Arrhenius Equation:

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Postby TanveerDhaliwal3G » Tue Mar 10, 2020 6:44 pm

Can someone explain the reasoning behind how the solutions manual does 7E.3 and why it is completely different from example 7E.1 even though it is the same type of problem.

Niharika 1H
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Joined: Thu Jul 25, 2019 12:16 am

Re: 7E.3

Postby Niharika 1H » Tue Mar 10, 2020 9:34 pm

For 7E.3, we have to use a ratio of the Arrhenius equation, one at which the activation energy is 125 kJ/mol, and one at which it is 75kJ/mol. 7E.1 takes a different approach, as it says that the reverse and forward reactions rates increase using a catalyst.

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