The rate constant of a chemical reaction increased from 0.100s−1 to 3.30s−1 upon raising the temperature from 25.0∘C to 47.0∘C.?
A. Calculate the value of (1/T2−1/T1) where T1 is the initial temperature and T2 is the final temperature?
B. Calculate the value of ln(k1/ k2) where k1 and k2 correspond to the rate constants at the initial and the final temperatures as defined in part A?
回答 (1)
A.
k₁ = 0.100 s⁻¹, T₁ = (273.2 + 25.0) K = 298.2 K
k₂ = 3.30 s⁻¹, T₂ = (273.2 + 47.0) K = 320.2 K
(1/T₂ - 1/T₁)
= (1/320.2 - 1/298.2) K⁻¹
= -2.304 × 10⁻⁴ K⁻¹
B.
ln(k₁/k₂)
= ln(0.100/3.30)
= -3.50
收錄日期: 2021-04-18 16:04:45
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