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29 December, 14:13

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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  1. 29 December, 16:52
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    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
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