If a first order reaction takes 40 minutes for 75 % completion, then time required for 50 % completion is

Question

If a first order reaction takes 40 minutes for 75 % completion, then time required for 50 % completion is (A) 5 min (B) 10 min (C) 15 min (D) 20 min

Tardigrade Admin · Accepted Answer

k=(2.303/t) log (a/a-x) = (2.303/40) log ((100/25)) (log 4 = 0.60206) k = 0.035 min-1 Time for 50 % completion (t1/2 ) = (0.693/k) t1/2 = (0.693/0.035) = 19.8 ≈ 20 min

Q. If a first order reaction takes $40$ minutes for $75%$ completion, then time required for $50%$ completion is

$5 min$

$10 min$

$15 min$

$20 min$

$k = \frac{2.303}{t} l o g \frac{a}{a - x}$
$= \frac{2.303}{40} l o g (\frac{100}{25}) (l o g 4 = 0.60206)$
$k = 0.035 mi n^{- 1}$
Time for $50%$ completion $(t_{1/2}) = \frac{0.693}{k}$
$t_{1/2} = \frac{0.693}{0.035} = 19.8 \approx 20 min$

Q. If a first order reaction takes 40 minutes for 75% completion, then time required for 50% completion is

5min

10min

15min

20min