Question

Calculate the free energy of transport (Delta G) for the movement of sodium ions (Na+ ) from the extracellular space into the cell under the following conditions: [Na+ ]out = 140 mM; [Na+ ]in = 12.0 mM; membrane potential =−60.0 mV; T = 37.0C. Use R = 8.314 J/mol K; F = 96,500 J/V mol. 2. Calculate the free energy of transport (Delta G) for the movement of potassium ions (K+ ) from the extracellular space into the cell under the following conditions: [K+ ]out = 4.50 mM; [K+ ]in = 145 mM; membrane potential =−60.0 mV; T = 37.0C. Use R = 8.314 J/mol K; F = 96,500 J/V mol. 3. The sodium-potassium ATPase pumps 3 moles of Na+ out of the cell and 2 moles of K+ into the cell for every mole of ATP hydrolyzed to ADP and Pi . Using the values you calculated above, and assuming that the free energy of hydrolysis of ATP is – 50.5 kJ/mol, what is the net free energy change associated with the activity of the sodium-potassium ATPase? Is this process thermodynamically favorable under these conditions?

Answer 1

The movement of ions in the permeable membrane involves change in concentration and change in free energy.

Delta G will be negative if the concentration gradiant is less during the transport of ion (this means the energy is released). Likewise if the energy is consumed the delta G will be positive.

The amount of free energy is calculated by using a formula

Delta G = R x T X ln ([X]in/[X]out)

R = Universal gas constant 8.314 J/mol K
T = temperature in K
X = Concentration of ion

1)

1. Sodium ions concentration outside = 12 mM; Sodium ions inside = 140 mM.
2. Temperature 37 deg C = 273 + 37 = 310 K
3. R = 8.314 J/mol K

By applying above equation

Delta G = 8.314 X 310 X ln (12/140)
= 2577 X ln (0.085)
   = 2577 X (-2.465)
   = -6,352 J/mol
by converting into Kcal/mol we get
   = -1.51 Kcal/mol

2)

1. Potassium ions concentration outside = 4.5 mM; Potassium ions inside = 145 mM.
2. Temperature 37 deg C = 273 + 37 = 310 K
3. R = 8.314 J/mol K

By applying above equation

Delta G = 8.314 X 310 X ln (145/4.5)
= 2577 X ln (32)
   = 2577 X (3.46)
   = 8,931 J/mol
by converting into Kcal/mol we get
   = 2.13 Kcal/mol