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
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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
Answer:
During deposition, a gas changes directly into a solid
Explanation:
A. Theresa is going to be my girl this year, he promised himself as he left the gym full of students in his new fall clothes.
I don't recognize this problem, please make sure the input is complete.
Explanation:
Pressure of a gas is the combined force with which the molecules bombard a unit area of the wall of the container.
1 atm = 760mmHg
= 760torr
= 101325Pa
= 101325Nm⁻²
1mmHg = 1torr
1Pa = 1Nm⁻²
101.325 kPa and 101,325 Pa are the same
1000Pa = 1kPa
101,325 Pa and 1 atm
1atm and 101.325 kPa
