Question

Please help with this question about Gibb's free energy. The answer is: -47.6 kJ/mol. Please show steps on how to find this answer!

1) In a rat hepatocyte, the concentrations of ATP, ADP and Pi are 3.4 mM, 1.3 mM and 4.8 mM, respectively. ATP + H2O → ADP + Pi + H+ a) Calculate the actual Gibb's free energy change for hydrolysis of ATP in this cell. ([H2O) = 1M; [H+] = 1M; R = 8.315 JK-mol-1; T = 298K) (4 points) Table 10.2 Standard Gibbs free energies of hydrolysis for ATP, AMP, and pyrophosphate AG°'hydrolysis Reactants and products (kJ moll) -32 ATP + H2O → ADP + P + H -45 -13 ATP + H2O → AMP + PP: + H AMP + H2O → Adenosine + P + H PP: + H202P P:(inorganic phosphate) = HPO4 PP (pyrophosphate) - HP20,0 -29

Answer 1

ATP + H2O $\rightarrow$ ADP + Pi + H+

Given : ∆G° = -32 KJ/mol

∆G = ∆G° + RT lnQ ----(1)

Reaction quotient (Q) = product/ reactant

= [ADP] [Pi] [H+]/ [ATP] [H2O]

Convert all mM into M

Example = [ATP] = 3.4 mM = 3.4 x 10-3 M

Putting all values given :

Q = ( 1.3 x 10-3) (4.8 x 10-3) (1) /(3.4 x 10-3) (1)

= 1.83 x 10-3

Using (1)

∆G = (-32 KJ/mol) + (0.008315 KJ/K.mol) (298 K) ln (1.83 x 10-3)

So, ∆G = -47.6 KJ/mol