Question #24463

The half-life for a first-order reaction is 32 sec.What was the original concentration if,after 2.0 minuites, the concentration is 0.062 M ?

Expert's answer

The half life is the time required for the concentration to drop to one-half its original value. 2 minutes is 2*60= 160 sec.

If you have 160 sec and time of half life, you can calculate number of concentration decreasing to one-half of its previous value.

160 sec / 32 sec = 5 and it means that you need divide original concentration on 2 five times to get 0.062 M, so

[C]/2^5 = 0.062 M

[C]/32 = 0.062 M

[C] = 0.062 M*32

[C] = 1.984 M = 2.0 M

If you have 160 sec and time of half life, you can calculate number of concentration decreasing to one-half of its previous value.

160 sec / 32 sec = 5 and it means that you need divide original concentration on 2 five times to get 0.062 M, so

[C]/2^5 = 0.062 M

[C]/32 = 0.062 M

[C] = 0.062 M*32

[C] = 1.984 M = 2.0 M

## Comments

Assignment Expert27.05.20, 21:50Dear Upasana,

Questions in this section are answered for free. We can't fulfill them all and there

is no guarantee of answering certain question but we are doing our best. And if answer

is published it means it was attentively checked by experts. You can try it yourself by

publishing your question. Although if you have serious assignment that requires large amount

of work and hence cannot be done for free you can submit it as assignment and our

experts will surely assist you.

upasana20.05.20, 13:49The half-life for a first-order reaction is 32 sec.What was the original concentration if,after 2.0 minuites, the concentration is 0.062 M ?

## Leave a comment