At $380^{\circ} C$ , the half-life period for the first order decomposition of $H_{2} O_{2}$ is $360 m i n$ . The energy of activation of the reaction is $200 k J / m o l^{1}$ . Calculate the time required for $75$ % decomposition at $450^{\circ}$ .
Use : $[\frac{1}{653} - \frac{1}{723}] = 1.48 \times 10^{- 4}, e^{14.8} = 2.67 \times 10^{6}$ .

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Solution

Given that ,
Decomposition of $H_{2} O_{2}$ is first order reaction.
at, $T = 380^{\circ} = 380 + 273 = 653 K$
$t_{1 / 2} = 360$ min
$E a = 200 k J / m o l$

We know that Arrhenius equation for calculation of activation energy is given by:-
$K = A$ $e^{- E a / R T}$ $- (4)$
where,
$K$ = rate constant of reaction at temperature $(T K)$
$A$ = pre exponential factor (Arrhenius constant)
$E a$ = universal gas constant
$T$ = temperature in $K$

Let us suppose at temperture $T_{1} K$ and $T_{2} K$ the rate of constants of the reaction are $K_{1}$ and $K_{2}$ respectively,so,
$K_{1} = A$ $e^{- E a / R T_{1}}$ $- (i i)$
$K_{2} = A$ $e^{- E a / R T_{2}}$ $- (i i i)$

Now, $(i i i)$ and $(i i)$
$\Rightarrow \frac{K_{2}}{K_{1}} = \frac{e^{- E a / R T_{2}}}{e^{- E a / R T_{1}}}$

Taking log on both sides:-
$log (\frac{K_{2}}{K_{1}}) = log ⎛ ⎜ ⎝ e^{\frac{- E a}{R} (\frac{1}{T_{2}} - \frac{1}{T_{1}})} ⎞ ⎟ ⎠$

Now, as given, $t_{1 / 2} = 360$ min at $T = 653 K$
we have for first order reaction,
$K_{1} = \frac{0.693}{t_{1 / 2}} = \frac{0.693}{360}$ $m i n^{- 1}$
$T_{2} = 450^{\circ} C = 450 + 273 = 723 K$

Putting the values of $K_{1}, E_{a}, R, T_{2}, T_{1}$ :-

$log (\frac{K_{2}}{K_{1}}) = \frac{- E a}{K} (\frac{1}{T_{2}} - \frac{1}{T_{1}})$
$\Rightarrow log (\frac{K_{2}}{0.693} \times 360) = \frac{200 \times 1000}{8.314} \times [\frac{1}{723} - \frac{1}{653}]$
$\Rightarrow log (\frac{K_{2} \times 360}{0.693}) = \frac{+ 200 \times 1000}{8.314} \times 1.48 \times 10^{- 4}$
$\Rightarrow log (\frac{K_{2} \times 360}{0.693}) = \frac{+ 200 \times 148}{8314}$

Taking antilog on both sides

$\Rightarrow \frac{K_{2} \times 360}{0.693} = A n t i l o g (3.56)$
$\Rightarrow K_{2} = \frac{0.693}{360} \times 35.16$
$= 0.068$ $m i n^{- 1}$

Now, at $T = 723 K$ , we have $K_{2} = 0.068$ $m i n^{- 1}$
Now, for first order reaction we have:-
$K_{2} t = log (\frac{[A]_{0}}{[A]_{t}})$

where, $[A]_{0}$ = conc. of reactant at $t = 0$
$[A]_{t}$ = conc. of reactant at $t = t$

Given, $\frac{[A]_{t}}{[A]_{0}} = 0.25$ (Since $75$ % of reactant is decomposed)
$\Rightarrow t = \frac{1}{K_{2}} l n (\frac{1}{0.25})$
$= \frac{1}{0.068} l n (4)$
$t = 20$ min.

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At 380∘C, the half-life period for the first order decomposition of H2O2 is 360 min. The energy of activation of the reaction is 200 kJ/mol1. Calculate the time required for 75% decomposition at 450∘.Use : [1653−1723]=1.48×10−4,e14.8=2.67×106.