Example from lecture on Wednesday

[AlyssaBei_1F]

On Wednesday in lecture, Dr. Lavalle went over an example about ATP Hydrolysis with ATP + h20 as the reactants going to ADP + Pi. Can someone explain the ice table values, such as how he got them and the answer?

[Neil Hsu 2A]

In the example, ATP + H20 <> ADP + Pi and given to us are the concentrations of ATP in healthy and in dead tissue. The question is asking for the concentrations of ADP and Pi in dead tissue. For this example, we are assuming that there is no (or negligible amounts of) ADP and Pi in the healthy tissue, so both their initial concentrations will start at zero. When you set up the ICE table, it portrays the initial molar concentration in the first row, the change in molar concentration in the second row, and the equilibrium concentration in the last row. In the first row, the problem gives us the ATP concentration in healthy tissue 8.435 x 10^-3 M, and we put zero for ADP and Pi for reasons mentioned above. H2O is not included since it is a liquid. At this point, we know that the ATP concentration will decrease by an amount x, and since the the coefficients are all one (all molar ratios are 1:1), we know that ADP and Pi both will increase by an x amount. Therefore, in the change, ATP has a change of -x and the other products +x. The problem also gives us the equilibrium molarity of ATP (ATP in dead tissue) 7.214 x 10^-11 M. The equilibrium concentrations of the products are unknown, so doing initial+change, their equilibrium concentrations are both x. At this point we can solve for x: 8.435 x 10^-3 - x = 7.214 x 10^-11 so x = 8.435 x 10^-3. Therefore, the products have an equilibrium concentration of 8.435 x 10^-3M and ATP has one of 7.214 x 10^-11M.

[Noah Fox 1E]

In the problem he gave us the given reaction of ATP hydrolysis of ATP + H2O --> ADP + P(inorganic)
The initial concentrations were based on real numbers from both healthy and dead tissues. The equilibrium point isn't necesarilly an equilibrium point more as the end point of the reaction as the last of the ATP is used in the dying muscular tissue

Reactants ATP ADP P(inorganic)

Initial 8.435*10^-3 0 0

Change -X +X +X

Equilibrium 7.24 * 10^-11 X X

These two numbers (8.435*10^-3 and 7.24 * 10^-11) are given in the problem as the concentration of ATP in living and dead muscle tissues respectively. Using this we can fill the rest of our ICE table in

because X is the difference between (8.435*10^-3 and 7.24 * 10^-11) we find X is 8.43499993*10^-11

Reactants ATP ADP P(inorganic)

Initial 8.435*10^-3 0 0

Change -8.43499993*10^-11 8.43499993*10^-11

Equilibrium 7.24 * 10^-11 8.43499993*10^-11

K=[ADP][P(inorganic)]/[ATP]

K= (8.43499993*10^-11)^2/7.24 * 10^-11

Solve for K and you find the equilibrium concentration of ATP as muscle tissues transition from a state of life to a state of death.