Q7b Lyndon's Review


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nelms6678
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Q7b Lyndon's Review

Postby nelms6678 » Sat Mar 17, 2018 12:08 pm

7b) Derive an expression for the rate of formation of the product E using the pre-equilibrium approach.

The unique rate was 1/2 d[E]/dt, why do we put a 1/2?

Chem_Mod
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Re: Q7b Lyndon's Review

Postby Chem_Mod » Sat Mar 17, 2018 12:11 pm

Unique rate takes into account stoichiometric coefficients. Since the product E had the coefficient 2, unique rate is 1/2 d[E]/dt

Johann Park 2B
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Re: Q7b Lyndon's Review

Postby Johann Park 2B » Sat Mar 17, 2018 12:12 pm

Because we are using the unique rate of E, we put 1/(coefficient) in front of the rate. The coefficient in this case is 2, so we have 1/2 d[E]/dt.

In a general rate, we would not use the stoichiometric coefficient.

Jessica Lutz 2E
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Re: Q7b Lyndon's Review

Postby Jessica Lutz 2E » Sat Mar 17, 2018 12:26 pm

Adding on, we knew we needed to use the unique rate because it asked for the rate of formation of product E, instead of just the rate of the reaction as a whole.

Katelyn B 2E
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Re: Q7b Lyndon's Review

Postby Katelyn B 2E » Sat Mar 17, 2018 12:28 pm

Also in this problem, I still don't understand the difference between Lyndon's little k and big k. Is one supposed to represent the rate constant while the other represents the equilibrium constant? And if so, which is which?

Adrienne Dang 1B
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Re: Q7b Lyndon's Review

Postby Adrienne Dang 1B » Sat Mar 17, 2018 4:19 pm

The big K is the equilibrium constant for step 1, and that is used for this approach since we assume that this step in the reaction is working in equilibrium, as a result of step 2 being the slow reaction. The little k is the rate constant.

Emily Mei 1B
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Re: Q7b Lyndon's Review

Postby Emily Mei 1B » Sat Mar 17, 2018 4:27 pm

Unique rate is 1/a*(d[A]/dt)


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