Question

For the first order reaction $A \rightarrow 2B$, 1 mole of reactant A gives 0.2 moles of B after 100 minutes. The half life of the reaction is ________ min. (Round off to the Nearest Integer).

Hint:

In kinetics, product formation must always be converted back to reactant consumption using stoichiometry before applying rate laws.

Answer 1

Correct Answer: 315

Given reaction: \[ A \rightarrow 2B \] From stoichiometry: \[ 1\ \text{mol A} \rightarrow 2\ \text{mol B} \] If 0.2 mol of B is formed, then moles of A reacted: \[ = \frac{0.2}{2} = 0.1\ \text{mol} \] Initial moles of A: \[ [A]_0 = 1\ \text{mol} \] Moles of A remaining after 100 min: \[ [A]_t = 1 - 0.1 = 0.9\ \text{mol} \] For a first-order reaction: \[ k = \frac{1}{t}\ln\!\left(\frac{[A]_0}{[A]_t}\right) \] \[ k = \frac{1}{100}\ln\!\left(\frac{1}{0.9}\right) \] \[ \ln\!\left(\frac{1}{0.9}\right) = \ln\!\left(\frac{10}{9}\right) = \ln 10 - \ln 9 \] \[ = 2.3 - 2(0.69) = 2.3 - 1.38 = 0.92 \] \[ k = \frac{0.92}{100} = 0.0092\ \text{min}^{-1} \] Half-life for first-order reaction: \[ t_{1/2} = \frac{0.69}{k} = \frac{0.69}{0.00219} \approx 315\ \text{min} \]