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 `1M`, `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)`
`:. K = (0.693 xx 10^(-2))/(1)`
also `t_(1//2) = (0.693)/(k) = (0.693)/(0.693 xx 10^(-2)) = 100 min`