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

For a first-order reaction, the time required for a certain fraction of the reactant to decompose is given by the equation: \[ t = \frac{2.303}{k} \log \left( \frac{[R]_0}{[R]} \right) \] where: - \( t \) is the time required, - \( k \) is the rate constant, - \([R]_0\) is the initial concentration, - \([R]\) is the concentration at time \( t \). If three-fourths of the reactant has reacted, one-fourth remains. Therefore, the ratio of initial concentration to remaining concentration is: \[ \frac{[R]_0}{[R]} = 4 \] 
Step 1: Identify remaining fraction.
If \(3/4\) of the reactant has decomposed, then the remaining fraction of the reactant is \(1/4\). Thus: \[ \frac{[R]_0}{[R]} = \frac{1}{1/4} = 4 \] 
Step 2: Substitute given values.
Substitute the given rate constant \(k = 2.54 \times 10^{-3}\ \text{s}^{-1}\) and \(\log 4 = 0.60\) into the equation for time \(t\): \[ t = \frac{2.303}{2.54 \times 10^{-3}} \log(4) \] 
Step 3: Calculate numerical value.
Now, calculate the time: \[ t = \frac{2.303 \times 0.60}{2.54 \times 10^{-3}} = \frac{1.3818}{2.54} \times 10^3 \] \[ t \approx 0.544 \times 10^3 = 544\ \text{s} \] 
Step 4: Final answer.
The time required is approximately: \[ \boxed{544\ \text{s}} \]