We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.
<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.
glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate
In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.
ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol
Then, the equation to relate ΔG° to the equilibrium constant K is
ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62
Sorry but I don’t understand the question. Sorry I’m not any help
The building with incandescent light bulbs would have higher energy bills because less than 10% of the bulb is used for light while the rest is given off as heat. Fluorescent light bulbs use ¼ as much energy and provide the same amount of light.
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Answer:
The pressure equilibrium constant (Kp) = (P O₂)³/(P CO₂)²(P H₂O)⁴.
Explanation:
<em>2CO₂ (g) + 4H₂O (g) → 2CH₃OH (l) + 3O₂ (g).</em>
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The pressure equilibrium constant (Kp) = the product of the pressure of the products side components / the product of the pressure of the reactantss side components.
each one is raised to a power equal to its coefficient.
<em>∴ The pressure equilibrium constant (Kp) = (P O₂)³/(P CO₂)²(P H₂O)⁴.</em>
