From owner-chemistry@ccl.net Mon Feb 20 19:32:01 2006 From: "John Bushnell bushnell,+,chem.ucsb.edu" To: CCL Subject: CCL: G3B3 Atomization Energy Methane Message-Id: <-30945-060220192833-21829-2KJSg84hV53WOP19UJg3cw]|[server.ccl.net> X-Original-From: John Bushnell Content-Type: TEXT/PLAIN; charset=US-ASCII; format=flowed Date: Mon, 20 Feb 2006 15:19:16 -0800 (PST) MIME-Version: 1.0 Sent to CCL by: John Bushnell [bushnell**chem.ucsb.edu] I'm not at all familiar with the "G3B3" method, but right off hand I would think that this is very good agreement with experiment. You show an error of 13 kJ/mol at zero K. But the calculation involves the difference of some very large numbers. The difference of only 3 kJ/mol at 298 K seems fortuitously small in fact. If enthalpies of formation could be routinely calculated to this accuracy, we wouldn't have to do so many experiments. :-) Just my offhand impression... - John On Mon, 20 Feb 2006, Roger Kevin Robinson r.robinson .. imperial.ac.uk wrote: > Sent to CCL by: Roger Kevin Robinson [r.robinson,,imperial.ac.uk] > Hi, > > I've asked about this before but i still seem to be having some > trouble. Im just using Methane as an example. > > Using G3B3 methods. > > At 298K I get > > Name G3-Energy(G3B3) ZPE > > C -37.778738 > > H -0.499671 > > CH4 -40.455401 0.043410 > > Using this values to calculate Atomization Energy > > = (-37.778738 + 4* -0.499671) - -40.455401 - 0.043410= 0.634569 = > 1666.06091 kJ/mol > > this fits in well with an experimental value of 1663.3 > > to calculate Enthalpy of formation you need the atomization Energy at 0K > as far as im aware. > > Right at 0K > > > Name G3(0K) - (G3B3) ZPE > > C -37.780154 > > H -0.501087 > > CH4 -40.458277 0.043410 > > = ( -37.780154 +4 * -0.499671) - -40.455401 - 0.043410 = 0.630365 = > 1655.02331. > > But the experimental value is 1642.27 > > Does any one know where im making the mistake ? > > Thanks Roger >