(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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(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 $-\frac{d[X]}{dt} = 7.2 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.Substituting this into the relation: $\frac{1}{2} \frac{d[Y]}{dt} = \frac{1}{3} \left( -\frac{d[X]}{dt} \right)$.Therefore,the rate of formation of $Y$ is $\frac{d[Y]}{dt} = \frac{2}{3} \times (7.2 \times 10^{-3}) = 4.8 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

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$A$ 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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A
$3.6 \times 10^{-3}$
B
$4.8 \times 10^{-3}$
C
$2.4 \times 10^{-3}$
D
$1.2 \times 10^{-3}$

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$A$ 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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$A$ solid cylinder $30 \ cm$ in diameter at the top of an inclined plane $2.0 \ m$ high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom is .......... $m/s$.

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