Just to clarify:
If a 2nd order rate looked like rate = k [OH-] [H3Br]
the rate is second order for the reaction
but the rate is first order for each reactant?
2nd order rates
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Re: 2nd order rates
Yes, that is true. The total rate order involves summing the exponents of each reactant.
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Re: 2nd order rates
That's perfectly fine.
The total order of the reaction is calculated by summing up the exponents the individual reactants. (1+1=2)
The total order of the reaction is calculated by summing up the exponents the individual reactants. (1+1=2)
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Re: 2nd order rates
Yes. In order to get overall order, you simply add the powers of the reactants--so, two reactants of first order will form a second order rate law for the overall reaction.
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Re: 2nd order rates
Ahmed Mahmood 4D wrote:Yes. In order to get overall order, you simply add the powers of the reactants--so, two reactants of first order will form a second order rate law for the overall reaction.
Thank you! I’m actually starting to understand this. Thanks to your comment. Do you think we will be expected to know 3rd orders? Or would be just need to be able to identify them?
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Re: 2nd order rates
I’d look through the textbook/notes and memorize concepts and formulas relating to third order rxn. Anything that isn’t mentioned probably isn’t necessary.
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Re: 2nd order rates
Yes, because total rate order is found by adding the exponents of the reactants.
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