Answer:
[tex]r_{H_2} = \frac{-1}{2} r_{HI}[/tex]
Explanation:
Hello!
In this case, considering the given chemical reaction:
[tex]H_2(g) + I_2(g) \rightarrow 2HI(g)[/tex]
Thus, by applying the law of rate proportions, we can write:
[tex]\frac{1}{-1} r_{H_2} = \frac{1}{-1}r_{i_2} = \frac{1}{2} r_{HI}[/tex]
Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:
[tex]r_{H_2} = \frac{-1}{2} r_{HI}[/tex]
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