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

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

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

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Solution

The correct option is B $\frac{- d [A]}{d t} = \frac{1}{4} \frac{d [B]}{d t}$
For the given chemical equation, we have
$\frac{1}{V_{A}} \frac{- d [A]}{d t} = \frac{1}{V_{B}} \frac{d [B]}{d t}$
$\Rightarrow$ $\frac{1}{(1 / 2)} \frac{- d [A]}{d t} = \frac{1}{2} \frac{d [B]}{d t}$
Hence, $\frac{- d [A]}{d t} = \frac{1}{4} \frac{d [B]}{d t}$

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Similar questions

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

= \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}

B r O_{3 (a q)}^{-} + 5 B r_{(a q)}^{-} + 6 H_{(a q)}^{+} \to 3 B r_{2 (l)} + 3 H_{2} O_{(l)}

+ \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})

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