Question

The thermal decomposition of an acid is a first-order reaction with a rate constant of \(2.3 \times 10^{-5}\ \text{s}^{-1}\) at a certain temperature. Calculate how long it will take for three-fourths of the initial quantity of acid to decompose. (\(\log 4 = 0.6021\), \(\log 2 = 0.3010\))

Hint:

Use the first-order rate law formula: \( t = \frac{2.303}{k} \log \frac{[A]_0}{[A]} \) when amount decomposed is given.

Answer 1

For a first-order reaction: \[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right) \] Let the initial amount be \([A]_0\), and three-fourths decomposed means \([A] = \frac{1}{4}[A]_0\) So: \[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_0/4} \right) = \frac{2.303}{t} \log 4 \Rightarrow k = \frac{2.303 \times 0.6021}{t} \] Given \( k = 2.3 \times 10^{-5} \ \text{s}^{-1} \), so: \[ 2.3 \times 10^{-5} = \frac{2.303 \times 0.6021}{t} \Rightarrow t = \frac{2.303 \times 0.6021}{2.3 \times 10^{-5}} \approx \frac{1.387}{2.3 \times 10^{-5}} \approx 60217.4\ \text{seconds} \] \[ \boxed{t \approx 6.02 \times 10^4\ \text{seconds}} \]