Question

In a first order reaction the concentration of a substance gets dissociated by 99% of the initial concentration in 100 minutes. Calculate the velocity constant of the reaction.

Hint:

In first order kinetics, percentage dissociation is directly related to \(\log \frac{[R]_0}{[R]}\).

Answer 1

Step 1: Recall integrated rate law for first order.
\[ k = \frac{2.303}{t} \log \frac{[R]_0}{[R]} \] Step 2: Apply given values.
99% dissociated → remaining concentration = 1% of initial. \[ \frac{[R]_0}{[R]} = \frac{100}{1} = 100 \] \(t = 100 \ \text{min}\).
Step 3: Substitution.
\[ k = \frac{2.303}{100} \log (100) \] \[ k = \frac{2.303}{100} \times 2 = 0.04606 \ \text{min}^{-1} \] Step 4: Conclusion.
The velocity constant = \(0.046 \ \text{min}^{-1}\).