What is the delta G for alcohol dehydrogenase?
152826
Delta G refers to the change in free energy.\n\nWhen a reaction takes place there is a change in the free energy, like dehydrogenation of alcohols. The delta G value depends on reactants and products, i.e. delta G value of dehydrogenation of different alcohols will be different.\n\nEnzymes are biological catalysts. They act by lowering the energy of activation(quite a different parameter than delta G) of a reaction. \n\nEnzyme itself can't have delta G value.
199900
The easiest way is to simply find literature listing delta H for the reaction.  Otherwise you have to do one of 2 things.  \n\n1. Find the enthalpy of formation for the product of the reaction which should be what you're looking for.\n\n2. Use the equation G=H-TS, plugging in the Gibbs free energy, temperature, and entropy values for the reaction,then solving for H, the enthalpy
27991
Ethyl Glucuronide is a metabolite of alcohol. It is a byproduct of alcohol metabolism.
798769
1. Glycolysis\n2. lactate\n3. 8\n3. fermentation\n6. FADH2 and NADH
801751
Well, H202 would be the substrate....and I forget what enzyme degrades H2O2........its in the peroxisome i think...
774935
http://wine1.sb.fsu.edu/krebs/krebs.htm
750255
I think you mean LDH (lactate dehydrogenase)\n\nHere is a link that explains it.\n\nhttp://en.wikipedia.org/wiki/Lactate_dehydrogenase
816266
At constant temperature and pressure, a reaction will occur spontaneously if the Gibbs free energy change of the reaction is less than zero.\n\nThe Gibbs free energy change, delta-G, is given by:\n\ndelta-G = delta-H - T*delta-S\n\nwhere T is the absolute temperature, \n\ndelta-H is the enthalpy change for the reaction at the temperature and pressure of interest, and \n\ndelta-S is the entropy change for the reaction at the temperature and pressure of interest.\n\nWe are given the enthalpy and entropy changes for the reaction of interest in this question.  I will assume that the enthalpy change and entropy change for this reaction are independent of temperature. Note also that the values of both of these thermodynamic parameters are given in "per mole" units in this question.  I will assume these values are normalized to one mole of one of the reactants, and not one mole of the product.\n\nPlugging in the enthalpy and entropy values, and setting delta-G to be less than zero:\n\n0 > delta-G = 180500 J/mol - T*24.8 J/(mol*K)\n\nT  > (180500 J/mol)/(24.8 J/(mol*K))\n\nT > 7278 K\n\nThis is an extremely high temperature, and the assumption that we made above that the enthalpy and entropy changes were independent of temperature is probably not very good given the large temperature extrapolation being made here.
400981
first, you find out the equation of Keq and delta G. I think it should be Keq = e^ [-deltaG/(RT)].\nThis equation and constant R can be found in any general chemistry textbook. And T is usu. 25+273=298K\nthen you can find out deltaG for the equation.\n\nwe also have deltaG= [- delta G for Ag(NH3)2^+1(aq) + (-32.0)] - [-26.2 - 3.8]\n\nand there comes the answer!!! ;)
926583
Fermentation:\nC6H12O6 → 2C2H5OH + 2CO2 \n\nCellular Respiration:\nC6H12O6 + 6O2 → 6CO2 + 6H2O + Energy released (2830 kJ mol−1)\nmore specifically:\n-TCA Cycle\n--Isocitrate Dehydrogenase\n--alpha-ketogluterate Dehydrogenase
426332
Alcohol alcohol alcohol.
516682
I found this at the Ohio division of the EPA:\nhttp://www.epa.state.oh.us/ddagw/pdu/gw_map_aqtype.html
723847
I am assuming that the values of H° and G° are the standard enthalpy and Gibbs free energy changes for the formation of phsogene from CO(g) and Cl2(g).\n\nFor part (a), you should already know that:\n\ndelta-G = delta-H - T*delta-S\n\nSo, for the values you are given for 298 K,\n\n-206 kJ/mol = -220 kJ/mol - 298 K * delta-S°\n-46.98 J/(mol*K) = delta-S°\n\nFor part (b), we now have (assuming delta-H and delta-S are independent of temperature) that \n\ndelta-G_601K = -220kJ/mol - 601 K * (-4.698*10^-02 kJ/mol*K)\ndelta-G_601K = -191.8 kJ/mol
928379
(Triangle) or (Delta) H for this process will equal:\n\n60g * Cp of water (4.1855 J/(C*g) * (Delta T)\n\n60*4.1855*14* = 3515.82 J\n\nThe change in enthalpy or (triangle H) is 3515.82J
473544
I think you meant to ask about glucose 6 phosphate dehydrogenase.\n\nSee this site; it is easy to read:\n\nhttp://www.bbc.co.uk/health/conditions/g6pdd1.shtml
657704
gamma, delta, epsilon, etc.
624609
Alcohols would be the product of the reduction of aldehydes or ketones.  To get isoproanol, you would have to reduce acetone with either lithium aluminum hydride or sodium borohydride.\n\n(CH3)2C=O  --->  (CH3)2CH-OH