Question

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
A. 6.932 min
B. 100 min
C. `0.6932 xx 10^(-3)` min
D. `0.6932 xx 10^(-2)` min

Answer 1

Correct Answer - b
`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