The equation for ATP hydrolysis is
ATP yields ADP + Pi delta G*= -30.5 kJ/mol

( H20)

(a) Calculate ?G for ATP hydrolysis to rank the following conditions from most favorable to least favorable. Assume a temperature of 37.0C. R = 8.315 J/(mol. K).
ATP Hydrolysis most favorable
ATP hydroysis least favorable
The choices are:

a. muscle: [ATP]= 8.1mM; [ADP]= 0.9mM [Pi]= 8.1mM
b. brain: [ATP]= 2.6mM; [ADP]= 0.7mM [Pi]= 2.7mM
c. liver: [ATP]= 3.4mM; [ADP]= 1.3mM [Pi]= 4.8mM

(b) Calculate ?G for ATP hydrolysis in muscle at 18 degree C. Use the muscle concentrations from part a.
Delta g = kJ/mol

The  most favorable is the brain tissue, followed by the muscle tissue while the least favorable is the liver tissue.

The equation of the reaction is; ATP > ADP + Pi ΔG°= -30.5 kJ/mol

For muscle tissue;

Q = [ADP] [Pi]/[ATP]

Q = [0.9 × 10^-3] [8.1 × 10^-3]/[[8.1 × 10^-3]

Q = 9.0 × 10^-4

ΔG =ΔG° + RTlnQ

Where  ΔG°= -30.5 kJ/mol or  -30.5  × 10^3 J/mol from the equation

ΔG = -30.5  kJ/mol + (8.315 × 310 × ln(9.0 × 10^-4)) kJ/mol

ΔG =-48.6  kJ/mol

For the brain tissue;

Q = [0.7 × 10^-3] [2.7 × 10^-3]/[2.6 × 10^-3]

Q = 7.3 × 10^-4

ΔG =  -30.5  kJ/mol + (8.315 × 310 × ln(7.3 × 10^-4)) kJ/mol

ΔG = -49.1  kJ/mol

For the liver tissue;

Q = [1.3 × 10^-3] [4.8 × 10^-3]/[3.4 × 10^-3]

Q = 1.8 × 10^-3

ΔG =  -30.5  kJ/mol + (8.315 × 310 × ln( 1.8 × 10^-3)) kJ/mol

ΔG = -46.3  kJ/mol

The most favorable is the brain tissue, followed by the muscle tissue while the least favorable is the liver tissue.

For the muscle tissue at 18°;

ΔG =  -30.5  kJ/mol + (8.315 × 291 × ln(9.0 × 10^-4)) kJ/mol

ΔG = -47.5 kJ/mol

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Here's what I get  

Explanation:

ATP ⟶ ADP + Pi; ΔG° = -30.5 kJ·mol⁻¹

(a) Rank the favourability of ATP hydrolysis

To get ΔG under physiological conditions, we must use the equation

ΔG = ΔG° + RTlnQ where

(i) In muscle tissue

(ii) In brain tissue

(iii) In liver tissue

(b) ΔG in muscle at 18 °C

T= (18.0 + 273.15) = 291.15 K

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