If 70% of a first order reaction was completed in 52 minutes, 50% of the same reaction would be completed in approximately (log 3 = 0.47)

Question

If 70% of a first order reaction was completed in 52 minutes, 50% of the same reaction would be completed in approximately (log 3 = 0.47) (A) 30 minutes (B) 42 minutes (C) 40 minutes (D) 52 minutes

Tardigrade Admin · Accepted Answer

(0.693/t(1)2)=(2.303/52)log (100/(100-70)) t1/2 ≃ 30 minute

Q. If 70% of a first order reaction was completed in 52 minutes, 50% of the same reaction would be completed in approximately (log 3 = 0.47)

30 minutes

42 minutes

40 minutes

52 minutes

$\frac{0.693}{t _{\frac{1}{2}}} = \frac{2.303}{52} l o g \frac{100}{( 100 - 70 )}$
$t_{1/2} ≃ 30 min u t e$

Q. If 70% of a first order reaction was completed in 52 minutes, 50% of the same reaction would be completed in approximately (log 3 = 0.47)

30 minutes

42 minutes

40 minutes

52 minutes

t21​​0.693​=522.303​log(100−70)100​ t1/2​≃30minute

$\frac{0.693}{t _{\frac{1}{2}}} = \frac{2.303}{52} l o g \frac{100}{( 100 - 70 )}$
$t_{1/2} ≃ 30 min u t e$