Consider the following reaction: 2 NOBr(g) ➔2 NO(g) +Br2(g)When the reaction takes place in a 1.0-L flask, 8.0 x 10-4 mole of Br2is formed during the first5.0 sec of the reaction. What is the rate of disappearance of NOBr?
answer should be (Ans. 3.2 x 10-4 mol/L.s
rate law question?
2020-07-14 9:59 pm
回答 (1)
2020-07-14 10:10 pm
2 NOBr(g) → 2 NO(g) + Br₂(g)
Rate of formation of Br₂ = (8.0 × 10⁻⁴ mol/L) / (5.0 s) = 1.6 × 10⁻⁴ mol/L•s
Mole ratio NOBr : Br₂ = 2 : 1
Rate of disappearance of NOBr = (1.6 × 10⁻⁴ mol/L•s) × 2 = 3.2 × 10⁻⁴ mol/L•s
Rate of formation of Br₂ = (8.0 × 10⁻⁴ mol/L) / (5.0 s) = 1.6 × 10⁻⁴ mol/L•s
Mole ratio NOBr : Br₂ = 2 : 1
Rate of disappearance of NOBr = (1.6 × 10⁻⁴ mol/L•s) × 2 = 3.2 × 10⁻⁴ mol/L•s
收錄日期: 2021-04-18 18:34:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200714135940AA6gTiq