Answer:
The concentration of fructose-6-phosphate F6P ≅ 1.35 mM
Explanation:
Given that:
ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol
concentration of glucose-6-phosphate at equilibrium = 2.65 mM
Assuming temperature = 25.0°C
=( 25 + 273)K
= 298 K
We are to find the concentration of fructose-6-phosphate
Using the relation;
ΔG' = -RT In K_c
where;
R = 8.314 J/K/mol
1670 = - (8.314 × 298 ) In K_c
1670 = -2477.572 × In K_c
1670/ 2477.572 = In K_c
0.67 = In K_c
0.511
Now using the equilibrium constant
F6P = 0.511 × 2.65
F6P = 1.35415
F6P ≅ 1.35 mM
In balanced equation there are same number of atoms in each element on both sides of the equation. unbalanced equation is when there are different number of atoms in each element on the both sides
Because it throws the earth off balance and if it does it often enough then it will soon add up.
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