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(D) The rate of reaction is given by the expression: $-\frac{1}{2} \frac{d[A]}{dt} = \frac{1}{4} \frac{d[B]}{dt}$.Given that the rate of increase of c

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$2.5 	imes 10^{-4}$

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(D) The rate of reaction is given by the expression: $-\frac{1}{2} \frac{d[A]}{dt} = \frac{1}{4} \frac{d[B]}{dt}$.Given that the rate of increase of concentration of $B$ is $\frac{\Delta[B]}{\Delta t} = \frac{5 	imes 10^{-3} \ mol \ L^{-1}}{10 \ s} = 5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.Substituting this into the rate expression: $-\frac{d[A]}{dt} = \frac{2}{4} 	imes \frac{d[B]}{dt} = \frac{1}{2} 	imes (5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1})$.Therefore,the rate of disappearance of $A$ is $2.5 	imes 10^{-4} \ mol \ L^{-1} \ s^{-1}$.

Consider a gas phase reaction which occurs in a closed vessel: $2A \rightarrow 4B + C$. The concentration of $B$ is found to be increased by $5 \times 10^{-3} \ mol \ L^{-1}$ in $10 \ s$. The rate of disappearance of $A$ (in $mol \ L^{-1} \ s^{-1}$) is:

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