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A reaction rate increases by a factor of 655 in the presence of a catalyst at 37°C. The activation energy of the original pathway is 106 kJ/mol. What is the activation energy of the new pathway, all other factors being equal? Group of answer choices 89.3 kJ/mol 16,600 kJ/mol 89.3 J/mol 16,600 J/mol

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For a reaction rate that increases by a factor of 655 in the presence of a catalyst at 37°C, the activation energy of the new pathway  is mathematically given as

Ea2=89.28JKoule/mol

What is the activation energy of the new pathway, all other factors being equal?

Generally, the equation for the rate constant  is mathematically given as

K=Ae^{Ea1/Rt}

Therefore

[tex]\frac{1}{655}=\frac{1}{8.31*3.10}(Ea2-105000)[/tex]

Ea2=89286.083Joule/mol

In conclusion, the activation energy

Ea2=89.28JKoule/mol

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