The velocity of disappearance for $HI$ is $2.4 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$. Calculate the rate of formation of $H_2$.
The decomposition reaction is $2HI \rightarrow H_2 + I_2$.
The rate expression for the reaction is $-\frac{1}{2} \frac{d[HI]}{dt} = \frac{d[H_2]}{dt}$.
Given the rate of disappearance of $HI$ is $-\frac{d[HI]}{dt} = 2.4 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
Substituting this into the rate expression,the rate of formation of $H_2$ is $\frac{d[H_2]}{dt} = \frac{1}{2} \times (2.4 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}) = 1.2 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
The rate expression for the reaction is $-\frac{1}{2} \frac{d[HI]}{dt} = \frac{d[H_2]}{dt}$.
Given the rate of disappearance of $HI$ is $-\frac{d[HI]}{dt} = 2.4 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
Substituting this into the rate expression,the rate of formation of $H_2$ is $\frac{d[H_2]}{dt} = \frac{1}{2} \times (2.4 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}) = 1.2 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$.
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