Answer:
decrease temperature of the oxygen
Explanation:
Answer:
= 19
ΔG° of the reaction forming glucose 6-phosphate = -7295.06 J
ΔG° of the reaction under cellular conditions = 10817.46 J
Explanation:
Glucose 1-phosphate ⇄ Glucose 6-phosphate
Given that: at equilibrium, 95% glucose 6-phospate is present, that implies that we 5% for glucose 1-phosphate
So, the equilibrium constant
can be calculated as:
![= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Bglucose-6-phosphate%5D%7D%7B%5Bglucose-1-%5Bphosphate%5D%7D)


= 19
The formula for calculating ΔG° is shown below as:
ΔG° = - RTinK
ΔG° = - (8.314 Jmol⁻¹ k⁻¹ × 298 k × 1n(19))
ΔG° = 7295.05957 J
ΔG°≅ - 7295.06 J
b)
Given that; the concentration for glucose 1-phosphate = 1.090 x 10⁻² M
the concentration of glucose 6-phosphate is 1.395 x 10⁻⁴ M
Equilibrium constant
can be calculated as:
![= \frac{[glucose-6-phosphate]}{[glucose-1-[phosphate]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Bglucose-6-phosphate%5D%7D%7B%5Bglucose-1-%5Bphosphate%5D%7D)

0.01279816514 M
0.0127 M
ΔG° = - RTinK
ΔG° = -(8.314*298*In(0.0127)
ΔG° = 10817.45913 J
ΔG° = 10817.46 J
To solve for the enthalpy of reaction, we apply the Hess's Law.
ΔHrxn = ∑(ν×Hf of products) - ∑(ν×Hf of reactants)
where
v is the stoichiometric coefficient determined from the balanced reaction
Hf is the standard heat of formation; these are empirical values:
*For CH₄: Hf = <span>−74.87 kJ/mol
*For O</span>₂: Hf = 0
*For CO₂: <span>-393.5 kJ/mol
*For H</span>₂O: <span>-241.82 kJ/mol
</span>ΔHrxn = [(2*-241.82 kJ/mol)+(1*-393.5 kJ/mol)] - [(1*−74.87 kJ/mol)+(2*0 kJ/mol)] =<em> -802.27 kJ/mol</em>