Question

Methanol can be produced by the following reaction: CO(g) 2 H2(g) CH3OH(g). How is the rate of disappearance of hydrogen gas related to the rate of appearance of methanol

Answer 1

Answer:

$r_{H_2}=-2r_{CH_3OH}$

Explanation:

Hello!

In this case, for the reaction:

$CO(g)+ 2 H_2(g) \rightarrow CH_3OH(g)$

In such a way, via the rate proportions, that is written considering the stoichiometric coefficients, we obtain:

$\frac{1}{-1} r_{CO}=\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}$

Whereas the reactants, CO and H2 have negative stoichiometric coefficients; therefore the rate of disappearance of hydrogen gas is related to the rate of appearance of methanol as shown below:

$\frac{1}{-2} r_{H_2}=\frac{1}{1} r_{CH_3OH}\\\\r_{H_2}=\frac{-2}{1} r_{CH_3OH}\\\\r_{H_2}=-2r_{CH_3OH}$

Which means that the rate of disappearance of hydrogen gas is negative and the rate of appearance of methanol is positive.

Regards!