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