1.

The rate for a first order reaction is 0.6932 xx 10^(-2) mol l^(-1) "min"^(-1) and the initial concentration of the reactants is 1 M , T_(1//2) is equal to

Answer»

6.932 MIN
100 min
`0.6932 xx 10^(-3)` min
`0.6932 xx 10^(-2)` min

Solution :`R = K["reactant"]^(-1) therefore k = (0.693 xx 10^(-2))/(1)` also
`t_(1//2) = (0.693)/(k) = (0.693)/(0.693 xx 10^(-2)) = 100 ` min


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