Question

Given two separate reactions
(I) $\longrightarrow $ Products
(II) $\longrightarrow $ Products,
Reaction (I) follows first order kinetics while reaction (II) follows second order kinetics. Both the reactions have same half lives. When the initial concentration of $(A) $ is ‘n’ times that of $(B)$, then the rate of reaction (I) is $1.386$ times that of reaction (II) at the start of reaction. The value of n is ___

Answer 1

For (A), rate $= k _{ A }[ A ]$
$\left( t _{1 / 2}\right) A =\frac{0.693}{ k _{ A }}$
For (B), rate $= k _{ B }[ B ]^{2}$
$\left( t _{1 / 2}\right)_{ B }=\frac{1}{ a \cdot k _{ B }}$, where a is the conc. of $[ B ]$ at the start of reaction
Since half lives are same
$\frac{0.693}{ k _{ A }}=\frac{1}{ a \cdot k _{ B }}$
$\frac{ k _{ A }}{ k _{ B }}=0.693\, a$
Since initial concentration of (A) is $n$ times that of (B).
$[ A ]=na $
$\frac{ r _{ A }}{ r _{ B }}=\frac{ k _{ A }}{ k _{ B }} \times \frac{\text { na }}{ a _{2}}, \text { given } \frac{ r _{ A }}{ r _{ B }}=1.386$
$1.386=0.693 a \times \frac{ na }{ a ^{2}}$
$n =\frac{1.386}{0.693}=2$