What is the intracellular glucose concentration if the ΔG for the following reaction is -20.1 kJ/mol...

What is the intracellular glucose concentration if the ΔG for the following reaction is -20.1 kJ/mol at 37°C and concentrations for glucose-6-phosphate and phosphate are both 1 mM?

glucose-6-phosphate --------> glucose + Pi ΔG° = -13.8 kJ/mol

A) 1.9 M

B) 87 M

C) 1.9 mM

D) 27 mM

E) 87 mM

The Answer is E. I just need help understanding why it is E. please show all your work. I really need to understand the entire process. Please don't skip over small steps.

Solutions

Expert Solution

ΔG = ΔG⁰ + RTlnK

ΔG = -20.1KJ/mole = -20100J/mole

ΔG⁰ = -13.8KJ/mole = -13800J/mole

R = 8.314J/mole-K

T = 37+273 = 310K

ΔG = ΔG⁰ + RTlnK

-20100 = -13800 + 8.314*310*2.303logK

-20100 = -13800+5935.61logK

-20100+13800 = 5935.61logK

-6300 = 5935.61logK

logK = -6300/5935.61

logK = -1.06

K = 10^-1.06 = 0.087

glucose-6-phosphate --------> glucose + Pi

[glucose-6-phosphate] = 1mM = 0.001M

[pi] = 1mM = 0.001M

K = [glucose][Pi]/[glucose-6-phosphate]

0.087 = [glucose]*0.001/0.001

[glucose] = 0.087M = 87mM >>>>>answer

queen_honey_blossom answered 1 year ago

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