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Consider the transport of a potassium ion from the blood (where its concentration is about 4.5...

Consider the transport of a potassium ion from the blood (where its concentration is about 4.5 mM) into a Red Blood Cell that contains 140 mM K+ at a temperature of 298 K. The transmembrane potential is about 58 mV, inside negative relative to outside.

What is the free-energy change for transport of potassium ions into a Red Blood Cell?

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Answer #1

Ans:- We know free energy Change G = - nFEo

Where n = valency of Potassium ion = 1, F= 96500 coloumb , Eo = 58 mV = 58 10-3 V

Therefore G = - 1 96500 58 10-3 = -5597 J/mole = -5.597 KJ/mole

Thus the transport of potassium from outside (blood) to inside (RBC's) is spontaneous reaction with G of - 5.597 KJ/mole.

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