Reaction `A` follows first order kinetics and reaction `B` follows second order kinetics. If their half lives are equal, compare their rates (i) at the start of the reaction and (ii) after lapse of one half life.

Explanation

For the first order reaction `A`, half life `t_(50)(1) = (0.693)/(k_(1))` and for the second order reaction `B`, half life `t_(50) (2) = (1)/(k_(2)a)`
But `t_(50)(1) = t_(50) (2)` (given)
`:. (0.693)/(k_(1)) = (1)/(k_(2)a)`
`:. (k_(1))/(k_(2)a) = 0.693` ...(i)
a. Rate at the start of reaction `A, r_(1) = k_(1)a` ...first order and aret at the start of reaction `B, r_(2) = k_(2)a^(2)` ...second order
Hence `(r_(1))/(r_(2)) = (k_(1)a)/(k_(2)a^(2)) = (k_(1))/(k_(2)a)= 0.693` by Eq. (i)
Rate after one half life when the concentration of the reactant is reduced to `(a//2)`
for reaction `A,r'_(1) = k_(1)((a)/(2))`
and for reaction `B, r'_(2) = k_(2)((a)/(2))^(2) = (k_(2)a^(2))/(4)`
Hence, `(r'_(1))/(r'_(2)) = (2k_(1))/(k_(2)a) = 2 xx 0.693 = 1.386`