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 the answer might be a little late but it would be <span>Simple squamous</span>
61.24 is the molar mass of a gas which has a density of 0.00249 g/mL at 20.0 degrees celcius and 744.0 mm Hg.
Explanation:
given that:
density = 0.00249 g/ml () or 2.49 grams/litre
P = 744 mm Hg OR 0.978 atm
T = 20 Degrees or 293.15 Kelvin
R = 0.08206 Litre atm/mole K
molar mass =?
Formula used/
PV = nRT equation 1
here n is number of moles:
n =
putting the value of n and value of density in the equation 1:
PV = x RT
molar mass = x
= density x
=
= 61.24 is the molar mass of the gas.
A
Hope this helped thanks u