At $30^{\circ}\mathrm{C}$, the half life for the decomposition of $\mathrm{AB}_{2}$ is 200 s and is independent of the initial concentration of $\mathrm{AB}_{2}$. The time required for $80\%$ of the $\mathrm{AB}_{2}$ to decompose is (Given: $\log 2=0.30$; $\log 3=0.48$)

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Chemical Kinetics

3 0^{\circ} C

, the half life for the decomposition of

AB_{2}

is 200 s and is independent of the initial concentration of

AB_{2}

. The time required for

80%

of the

AB_{2}

to decompose is (Given:

lo g 2 = 0.30

;

lo g 3 = 0.48

)

Answer: (C)

Since the half-life is independent of initial concentration, this is a first-order reaction. For first-order kinetics,

k = \frac{0.693}{t _{1/2}} = \frac{0.693}{200}

. For 80% decomposition, 20% remains:

k = \frac{2.303}{t} lo g \frac{A _{0}}{0.2 A _{0}} = \frac{2.303}{t} lo g 5

. Since

lo g 5 = lo g 10 - lo g 2 = 1 - 0.30 = 0.70

, we have

\frac{0.693}{200} = \frac{2.303 \times 0.70}{t}

. Solving:

t = \frac{2.303 \times 0.70 \times 200}{0.693} = \frac{7}{3} \times 200 \approx 467

s. Answer: (C).

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option