Question

Please provide answers and explanation for the below problem, thanks.

A key step in the metabolism of glucose for energy is the isomerization of glucose-6-phosphate (G6P) to fructose-6 phosphate (F6P) as shown below. G6P F6P At 298 K, the equilibrium constant for the isomerization is 0.510. (a) Calculate ΔG° at 298 K. (b) Calculate ΔG When Q, the [F6P]/[G6P] ratio, equals 20.0. (c) Calculate ΔG When Q = 0.700. (d) Calculate Q in the cell iAG =-2.50 kJ/mol. kJ/mol kJ/mol kJ/mol

Answer 1

$2AgNO_{3} + K_{2}CrO_{4} \rightarrow Ag_{2}CrO_{4} + 2KNO_{3}$

2 moles            1 mole                1 mole             2 moles

Number of moles of silver nitrate = wt/MW = 3.14 X 10^-5 / 170 = 2 X 10^-7 moles

Concentration of Silver nitrate = number of moles / volume = 2 X 10^-7 / 15 X 10^-3 = 1.33 X 10^-5 M

Concentration of CrO₄^2- = 6 X 10^-5 M

Q_SP = [Ag⁺]² [CrO₄^2- ] = (1.33 X 10^-5 )² (6 X 10^-5)

                                   = 10.666 X 10^-15

Number of moles of Silver nitrate present are 2 X 10^-7

Number of moles of potassium chromate are 6 X 10^-5 mol/L (15X 10^-3 L) = 9 X 10^-7

Precipitate will not be formed because there is no sufficient amount of silver nitrate for the reaction to occur

########    Calculating K_SP

The amount of silver nitrate needeed = 18 X 10^-7 moles

Molarity of this substance = n/V = 18 X 10^-7 / 15 X 10^-3 = 1.2 X 10^-4 M

Molarity of potassium chromate = 6 X 10^-5 M

K_SP = [Ag⁺]² [CrO₄^2-] = (1.2 X 10^-4)² (6 X 10^-5)

                                 = 1.44 X 6 X 10^-13

                                            = 8.64 X 10^-13 = 0.864 X 10^-12

K_SP = 0.864e-12

Q_SP = 10.667e-15