# Answer to Question #24463 in Physical Chemistry for arundhati roy

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

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