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Step-by-Step Solution
Step 1: Identify the definition of t1/4
For a first-order reaction, $t_{1/4}$ is the time when the concentration of the reactant drops to $ \tfrac{3}{4} $ of its initial value. This is analogous to half-life ($t_{1/2}$) but instead we consider the reactant concentration to be $\tfrac{3}{4}$ of the original value rather than half.
Step 2: Express the rate law for a first-order reaction
For a first-order reaction with rate constant $K$, if the initial concentration is $[A]_0$ and the concentration after time $t$ is $[A]$, then:
$ [A] = [A]_0 \, e^{-Kt} $
Alternatively, using the common logarithm form:
$ \ln\left(\dfrac{[A]}{[A]_0}\right) = -Kt
\quad \Longleftrightarrow \quad
\log\left(\dfrac{[A]}{[A]_0}\right) = -\dfrac{K}{2.303}\,t. $
Step 3: Substitute the condition for t1/4
At time $t_{1/4}$, $[A] = \tfrac{3}{4}[A]_0$. So,
$ \dfrac{[A]}{[A]_0} = \dfrac{3}{4}. $
Hence, from the logarithmic form:
$ \log\left(\dfrac{3}{4}\right) = -\dfrac{K}{2.303}\,t_{1/4}. $
Step 4: Solve for t1/4
Rearrange to find $t_{1/4}$:
$ t_{1/4} = -\dfrac{2.303}{K}\,\log\left(\dfrac{3}{4}\right)
\;=\;
\dfrac{2.303}{K}\,\log\left(\dfrac{4}{3}\right). $
Step 5: Simplify the logarithmic term
We know:
$ \log\left(\dfrac{4}{3}\right) = \log(4) - \log(3). $
Using approximate common log values, $\log(2) \approx 0.3010$ and $\log(3) \approx 0.4771$, we get:
$ \log(4) = \log(2^2) = 2\,\log(2) \approx 2 \times 0.3010 = 0.6020. $
So,
$ \log\left(\dfrac{4}{3}\right) \approx 0.6020 - 0.4771 = 0.1249. $
Thus,
$ t_{1/4} = \dfrac{2.303}{K}\times 0.1249 \approx \dfrac{0.29}{K}. $
Step 6: Conclude the expression for t1/4
Therefore, the time required for the concentration to drop to $\tfrac{3}{4}$ of its initial value for a first-order reaction can be written as:
$ \boxed{ t_{1/4} = \dfrac{0.29}{K} }. $
