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.

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

it should be all of them because it asks if you included all of those in your answer

explanation:

i think it's c. i'm not sure but i think so.