In the reaction CL₂(g) + 3F₂(g) $\rightarrow$ 2ClF₃(g), if the average rate of disappearance of F₂ is -0.250 M/s over the first 6.00 s, how much ClF₃ will form during this time?

Question

Quizplus · Accepted Answer

The rate of disappearance of $ \text{F}_2 $ is $ -0.250 \, \text{M/s} $. Since the stoichiometry of the reaction shows that 3 moles of $ \text{F}_2 $ produce 2 moles of $ \text{ClF}_3 $, the rate of formation of $ \text{ClF}_3 $ is $ \frac{2}{3} $ of the rate of disappearance of $ \text{F}_2 $. Thus, $ \frac{2}{3} \times 0.250 \, \text{M/s} = 0.167 \, \text{M/s} $. Over 6.00 s, the amount of $ \text{ClF}_3 $ formed is $ 0.167 \, \text{M/s} \times 6.00 \, \text{s} = 1.00 \, \text{M} $.

In the Reaction CL2(g) + 3F2(g) →\rightarrow→ 2ClF3(g), If the Average Rate of Disappearance of F2 Is

In the Reaction CL₂(g) + 3F₂(g) $\rightarrow$ 2ClF₃(g), If the Average Rate of Disappearance of F₂ Is