The rate constant for the decomposition of acetaldehyde, CH₃CHO, to methane, CH₄, and carbon monoxide, CO, in the gas phase is 1.32 ✕ 10⁻² L/mol/s at 707 K and 3.20 L/mol/s at 851 K. Determine the activation energy (in kJ/mol) for this decomposition.

Question

Chemistry

Russell Lyons

7 October, 01:18

The rate constant for the decomposition of acetaldehyde, CH₃CHO, to methane, CH₄, and carbon monoxide, CO, in the gas phase is 1.32 ✕ 10⁻² L/mol/s at 707 K and 3.20 L/mol/s at 851 K. Determine the activation energy (in kJ/mol) for this decomposition.

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Answers (1)

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Lilly Klein · Answer 1

The correct answer is 190.74 KJ/mol

Explanation:

In order to determine the activation energy of a reaction (Ea) from two reaction rates (k1 and k2) at two different temperatures (T1 and T2), we have to use the following equation:

ln (k2/k1) = Ea/R x (1/T1 - 1/T2)

Where R is the gas constant (8.3145 J/mol. K).

We have: k1 = 1.32 x 10⁻² L/mol/s; T1 = 707 K

k2 = 3.20 L/mol/s; T2 = 851 K

We introduce the data in the equation and determine Ea:

ln (3.20 L/mol/s/1.32 x 10⁻² L/mol/s) = Ea / (8.3145 J/K. mol) x ((1/707 K) - (1/851 K))

⇒Ea = ln (3.20 L/mol/s/1.32 x 10⁻² L/mol/s) / ((1/707 K) - (1/851 K)) x (8.3145 J/K. mol)

⇒Ea = 190,743 J/mol

We divide by 1000 to obtain Ea in KJ/mol: 190.74 KJ/mol