Question

The rate of a reaction is given by $\frac{dx}{dt}=\frac{K(a-x)}{1+bx}$, where a is the initial concentration of the reactant and K, b are constants, x is the concentration of product at time t. What is half-life of this reaction?

• A. $\frac{log2+ab(log2-0.5)}{K}$
• B. $\frac{log2-ab(2log2-0.5)}{K}$
• C. $\frac{log2+2ab(2log2-0.5)}{K}$
• D. None of these

Answer 1

The correct option is A $\frac{log2+ab(log2-0.5)}{K}$

$\frac{dx}{dt}=K\left[\frac{a-x}{1+bx}\right]$
On intregating
or $\int dx\frac{(1-bx)}{(a-x)}=K\int dt$
$\int \frac{dx[1-b(a-x)+ab]}{(a-x)}=K\int dt$
or $\int dx\frac{(1+ab)}{(a-x)}-\int dx\cdot b=Kdt$
or $-(1+ab)log(a-x)-bx=Kt+C$
at $t=0,x=0$
$\therefore C=-(1+ab)loga$
or $-(1+ab)log(a-x)-bx=Kt-(1+ab)loga$
or $Kg=(1+ab)log\left[\frac{a}{(a-x)}\right]-bx$
at $t=t_{1/2},x=\frac{a}{2}$
$\therefore K{t}_{1/2}=(1+ab)log\left[\frac{a}{1-\left(a/2\right)}\right]-\frac{ba}{2}$
$=(1+ab)log2-\frac{ab}{2}=log2+ablog2-ab\times 0.5$
$t_{1/2}=\frac{1og2+ab(log2-0.5)}{K}$