Question

If rate constant of a first order reaction is 3x10-2 s-1 then the half life period of the reaction will be
O 46.2 s
O 23.1 s.
O 75
O 90 S​

Answer 1

Answer:

for example =>

The rate constant for a first order reaction is 60 s

−1

. How much time will it take to reduce the initial concentration of the reactant to its 1/16

th

value?

:-

Given that the rate constant for a first order reaction is 60 s

−1

.

Let a M be the initial concentration.

Final concentration will be

16

a

M.

Rate constant, k=60/s

t=

k

2.303

log(

[A]

[A]

0

)

t=

60

2.303

log

⎝

⎛

16

a

a

⎠

⎞

t=4.6×10

−2

seconds

Answer 2

[ For MCQs, follow this method.]

$\orange{ \boxed{ \boxed{ \begin{array}{cc} \sf \: \to \: given : \\ \\ \hookrightarrow \sf \:\bf \: rate \: \: constant \: , \: k = 3 \times {10}^{ - 2} \: {s}^{ - 1} \\ \\ \hookrightarrow \bf \:half \: life \: period \: , t_{ \frac{1}{2} } = \: to \: find \\ \\ \red{ \sf \: we \: know \: that : } \\ \\ \blue{ \rm \: for \: first \: order \: reaction : } \\ \\ \boxed{\bf \: t_{ \frac{1}{2} } = \frac{0.693}{k} } \\ \\ \bf \: \implies \bf \:t_{ \frac{1}{2}} = \frac{0.693}{k } \\ \\ \implies \bf \:t_{ \frac{1}{2} } = \frac{0.693}{3 \times {10}^{ - 2} } \\ \\ \implies \bf \:t_{ \frac{1}{2}} = 23.1 \: s \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( \sf \: answer) \end{array}}}}$