Question

the rate constant for the decomposition of acetaldehyde, ch3cho, to methane, ch4, and carbon monoxide, co, in the gas phase is 1.10 × 10−2 m −1 s−1 at 703.7 k and 4.90 x 10-2 m−1 s−1 at 865 k. determine the activation energy for this decomposition. give your answer in kilojoules per mole and use 3 sig figs in your answer.

Answer 1

The rate constant for the decomposition of acetaldehyde, CH₃CHO, to methane, CH₄, and carbon monoxide, CO, in the gas phase is 1.10 × 10² m⁻¹ s⁻¹ at 703.7 k and 4.90 x 10⁻² m⁻¹ s⁻¹ at 865 k.The activation energy for this decomposition in KJ/mole is 154.303

To determine the activation energy of a reaction (Ea) from two reaction rates (k₁ and k₂) at two different temperatures (T₁ and T₂).

we use the equation -

k₂ = Eₐ x ( 1   -  1  )

k₁     R       T₁    T₂

Where,

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

We have -

k₁= 1.10 x 10⁻²  m⁻¹ s⁻¹

T₁= 703.7 K

k₂= 4.90 x 10⁻² m⁻¹ s⁻¹  

T₂= 865 K

Putting the values in the equation -

4.90 x 10⁻² m⁻¹ s¹ =          Ea              x (     1        -       1    )

1.10 x 10⁻² m⁻¹ s⁻¹   (8.3145 J/K.mol)      703.7 K      865 K

Ea = (4.90 x 10⁻² m⁻¹ s⁻¹) / (     1       -        1     ) x (8.3145 J/K.mol)

         1.10 x 10⁻² m⁻¹ s⁻¹        703.7 K    851 K

Ea = 154303 J/mol

When divided by 1000,

Ea in KJ/mol = 154.303 KJ/mol

To learn more about activation energy,

brainly.com/question/8911926

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