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Scrat [10]
3 years ago
10

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. 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. 12.5 kj/mol 11.5 kj/mol 23.8 kj/mol 58.2 kj/mol 42.0 kj/mol
Chemistry
1 answer:
g100num [7]3 years ago
3 0

The relation between rate constants at different temperatures, temperature and activation energy is known as Arrhenius equation

<u>Arrhenius equation</u>

K=Ae^-{\frac{Ea}{RT} }

For two temperatures

Ea=R(\frac{ln\frac{k1}{k1} }{(\frac{1}{T1})-(\frac{1}{T2})})

Where

Ea = ? = activation energy

k1 = 3.36 × 10⁴

T1=344 k

k2=7.69

T2=219K

R= gas constant = 8.314 J /molK

Putting values

Ea= (8.314)(\frac{ln(\frac{7.69}{33600})}{(\frac{1}{344})(\frac{1}{219})}

Ea = (-69.69)/(-0.00166) = 41981.93 J/mol

Or

Activation energy is 42.0 kJ /mol

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