Question

A first-order reaction has a rate constant \( 1.25 \times 10^{-3} \, s^{-1} \). How long will 5 g of this reactant take to reduce to 2.5 g?
$[ log 2 = 0·301, log 3 = 0·4771, log 4 = 0·6021 ]$

Hint:

For first-order reactions, the half-life formula: \[ t_{1/2} = \frac{0.693}{k} \] can be used to estimate the reaction time.

Answer 1

For a first-order reaction, the integrated rate law is: \[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \] where: - \( k = 1.25 \times 10^{-3} \, s^{-1} \) - \( [R]_0 = 5 \) g - \( [R] = 2.5 \) g Substituting the values: \[ 1.25 \times 10^{-3} = \frac{2.303}{t} \log \left( \frac{5}{2.5} \right) \] Since \( \log 2 = 0.301 \), we get: \[ 1.25 \times 10^{-3} = \frac{2.303}{t} \times 0.301 \] Solving for \( t \): \[ t = \frac{2.303 \times 0.301}{1.25 \times 10^{-3}} \] \[ t = \frac{0.693}{1.25 \times 10^{-3}} \] \[ t = 554.5 \, s \text{ or } 5.54 \times 10^2 \, s \]