For the reaction 4 NH₃ (g)+ 7 O₂ (g) $\rightarrow$ 4 NO₂(g) + 6 H₂(g), if H₂ is produced at a rate of 3.5 * 10^-3 M/s, O₂ is consumed at a rate of

Question

Quizplus · Accepted Answer

Given the stoichiometry of the reaction, the rate of production of $\text{HS}_2\text{O}_2$ is directly related to the rate of consumption of $\text{O}_2$. Since 6 moles of $\text{HS}_2\text{O}_2$ are produced for every 7 moles of $\text{O}_2$ consumed, the rate of consumption of $\text{O}_2$ can be calculated using the ratio of their stoichiometric coefficients. Given the rate of production of $\text{HS}_2\text{O}_2$ is $3.5 \times 10^{-3} \text{M/s}$, the rate of consumption of $\text{O}_2$ is $\frac{7}{6} \times 3.5 \times 10^{-3} \text{M/s} = 4.1 \times 10^{-3} \text{M/s}$.

For the Reaction 4 NH3 (G)+ 7 O2 (G) →\rightarrow→ 4 NO2(g) + 6 H2(g), If H2 Is Produced at 4

For the Reaction 4 NH₃ (G)+ 7 O₂ (G) $\rightarrow$ 4 NO₂(g) + 6 H₂(g), If H₂ Is Produced at 4