## Q7b Lyndon's Review

$K = \frac{k_{forward}}{k_{reverse}}$

nelms6678
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### Q7b Lyndon's Review

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

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

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

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

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?

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### Re: Q7b Lyndon's Review

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

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