Question

19. 99.9% at a first order reaction was completed
in 32 min. When will 96.875% of the reaction
complete?
1) 48 min
2) 16 min
3) 32 min
4) 24 min​

Answer 1

Answer:

Assume that initial concentration is 100 .

Remaining after 32 min will be 100 - 99.9 = 0.1

Now let after time t , 100 - 96.875 = 3.125 remains

Since

$since \: a = a_0{e}^{ - \lambda \: t} \\ so\\ \frac{ln \: a_{1}}{ ln \: a_{2} } = \frac{ { - \lambda \: t_{1}} }{{ - \lambda \: t} } \\ \\ \frac{ln \: 0.1}{ ln(3.125) } = \frac{32}{t} \\ \\ t= 32 \times \frac{ log( \frac{3125}{1000} ) }{ log(0.1) } \\ \\ take \: base \: 10 \\ \\ t_2 = - 32 \times log( \frac{25}{8} ) \\ = 32(3 log(2) - 2 log(5) ) \\ = 32 (3 \times 0.30 - 2 \times 0.7) \\ = 32(1.4- 0.90) \\ = 32 \times 0.5 \\ = 16 \: min$