Question

Two reactants A and B separately shows two chemical reactions. Both reactions are made with same initial concentration of each reactant. Reactant A follows first order kinetics, whereas reactant B follows second order kinetics. If both have same half-lives, compare their rates after the lapse of one half-life:

Answer 1

For the reactant A, the rate law expression is as given below :
$rate=K_A[A{]}^1$ .....(i)

and the expression for the half life period is,
$(t_{1/2}{)}_A=\frac{0.693}{K_A}$ ....(ii)

For the reactant B, the rate law expression is as given below :
$rate=K_B[B{]}^2$ ....(iii)

and the expression for the half life period is,
$(t_{1/2}{)}_B=\frac{1}{a\cdot K_B}$ .....(iv)

where, a is initial concentration.
After lapse of I half, the new rates are $r_a^{\prime }$ and $r_B^{\prime }$
$r_B^{\prime }=K_A\times \frac{a}{2};r_B^{\prime }=K_B\times (a/2{)}^2$
$\therefore \frac{r_A^{\prime }}{r_B^{\prime }}=\frac{K_A}{K_B}\times \frac{2}{a}$ .....(vii)

By eqs. (vi) and (vii)
$\frac{r_A^{\prime }}{r_B^{\prime }}=0.693\times a\times \frac{2}{a}=1.386$
Hence, the ratio of the rates of two reactants A and B after the lapse of on half life is 1.386