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A reaction is followed and found to have a rate constant of 3.36 × 104 M-1s-1 at 344 K and a rate constant of 7.69 M-1s-1 at 219 K. Determine the activation energy for this reaction.23.8 kJ/mol11.5 kJ/mol12.5 kJ/mol42.0 kJ/mol58.2 kJ/mol

Answer :

jufsanabriasa

Answer:

Ea = 42,0 kJ/mol

Explanation:

It is possible to solve this question using Arrhenius formula:

[tex]ln\frac{k2}{k1} = \frac{-Ea}{R} (\frac{1}{T2} -\frac{1}{T1} )[/tex]

Where:

k1: 3,36x10⁴ M⁻¹ s⁻¹

T1: 344K

Ea = ???

R = 8,314472x10⁻³ kJ/molK

k2 : 7,69 M ⁻¹ s⁻¹

T2: 219K

Solving:

[tex]-8,382 = \frac{-Ea}{8,314472x10^{-3}kJ/molK} (1,659x10^{-3}K^{-1})[/tex]

[tex]-8,382 = -Ea*0,19956mol/kJ[/tex]

[tex]-8,382 = -Ea*0,19956mol/kJ[/tex]

[tex]-42,0 kJ/mol = -Ea[/tex]

Ea = 42,0 kJ/mol

I hope it helps!

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