Need help on finding rate of reaction! Chemistry.?

[匿名]

2018-01-24 9:03 am

更新1:

The value of the rate constant at 302°C is 2.45 × 10-4 L/mol s and at 508°C the rate constant is 0.166 L/mol s. The value of R is 8.3145 J/K mol. Calculate the value of the rate constant for this reaction at 375°C. what I did is to use Arrhenius equation. I find the Ea=11.807 Joules But I'm not sure where to go from there. Which of value should I use to find the K value for reaction at 375? the .166 one or 2.45e-4 one? Thank you!

更新2:

Sorry! I calculated my Ea incorrectly. It's 118.143 kj / or 118143 Joules.

回答 (1)

wanszeto

2018-01-24 2:50 pm

✔ 最佳答案

When T₁ = (273 + 302) K = 575 K, k₁ = 2.45 × 10⁻⁴ L / (mol s)
When T₂ = (273 + 508) K = 781 K, k₂ = 0.166 L / (mol s)
When T₃ = (273 + 375) K = 648 K, k₃ = ? L / (mol s)

One form of Arrhenius equation: ln(k₁/k₂) = (Eₐ/R) [(1/T₂) - (1/T₁)]
ln[(2.45 × 10⁻⁴)/0.166] = (Eₐ/8.3145) [(1/781) - (1/575)]
Activation energy, Eₐ = 8.3145 × ln[(2.45 × 10⁻⁴)/0.166] / [(1/781) - (1/575)] = 118150 J/mol

ln(k₃/k₂) = (Eₐ/R) [(1/T₂) - (1/T₃)]
ln(k₃) - ln(k₂) = (Eₐ/R) [(1/T₂) - (1/T₃)]
ln(k₃) - ln(0.166) = (118150/8.3145) [(1/781) - (1/648)]
ln(k₃) = (118150/8.3145) [(1/781) - (1/648)] + ln(0.166)
k₃ = e^{(118150/8.3145) [(1/781) - (1/648)] + ln(0.166)} = 3.97 × 10⁻³ L / (mol s)

The rate constant for the reaction at 375°C = 3.97 × 10⁻³ L / (mol s)

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