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Fructose-1,6 diphosphate = glyceraldehyde - 3-phosphate + dihydroxy acetone phosphate 10 M T = 25°c or T: 298k+(273 +25) (a) Gibbs free energy change (AG) is given by: AG = -RT enke = – 8.314 J KA met x 298* en (8-9x105) = -8.314 I met x 298 × (-9.33) AGO = +23115.7 J/mell AG = 23.1 kJ/met. To find in which direction reaction will occur We have to find reaction quoteint 'Q. If & 7 kc Reaction proceed to the left. a <ke Reaction proceed to the right.
Q = kc System is at equilibrium. Q = [glyceraldehyde - 3-P] [dihydroxyacetone P] Fructose 1, 6 diphosphate] = 105m x 105 = 100 M 0.01 M Q = 1x108 M La <ke] AG = DG + RT eng = 23115. 7 5 +8.314 Imel K x 298K en (1x108) mol
AG = 2311507 Ime - 45638.6 Ilmar AG = -2252219 J/mel08 -22522.9 EXO SKF [AG = -22.5 kJ/mel. AG Co and a <ke, the reaction proceed espontaneously to the right (forward direction)
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