Lecture Example of Calculating the equilibrium composition

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Adams Dis 1A
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Joined: Wed Nov 14, 2018 12:23 am

Lecture Example of Calculating the equilibrium composition

Postby Adams Dis 1A » Wed Jan 09, 2019 9:14 pm

From the example from the lecture on 1-9-18 about calculating the equilibrium composition using the ice table, I am still a bit confused on why there was the subtraction of the 8.434x10^-3, on the ATP side, and the addition of it on the ADP side to get the answer.
May someone please clarify this particular example?

Rian Montagh 2K
Posts: 60
Joined: Fri Sep 28, 2018 12:15 am

Re: Lecture Example of Calculating the equilibrium composition

Postby Rian Montagh 2K » Wed Jan 09, 2019 9:22 pm

The subtraction occurred on the ATP side because we were given the initial and final amounts of ATP, with the final amount being smaller than the initial amount. This means that 8.434x10^-3 was subtracted from the given initial amount to get to the given final amount.

Since matter is conserved, all the matter that was subtracted from the ATP side is not held in the ATP molecules and is now held in the ADP and Pi side (the products). That's why we add 8.434x10^-3 to the ADP and Pi side.

megan blatt 2B
Posts: 61
Joined: Fri Sep 28, 2018 12:28 am

Re: Lecture Example of Calculating the equilibrium composition

Postby megan blatt 2B » Wed Jan 09, 2019 9:32 pm

You would add the 8.434x10^-3 on the ADP side because it is a product and you would subtract this number on the ATP side because it is a reactant. The basic idea of a reaction is that the reactants are undergoing a chemical change and forming the products. You use the same number for both ADP and ATP because they have a 1 to 1 mole ratio in the chemical reaction.


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