Question

Now for some calculations. ATP formation is a nonspontaneous process. It requires energy input. Under standard state conditions the delta Go’ is ~30 kJ/mol. If the best available measures in a particular tissue indicate cellular levels of ATP= 2.5 mM, ADP = 0.6 mM and Pi =0.2 mM, what value will you report as the delta G naught’ for ATP synthesis? Show all work

Answer 1

Solution :-

ADP + PI --- > ATP

Using the given concentrations we can find the reaction quotient (Q)

Q= [product]/ [reactant]

Q= [ATP]/[ADP][PI]

Q= [2.5 mM] / [0.6 mM][0.2 mM]

Q= 20.83

Now using the Q we can find the delta Go

Delta Go = - RT* ln Q

               = - 8.314 J per mol K * 298 K * ln [20.83]

          = -7523 J

Converting joules to kJ

-7523 J * 1 kJ / 1000 J = -7.523 kJ

Therefore the value of the delta Go = -7.523 kJ/mol