1.

The rate of a first order reaction is 0.69xx10^(-2) min^(-1) and the initial concentration is 0.5mol L^(-1) . The half life period is

Answer»

3000 s
0.33 s
50 s
100 s

Solution :(A) Rate = K[C]
`0.69xx10^(-2) = k[0.5]`
or `k = (0.60xx10^(-2))/(0.5)`
`t_(1//2) = (0.693)/(k) = (0.693xx0.5)/(0.693xx10^(2))`
50 min = 3000 sec


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