Question

1. in the chemical reaction: \\( \mathrm{ch}_4 + 2 \mathrm{o}_2 \

ightarrow \mathrm{co}_2 + 2 \mathrm{h}_2\mathrm{o}_{(\mathrm{g})} \\) the rate of consumption of oxygen gas is observed to be 4 mol/(l·min). what is the rate of production of carbon dioxide gas?

a. 1 mol/(l·min)

b. 2 mol/(l·min)

c. 4 mol/(l·min)

d. 8 mol/(l·min)

e. 9 mol/(l·min)

Explanation:

Step1: Identify stoichiometric ratio

From the reaction \( \ce{CH_{4(g)} + 2O_{2(g)} -> CO_{2(g)} + 2H_{2}O_{(g)}} \), the mole ratio of \( \ce{CO_2} \) to \( \ce{O_2} \) is \( 1:2 \) (1 mole \( \ce{CO_2} \) per 2 moles \( \ce{O_2} \)).

Step2: Relate reaction rates

Reaction rate relationship: \( \frac{\text{Rate of } \ce{CO_2} \text{ production}}{\text{Stoichiometric coefficient of } \ce{CO_2}} = \frac{\text{Rate of } \ce{O_2} \text{ consumption}}{\text{Stoichiometric coefficient of } \ce{O_2}} \)
Let \( r_{\ce{CO_2}} \) be rate of \( \ce{CO_2} \) production, \( r_{\ce{O_2}} = 4 \, \text{mol/(L·min)} \) (consumption rate).
\( \frac{r_{\ce{CO_2}}}{1} = \frac{4}{2} \)
\( r_{\ce{CO_2}} = 2 \, \text{mol/(L·min)} \)

Answer:

b. \( 2 \, \text{mol/(L·min)} \)