JEE MAIN - Chemistry (2022 - 26th July Evening Shift - No. 4)
At $$30^{\circ} \mathrm{C}$$, the half life for the decomposition of $$\mathrm{AB}_{2}$$ is $$200 \mathrm{~s}$$ 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: $$\log 2=0.30$$ $$\quad \log 3=0.48$$
200 s
323 s
467 s
532 s
Explanation
Since, half-life is independent of the initial concentration of $$A B_{2}$$. Hence, the reaction is "First Order".
$$k=\frac{2.303 \log 2}{t_{1 / 2}}$$
$$\frac{2.303 \log 2}{t_{1 / 2}}=\frac{2.303}{t} \log \frac{100}{(100-80)}$$
$$\frac{2.303 \times 0.3}{200}=\frac{2.303}{t} \log 5$$
$$t=467 \mathrm{~s}$$
$$k=\frac{2.303 \log 2}{t_{1 / 2}}$$
$$\frac{2.303 \log 2}{t_{1 / 2}}=\frac{2.303}{t} \log \frac{100}{(100-80)}$$
$$\frac{2.303 \times 0.3}{200}=\frac{2.303}{t} \log 5$$
$$t=467 \mathrm{~s}$$
Comments (0)
