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∆G°overall = ∆G1° + ∆G2°
-16.7 kJ/mol = ∆G1° + (-31.0 kJ/mol)
∆G1° = (-16.7 + 31.0)kJ/mol
= +13.3 kJ/mol
If ∆G = 0,
Then ∆G° = -RTln(Keq)
Here, Keq = [glucose-6-phosphate]/[glucose][Pi]
T = 37°C = 37 + 273 = 310 K
R = 8.314 J/K mol
∆G° = +13.3 kJ/mol = +13300 J/mol
+13300 J/mol = -(8.314 J/K mol)(310 K)ln(Keq)
ln(Keq) = +13300/(-2577.34) = -5.160
Keq = e-5.160 = 5.74 x 10-3
If only reaction 1 occurs, assuming reaction of 'x' mol each of glucose and phosphate, ICE table will be:
|Equilibrium (M)||1 - x||1 - x||x|
Keq = (x)/(1-x)(1-x) = 5.74 x 10-3
x/(1-x)2 = 0.00574
x = 0.00574 - 0.01148x - 0.00574x2
0.00574x2 + 1.01148x + 0.00574 = 0
On solving, x = 0.00567 M
So, the concentrations are:
[Glucose]= [phosphate] = 1-x
= 1- 0.00567 M = 0.994 M
[Glucose-6-phosphate] = x = 0.00567 M
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