I'm really confused?
2H2O2⟶2H2O+O2
Experiment no. c(H2O2) rate
1 14.45 3.35×10−3
2 28.90 6.65×10−3
3 57.80 13.35×10−3
So from previous parts of the question, I found the order of reaction (1st), the rate equation for the reaction (r=k[A]), the rate constant (2.3×10−4 s−1, which is correct according to the answers), and now I need to find the rate of decomposition of H2O2 when [H2O2] is exactly 10.0 mmol L−1.
I thought it's just 2.3×10−4×10.0, which is 2.3×10−3, but the answers provided is double that (so 4.6×10−3), so I must have gotten the rate equation wrong?
if it's r=2k[A], then why did I get k right in the first place?
Calculate the rate of decomposition of H2O2 at 25 °C when [H2O2] is exactly 10.0 mmol/L?
2016-06-14 3:20 pm
回答 (1)
2016-06-14 3:50 pm
I think your calculation is correct, but the answer provided (4.6 × 10⁻³) is incorrect.
Look at the data given.
1. When [H₂O₂] = 14.45 mmol/L, Rate = 3.35 × 10⁻³ mmol/(L s)
2. When [H₂O₂] = 28.90 mmol/L, Rate = 6.65 × 10⁻³ mmol/(L s)
3. When [H₂O₂] = 57.80 mmol/L, Rate = 13.35 × 10⁻³ mmol/(L s)
Obviously, the higher the [H₂O₂], the higher the reaction rate is.
When [H₂O₂] = 10.0 mmol/L < 14.45 mmol/L, It should be Rate < 3.35 × 10⁻³ mmol/(L s).
Then, 4.6 × 10⁻³ mmol/(L s) [< 3.35 × 10⁻³ mmol/(L s)] is impossible !
Look at the data given.
1. When [H₂O₂] = 14.45 mmol/L, Rate = 3.35 × 10⁻³ mmol/(L s)
2. When [H₂O₂] = 28.90 mmol/L, Rate = 6.65 × 10⁻³ mmol/(L s)
3. When [H₂O₂] = 57.80 mmol/L, Rate = 13.35 × 10⁻³ mmol/(L s)
Obviously, the higher the [H₂O₂], the higher the reaction rate is.
When [H₂O₂] = 10.0 mmol/L < 14.45 mmol/L, It should be Rate < 3.35 × 10⁻³ mmol/(L s).
Then, 4.6 × 10⁻³ mmol/(L s) [< 3.35 × 10⁻³ mmol/(L s)] is impossible !
收錄日期: 2021-04-18 15:00:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160614072018AApiNQO