In 20 minutes 80% of N 2 O 5 is decomposed. Find rate constant | TestHub

Passage / Comprehension

The order of reaction is an experimentally determined quantity. It may be zero, positive, negative, or fractional. The kinetic equation for an nᵗʰ-order reaction is:

$k \times t = \frac{1}{( n - 1 )} [\frac{1}{( a - x ) ^{n - 1}} - \frac{1}{a ^{n - 1}}]$

The half-life of an nᵗʰ-order reaction depends on the initial concentration according to the following relation:

$t_{1/2} \propto \frac{1}{a ^{n - 1}}$ .

The unit of the rate constant varies with the order, but the general relation for the unit of an nᵗʰ-order reaction is:

$U ni t o f k = [\frac{1}{( co n c . )}]^{n - 1} . (T im e)^{- 1}$

The differential rate law for an nᵗʰ-order reaction may be given as:

$\frac{d x}{d t} = k [A]^{n}$

where A denotes the reactant.

In 20 minutes $80%$ of $N_{2} O_{5}$ is decomposed. Find rate constant

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Answer:

Solution:

$r a t e = k_{1} [N O_{2}] [N O_{3}]$

$K_{e q} = \frac{[ N O _{2} ] [ N O _{3} ]}{[ N _{2} O _{5} ]}$ .

$r a t e = k_{1} [N O_{2}] K_{e q} \frac{[ N _{2} O _{5} ]}{[ N O _{2} ]}$ .

$r a t e = k^{'} [N_{2} O_{5}]$

$k^{'} = \frac{2.303}{20} lo g (\frac{100}{20}) = \frac{2.303}{20} \times 0.6990$

$k^{'} = 0.08 mi n^{- 1}$

Chemistry - Chemical Kinetics Question with Solution | TestHub

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