A reaction,3 X_{(g)} rightarrow 2 Y_{(g)} + Z | vedclass

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(B) The rate of reaction is given by the expression: $-\frac{1}{3} \frac{d[X]}{dt} = \frac{1}{2} \frac{d[Y]}{dt} = \frac{d[Z]}{dt}$ Given that the rat

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$4.8 	imes 10^{-3}$

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(B) The rate of reaction is given by the expression: $-\frac{1}{3} \frac{d[X]}{dt} = \frac{1}{2} \frac{d[Y]}{dt} = \frac{d[Z]}{dt}$ Given that the rate of disappearance of $X$ is $-\frac{d[X]}{dt} = 7.2 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$. From the rate expression,we have: $\frac{1}{2} \frac{d[Y]}{dt} = -\frac{1}{3} \frac{d[X]}{dt}$ Therefore,the rate of formation of $Y$ is: $\frac{d[Y]}{dt} = \frac{2}{3} 	imes (-\frac{d[X]}{dt}) = \frac{2}{3} 	imes 7.2 	imes 10^{-3} = 4.8 	imes 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

$A$ reaction,$3 X_{(g)} \rightarrow 2 Y_{(g)} + Z_{(g)}$ takes place in a closed vessel. What is the rate of formation of $Y$ (in $mol \ L^{-1} \ s^{-1}$) if the rate of disappearance of $X$ is $7.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$?

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