The value of ΔG°′ΔG°′ for the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) is +1.67 kJ/mol+1.67 kJ/mol . If the concentration of glucose-6-phosphate at equilibrium is 2.95 mM2.95 mM , what is the concentration of fructose-6-phosphate? Assume a temperature of 25.0°C25.0°C .

The concentration of fructose-6-phosphate is 1.50x10⁻³M

Explanation:

Gibbs Free energy, ΔG°, is defined as:

ΔG° = -RTln K

Where ΔG° = 1670J/mol

R is gas constant = 8.314J/molK

T is absolute temperature (25°C + 273.15 = 298.15K)

Thus, equilibrium constant, k, is:

1670J/mol = -8.314J/molK* 298.15K ln K

-0.6737 = ln K

0.5098 = K

K of the reaction:

G6P ⇄ F6P

is:

K = 0.5098 = [F6P] / [G6P]

As [G6P] is 2.95x10⁻³M:

0.5098 = [F6P] / [2.95x10⁻³M]

[F6P] = 1.50x10⁻³M

particle   p is an electron and particle q is a neutron

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer