order reactions, the time for a certain percentage of completion can be calculated by using the rate law and the relationship between the time and half-life.

Answer 1

The Correct Option is C

Step 1: Understand the first-order reaction kinetics.
For a first-order reaction, the rate law is given by:
\[ \ln \left( \frac{[A_0]}{[A_t]} \right) = kt \]
where:
- \( [A_0] \) is the initial concentration,
- \( [A_t] \) is the concentration at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.
The half-life \( t_{1/2} \) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]

Step 2: Use the first-order reaction formula.
For completion of 80%, the remaining concentration is 20%. Thus, \( [A_t] = 0.2 [A_0] \).
Now, applying this in the rate equation:
\[ \ln \left( \frac{[A_0]}{0.2 [A_0]} \right) = kt \]
\[ \ln (5) = kt \]
\[ kt = \ln 5 = 1.609 \]

Step 3: Relate to the half-life.
The time \( t_{1/2} \) corresponds to the half-life of the reaction, which is the time taken for 50% completion. The total time \( t \) for 80% completion is given by:
\[ t = y \cdot t_{1/2} \]
Thus, \( kt = y \cdot 0.693k \), where \( 0.693k \) is the half-life.

Step 4: Calculate the value of \( y \).
From the equation \( 1.609 = y \cdot 0.693 \), solve for \( y \):
\[ y = \frac{1.609}{0.693} = 2.32 \]

Step 5: Conclusion.
Thus, the value of \( y \) is \( 2.32 \), which corresponds to option (C).