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
Since a percentage is out of 100, do the % / 100
Divide the percent by 100
Answer:
1.09 × 10⁻⁷ m
UV region
Explanation:
Step 1: Given and required data
Energy of the photon of light (E): 1.83 × 10⁻¹⁸ J
Planck's constant (h): 6.63 × 10⁻³⁴ J.s
Speed of light (c): 3.00 × 10⁸ m/s
Step 2: Calculate the wavelength (λ) of this photon of light
We will use the Planck-Einstein's relation.
E = h × c/λ
λ = h × c/E
λ = 6.63 × 10⁻³⁴ J.s × (3.00 × 10⁸ m/s)/1.83 × 10⁻¹⁸ J = 1.09 × 10⁻⁷ m
This wavelenght falls in the UV region of the electromagnetic spectrum.
