Question

The equilibrium constant K for the reaction fructose-1,6-diphosphate, is 8.9 x 10-5 Mat 25°C and we can assume the behavior to be ideal. glyceraldehyde-3-phosphate + dihydroxyacetone phosphate a. Calculate AGº for the process (standard state: 1M) b. Suppose that we have a mixture that is initially 0.01M in fructose-1,6-diphosphate and 105 M in both glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. What is AG? in which direction will the reaction occur?

Answer 1

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)