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### Calculating the Equilibrium Composition in ATP Hydrolysis

Posted: **Wed Jan 09, 2019 4:25 pm**

by **Tamera Scott 1G**

In today's lecture, Lavelle mentioned that in the example of the ATP hydrolysis, to calculate the change in molarity you need to subtract the equilibrium molarity from the initial molarity of the ATP. How did that number apply to the ADP and Pi equilibrium concentrations?

### Re: Calculating the Equilibrium Composition in ATP Hydrolysis

Posted: **Wed Jan 09, 2019 4:28 pm**

by **Charles Hood Disc 1C**

That number will be equal to the concentration of ADP and Pi. This is due to the fact that the coefficients for ATP and ADP and Pi are all the same (1) so when you lose a certain concentration of ATP, you lose the same concentration of ADP and Pi.

### Re: Calculating the Equilibrium Composition in ATP Hydrolysis

Posted: **Wed Jan 09, 2019 4:54 pm**

by **Karyn Nguyen 1K**

How would we solve the problem if we didn't assume ADP and P are 0?

### Re: Calculating the Equilibrium Composition in ATP Hydrolysis

Posted: **Wed Jan 09, 2019 5:13 pm**

by **Saman Andalib 1H**

The value of Pi and ADP are equivalent to the reduction in the amount of ATP because they are represented with a stoichiometric coefficient of 1 in the equation ATP <--> ADP + Pi. Similar to when we calculated the amount of product created in a limiting reagent problem.

### Re: Calculating the Equilibrium Composition in ATP Hydrolysis

Posted: **Wed Jan 09, 2019 8:11 pm**

by **deepto_mizan1H**

One of the reasons Dr Lavelle subtracted the value of the equilibrium value from the initial value was to determine the change in molarity. Since the values were 0 for the others (ADP and Pi), and the equation was balanced with coefficients of 1, we were able to understand the change was equal to the change of the others and find the value pretty easily. When the others are not 0 there will be another way to calculate them.