For a given reaction, the rate of appearance of B is four times t… - Solution | JEE Chemistry

MCQEasyJEE 2024Rate of Reaction

For a given reaction, the rate of appearance of $B$ is four times the rate of disappearance of $A$ . The balanced reaction is:

Answer

Correct answer:A

Standard Method

Given: The rate of appearance of $B$ is four times the rate of disappearance of $A$ .

Find: The balanced reaction consistent with this rate relation.

For a reaction

aA \rightarrow bB

the rate relation is

-\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt}

So,

\frac{d[B]}{dt} = \frac{b}{a}\left(-\frac{d[A]}{dt}\right)

Given that the rate of appearance of $B$ is four times the rate of disappearance of $A$ ,

\frac{d[B]}{dt} = 4\left(-\frac{d[A]}{dt}\right)

Hence,

\frac{b}{a} = 4

So the stoichiometric ratio must be

a : b = 1 : 4

Therefore, the balanced reaction is $A \rightarrow 4B$ .

The correct option is A.

Using the rate relation directly as $4A \rightarrow B$ . This is wrong because the species appearing faster must have the larger stoichiometric coefficient. Instead, compare the ratio $\frac{b}{a} = 4$ .
Ignoring the stoichiometric definition of reaction rate. This is wrong because rates of disappearance and appearance are scaled by coefficients. Instead, use $-\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt}$ .

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