Question

Rate constant k for the first order reaction is $2.54 \times 10^{-3} s^{-1}$. Calculate the time required for three-fourth of the reactant to decompose. $(\log 4 = 0.60)$

Hint:

"Decomposed" means reacted. In the formula, $[R]$ represents the amount left remaining, not the amount reacted.

Answer 1

Step 1: Concept
The integrated rate law is used to find the time ($t$) needed to reach a specific fraction of completion in a first-order reaction.

Step 2: Meaning
If three-fourths ($3/4$) of the reactant has decomposed, it means exactly one-fourth ($1/4$) of the initial amount remains unreacted.

Step 3: Analysis
1. The remaining concentration $[R] = [R_0] - \frac{3}{4}[R_0] = \frac{1}{4}[R_0]$.
2. Rearranging the first-order rate equation for time $t$:
$t = \frac{2.303}{k} \log\left(\frac{[R_0]}{[R]}\right)$
3. Substitute $[R]$ and the given rate constant $k = 2.54 \times 10^{-3} s^{-1}$:
$t = \frac{2.303}{2.54 \times 10^{-3}} \log\left(\frac{[R_0]}{[R_0]/4}\right) = \frac{2.303}{2.54 \times 10^{-3}} \log(4)$
4. Substitute $\log 4 = 0.60$:
$t = \frac{2.303 \times 0.60}{2.54 \times 10^{-3}} = \frac{1.3818}{0.00254} \approx 544~s$

Step 4: Conclusion
The calculation precisely yields the time taken for $75\%$ completion.

Final Answer: $544 \text{ s}$