In a chemical reaction A ⟶ B , it is found that the rate of the reaction doubles | 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 a chemical reaction $A ⟶ B$ , it is found that the rate of the reaction doubles when the concentration of A is increased four times. The units of rate constant of the reaction is

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Solution:

The decomposition of gaseous N₂O₅ takes place as per the given reaction:

N₂O₅ ⟶ 2NO₂ + $\frac{1}{2}$ O₂

The mechanism for this reaction has been suggested as:

$k_{e q}$

N₂O₅ ⇌ NO₂ + NO₃ (fast equilibrium)

NO₂ + NO₃ $k_{1}$ NO₂ + NO + O₂ (slow)

$NO + N O_{3} k_{2} 2 N O_{2} (f a s t)$

Chemistry - Chemical Kinetics Question with Solution | TestHub

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