Question

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

Hint:

For first-order reactions, decomposition of $3/4$ corresponds to two half-lives: $t_{3/4} = 2t_{1/2}$, where $t_{1/2} = 0.693/k$.

Answer 1

Concept: For a first-order reaction: \[ t = \frac{2.303}{k} \log \left( \frac{[R]_0}{[R]} \right) \] If three-fourth of the reactant has reacted, then one-fourth remains: \[ \frac{[R]_0}{[R]} = 4 \]
Step 1: Identify remaining fraction. If $3/4$ decomposes, remaining fraction = $1/4$. So: \[ \frac{[R]_0}{[R]} = \frac{1}{1/4} = 4 \] 
Step 2: Substitute given values. Given: \[ k = 2.54 \times 10^{-3}\ \text{s}^{-1}, \quad \log 4 = 0.60 \] \[ t = \frac{2.303}{2.54 \times 10^{-3}} \log(4) \] 
Step 3: Calculate numerical value. \[ t = \frac{2.303 \times 0.60}{2.54 \times 10^{-3}} \] \[ t = \frac{1.3818}{2.54} \times 10^3 \] \[ t \approx 0.544 \times 10^3 = 544\ \text{s} \] 
Step 4: Final answer. The time required is approximately: \[ \mathbf{544\ s} \]