For the reaction $\frac{1}{2} A \to 2 B$ , the rate of disappearance of ‘A’ is related to the rate of appearance of ‘B’ by the expression:

- \frac{d [A]}{d t} = \frac{1}{2} \frac{d [B]}{d t}

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- \frac{d [A]}{d t} = \frac{1}{4} \frac{d [B]}{d t}

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- \frac{d [A]}{d t} = \frac{d [B]}{d t}

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- \frac{d [A]}{d t} = 4 \frac{d [B]}{d t}

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Solution

The correct option is B $- \frac{d [A]}{d t} = \frac{1}{4} \frac{d [B]}{d t}$

$\frac{1}{2} A \to 2 B - \frac{d [A]}{d t} \times \frac{2}{1} = \frac{1}{2} \frac{d [B]}{d t} - \frac{d [A]}{d t} = \frac{1}{4} \frac{d [B]}{d t}$

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Similar questions

\frac{1}{2} A \to 2 B

For the reaction, 2N₂O₅ → 4NO₂ + O₂, which of the relationship is correct?

= \frac{1}{2} \frac{d [C]}{d t} = - \frac{1}{3} \frac{d [D]}{d t} = \frac{1}{4} \frac{d [A]}{d t} = - \frac{d [B]}{d t}

+ \frac{1}{2} \frac{d [C]}{d t} = - \frac{1}{3} \frac{d [D]}{d t} = + \frac{1}{4} \frac{d [A]}{d t} = - \frac{d [B]}{d t}

Consider the reaction A $\to$ B. Which of the following defines the rate of this reaction?

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