Question

The rate constant for the decomposition of acetaldehyde, CH3CHO, to methane, CH4, and carbon monoxide, CO,...

The rate constant for the decomposition of acetaldehyde, CH3CHO, to methane, CH4, and carbon monoxide, CO, in the gas phase is 1.1 × 10−2 L/mol/s at 703 K and 4.95 L/mol/s at 865 K. Determine the activation energy for this decomposition.

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Answer #1

Solution :-

Rate constant K1 =1.1*10^-2 L/mol/s

Temperature T1 = 703 K

Rate constant K2 = 4.95 L /mol/s

Temperature T2 = 865 K

Activation energy Ea= ?

Using the Arrhenius equation we can calculate the activation energy as follows

ln[K2/K1] = Ea/R [ (1/T1)-(1/T2)]

lets put the values in the formula (R= 8.314 J per K mol )

ln[4.95/1.1*10^-2]= Ea/ 8.314 J per K mol [ (1/703)-(1/865)]

6.109 = (Ea/8.314 J per mol K) * 0.0002664

Ea = 6.109 * 8.314 J per K mol / 0.0002664

Ea = 190654 J per mol

190654 J per mol * 1 kJ / 1000 J = 190.6 kJ per mol

Therefore activation energy is 190.6 kJ per mol

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