CBSE Class 12 · 2006 · Foreign · Chemical KineticsShort Answer

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Show that in the case of a first order reaction, the time required for 99.9% of the reaction to be complete is 10 times that required for half of the reaction to take place. (log 2 = 0.3010)

Answer: t₉₉.₉% = (2.303/k)×3 ≈ 10 × (2.303/k)×0.3010 = 10 t₁/₂.

Solution

t_{99.9%} = \frac{2.303}{k} lo g \frac{100}{0.1} = \frac{2.303}{k} lo g 1000 = \frac{2.303}{k} \times 3

t_{1/2} = \frac{2.303}{k} lo g 2 = \frac{2.303}{k} \times 0.3010

Ratio:

\frac{t _{99.9%}}{t _{1/2}} = \frac{3}{0.3010} \approx 10

t_{99.9%} \approx 10 t_{1/2}

CBSE Class 12 Board · 2006 · Foreign · Subjective

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