For a first order reaction, $A \to P, t_{1/2}$ (half life) is 10 days. The time required for $\frac{1^{th}}{4}$ conversion of A (in days) is :
(ln 2=0.693, ln 3=1.1)
To solve this problem, we need to determine the time required for a first-order reaction, where reactant A converts to product P, to achieve a \frac{1}{4} conversion. Given that the half-life (t_{1/2}) is 10 days, and using the relationship between reaction rate constants and half-life for a first-order reaction, we can proceed as follows:
First, recall the formula for half-life of a first-order reaction:
t_{1/2} = \frac{0.693}{k}, where k is the rate constant.
Rearranging this formula to solve for k, we get:
k = \frac{0.693}{t_{1/2}}.
Plug in the given half-life into the equation:
k = \frac{0.693}{10 \, \text{days}} = 0.0693 \, \text{day}^{-1}.
The formula to find the time for a specific conversion in a first-order reaction is:
t = \frac{1}{k} \ln \frac{[A]_0}{[A]}
where [A]_0 is the initial concentration and [A] is the remaining concentration.
For a \frac{1}{4} conversion, [A] = \frac{3}{4}[A]_0. The conversion fraction is essentially the remaining fraction.
Substitute these values into the formula:
t = \frac{1}{0.0693} \ln \frac{[A]_0}{\frac{3}{4}[A]_0} = \frac{1}{0.0693} \ln \frac{4}{3}.
Using the approximation \ln 3 = 1.1 and knowing \ln 2 = 0.693, we have:
\ln \frac{4}{3} = \ln 4 - \ln 3 = (2 \times 0.693) - 1.1 = 1.386 - 1.1 = 0.286.
Finally, calculate t:
t = \frac{1}{0.0693} \times 0.286 = 4.13 days, which rounds to 4.1 days.
Thus, the time required for \frac{1}{4} conversion of A is approximately 4.1 days.
