Rate Determining


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Pooja Nair 1C
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Joined: Thu Jul 13, 2017 3:00 am

Rate Determining

Postby Pooja Nair 1C » Fri Mar 16, 2018 4:12 pm

How do you determine whether a step in a mechanism is slow or fast if they don't already give it to you?

Karen Ung 2H
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Re: Rate Determining

Postby Karen Ung 2H » Fri Mar 16, 2018 4:40 pm

According to my TA, information on whether the steps in a mechanism are slow or fast will be given.

704887365
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Joined: Sat Jul 22, 2017 3:00 am

Re: Rate Determining

Postby 704887365 » Fri Mar 16, 2018 7:47 pm

If they give you graphs with activation energy, the graph with the higher activation energy will be slower than the other bc it will take longer and more energy to carry out the reaction.

Michael Downs 1L
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Re: Rate Determining

Postby Michael Downs 1L » Fri Mar 16, 2018 8:56 pm

If the slow step is listed first, then you don't have to worry about calculating any other steps listed below it

Jennifer Ho 1K
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Re: Rate Determining

Postby Jennifer Ho 1K » Sat Mar 17, 2018 2:07 am

The slow step should be given, and the other steps are assumed to be fast.

GabrielGarciaDiscussion1i
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Re: Rate Determining

Postby GabrielGarciaDiscussion1i » Sat Mar 17, 2018 2:49 am

One way to help determine is the the number of molecules needed to combine. So more often than not, a bimolecular reaction will be slower than a unimolecular. Then termolecular being slower than bimolecular. THIS IS NOT FOOL PROOF OR ANY KIND OF LAW TO USE 100% OF THE TIME, just something to consider. Also, look at the Activation energies.

Clarissa Molina 1D
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Re: Rate Determining

Postby Clarissa Molina 1D » Sat Mar 17, 2018 3:25 am

Do you always ignore the reverse of the slow step? Why?

GabrielGarciaDiscussion1i
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Re: Rate Determining

Postby GabrielGarciaDiscussion1i » Sat Mar 17, 2018 11:31 am

Clarissa Molina 1D wrote:Do you always ignore the reverse of the slow step? Why?


Well the idea is that when it comes to the slow step, the reaction produces products slowly in comparison to the other reactions so that the slow reaction can only proceed forward for a significant amount of time since there is not a significant enough amount of product to proceed with the reverse rxn. It reaches equilibrium so slowly that we can negate the reverse k for the slow reaction.


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