Tag: kinetic study of some first order reactions

For $2H _2O _2(aq)\rightarrow 2H _2O(l)+O _2(g)$ rate of decomposition for $H _2O _2=k[H _2O _2]$. It is a first order reaction.  It proceeds through following mechanism.

Option (A),(B),(D) are correct.
(C) : The half-life of the reaction is independent of the initial concentration of the reactant. Half-life for first order reaction is :$t _{1/2} = 0.693/k$
A first-order reaction has a rate proportional to the concentration of one reactant.
First-order rate constants have units of $sec^{-1}$. In other words, a first-order reaction has a rate law in which the sum of the exponents is equal to 1. 

As we know,
$r=K[N _2O _5]=6.2\times 10^{-4}\times 1.25=7.75\times 10^{-4} mol L^{-1} s^{-1}$.

$[H _2SO _4]\, =\, 0.1\, M\, =\, 0.1\, \times\, 2\, =\, 0.2 N$

$[HCl]$ = $0.1 N$

In case of $[H _2SO _4]$ 

$r _1\, =\, k[H^{\oplus}][Ester]$ 

$\displaystyle k _{H _2SO _4}\, = \frac{r _1}{2\, N\, \times\, [Ester]}$ 

In case of HCl, $r _1\, =\, k[H^{\oplus}]\, [Ester]$ 

$\displaystyle k _{HCl}\, =\, \frac{r _2}{1\, N\, [Ester]}$ 

As we know,
for a reaction:
$2N _2O _5\, \rightarrow\, 4NO _2\, +\, O _2$
$\displaystyle -\, \frac{1}{2}\, \frac{d[N _2O _5]}{dt}\, =\, \frac{1}{4}\, \frac{d[NO _2]}{dt}\, =\, \frac{d[O _2]}{dt}$
So
$2k _1\, =\, k _2\, =\, 4k _3$

Given,
Since $t _{1/2}$ does not depends upon the sugar concentration means it is first order w.r.t [sugar]
$\therefore t _{1/2}\, \propto\, [sugar]^{1}$
$t _{1/2}\, \times\, a^{n\, -\, 1}\, =\, k$
$\displaystyle \frac{(t _1/2) _1}{(t _{1/2) _2}}\, =\, \frac{[H^{\oplus}] _1^{1-n}}{[H^{\oplus}] _2^{1-n}}$

$\displaystyle \frac{500}{50}\, =\, \left ( \frac{10^{-5}}{10^{-6}}\right )^{1-n}$

10 = $(10)^{1-n}\, \Rightarrow\, n\, =\, 0$