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Question

(A) The rate of reaction is given by the expression: $	ext{Rate} = -\frac{1}{4} \frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt} = \frac{1}{6} \frac{d[

Vedclass · Accepted Answer

$1 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$ and $30 	imes 10^{-2} \ mol \ L^{-1}$

Vedclass · Answer

(A) The rate of reaction is given by the expression: $	ext{Rate} = -\frac{1}{4} \frac{d[A]}{dt} = -\frac{1}{3} \frac{d[B]}{dt} = \frac{1}{6} \frac{d[C]}{dt} = \frac{1}{9} \frac{d[D]}{dt}$.Given,rate of disappearance of $A$ $(-\frac{d[A]}{dt})$ $= 4 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$.Therefore,$	ext{Rate of reaction} = \frac{1}{4} 	imes (4 	imes 10^{-2}) = 1 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$.Now,$	ext{Rate of disappearance of } B = 3 	imes 	ext{Rate of reaction} = 3 	imes 1 	imes 10^{-2} = 3 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$.Amount of $B$ consumed in $10 \ s = 	ext{Rate of disappearance of } B 	imes 	ext{Time} = (3 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}) 	imes 10 \ s = 30 	imes 10^{-2} \ mol \ L^{-1}$.

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