Question

A → B The above reaction is of zero order Half life of this reaction is 50 min The time taken for the concentration of A to reduce to one-fourth of its initial value is ____ min (Nearest integer)

Hint:

For zero-order reactions, the half-life is directly proportional to the initial concentration. Remember the integrated rate law and the half-life formula for zero order reactions.

Answer 1

Correct Answer: 75

For a zero-order reaction, the integrated rate law is given by:
\[[A]_t = [A]_0 - kt,\]
where \([A]_t\) is the concentration of \(A\) at time \(t\), \([A]_0\) is the initial concentration of \(A\), and \(k\) is the rate constant.
The half-life (\(t_{1/2}\)) of a zero-order reaction is given by:
\[t_{1/2} = \frac{[A]_0}{2k}.\]
Given that \(t_{1/2} = 50 \, \text{min}\), we can find the rate constant \(k\):
\[k = \frac{[A]_0}{2 \cdot t_{1/2}} = \frac{[A]_0}{2 \cdot 50} = \frac{[A]_0}{100}.\]
We are asked to find the time taken for the concentration of \(A\) to reduce to one-fourth of its initial value. Let this time be \(t\). So, \([A]_t = \frac{[A]_0}{4}\). Substituting this into the integrated rate law:
\[\frac{[A]_0}{4} = [A]_0 - kt.\]
\[\frac{3[A]_0}{4} = kt.\]
Substituting the value of \(k\) we found earlier:
\[\frac{3[A]_0}{4} = \frac{[A]_0}{100} \cdot t.\]
Simplify:
\[t = \frac{3[A]_0}{4} \cdot \frac{100}{[A]_0} = 3 \cdot 25 = 75 \, \text{min}.\]
Final Answer:
The time taken is \(t = 75 \, \text{min}\).

Answer 2

The correct answer is 75.
Order = zero order
t1/2 = 50min Rο - Rο/2 = Kt1/2
Rο/2 = K×50 K = Rο/2×1/50
Rο-Rο/4 = Rο/2×1/50 t
⇒ t = 75 min