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【求助】请教高斯计算中能量校正的问题
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初学高斯，想进行能量校正，请问我的操作是否正确？ 下面是H2O计算out文件的部分内容： Harmonic frequencies (cm**-1), IR intensities (KM/Mole), Raman scattering activities (A**4/AMU), depolarization ratios for plane and unpolarized incident light, reduced masses (AMU), force constants (mDyne/A), and normal coordinates: 1 2 3 A1 A1 B2 Frequencies -- 1613.2798 3881.7625 3990.2135 Red. masses -- 1.0826 1.0452 1.0823 Frc consts -- 1.6602 9.2791 10.1530 IR Inten -- 67.8281 12.7620 62.8282 Atom AN X Y Z X Y Z X Y Z 1 8 0.00 0.00 0.07 0.00 0.00 0.05 0.00 0.07 0.00 2 1 0.00 -0.43 -0.56 0.00 0.58 -0.40 0.00 -0.56 0.43 3 1 0.00 0.43 -0.56 0.00 -0.58 -0.40 0.00 -0.56 -0.43 ------------------- - Thermochemistry - ------------------- Temperature 298.150 Kelvin. Pressure 1.00000 Atm. Atom 1 has atomic number 8 and mass 15.99491 Atom 2 has atomic number 1 and mass 1.00783 Atom 3 has atomic number 1 and mass 1.00783 Molecular mass: 18.01056 amu. Principal axes and moments of inertia in atomic units: 1 2 3 EIGENVALUES -- 2.17514 4.15411 6.32926 X 0.00000 0.00000 1.00000 Y 1.00000 0.00000 0.00000 Z 0.00000 1.00000 0.00000 This molecule is an asymmetric top. Rotational symmetry number 2. Rotational temperatures (Kelvin) 39.81982 20.85013 13.68467 Rotational constants (GHZ): 829.71138 434.44674 285.14265 Zero-point vibrational energy 56734.4 (Joules/Mol) 13.55985 (Kcal/Mol) Vibrational temperatures: 2321.15 5584.98 5741.02 (Kelvin) Zero-point correction= 0.021609 (Hartree/Particle) Thermal correction to Energy= 0.024445 Thermal correction to Enthalpy= 0.025389 Thermal correction to Gibbs Free Energy= 0.003979 Sum of electronic and zero-point Energies= -76.413073 Sum of electronic and thermal Energies= -76.410237 Sum of electronic and thermal Enthalpies= -76.409293 Sum of electronic and thermal Free Energies= -76.430703 E (Thermal) CV S KCal/Mol Cal/Mol-Kelvin Cal/Mol-Kelvin Total 15.339 6.012 45.062 Electronic 0.000 0.000 0.000 Translational 0.889 2.981 34.608 Rotational 0.889 2.981 10.446 Vibrational 13.562 0.050 0.007 Q Log10(Q) Ln(Q) Total Bot 0.147901D-01 -1.830028 -4.213794 Total V=0 0.128648D+09 8.109403 18.672590 Vib (Bot) 0.115014D-09 -9.939250 -22.885968 Vib (V=0) 0.100042D+01 0.000181 0.000416 Electronic 0.100000D+01 0.000000 0.000000 Translational 0.300432D+07 6.477746 14.915562 Rotational 0.428032D+02 1.631476 3.756612 ***** Axes restored to original set ***** ------------------------------------------------------------------- Center Atomic Forces (Hartrees/Bohr) Number Number X Y Z ------------------------------------------------------------------- 1 8 -0.000000193 0.000000000 -0.000000136 2 1 0.000000105 0.000000000 0.000000056 3 1 0.000000088 0.000000000 0.000000080 ------------------------------------------------------------------- Cartesian Forces: Max 0.000000193 RMS 0.000000097 GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad Berny optimization. Internal Forces: Max 0.000000186 RMS 0.000000118 Search for a local minimum. Step number 1 out of a maximum of 2 All quantities printed in internal units (Hartrees-Bohrs-Radians) Second derivative matrix not updated -- analytic derivatives used. The second derivative matrix: R1 R2 A1 R1 0.55672 R2 -0.00588 0.55672 A1 0.02929 0.02929 0.15377 Eigenvalues --- 0.14950 0.55511 0.56260 Angle between quadratic step and forces= 90.00 degrees. Linear search not attempted -- first point. Iteration 1 RMS(Cart)= 0.00000067 RMS(Int)= 0.00000000 Variable Old X -DE/DX Delta X Delta X Delta X New X (Linear) (Quad) (Total) R1 1.80999 0.00000 0.00000 0.00000 0.00000 1.80999 R2 1.80999 0.00000 0.00000 0.00000 0.00000 1.80999 A1 1.83194 0.00000 0.00000 0.00000 0.00000 1.83193 Item Value Threshold Converged? Maximum Force 0.000000 0.000015 YES RMS Force 0.000000 0.000010 YES Maximum Displacement 0.000001 0.000060 YES RMS Displacement 0.000001 0.000040 YES Predicted change in Energy=-1.293968D-13 Optimization completed. -- Stationary point found. ---------------------------- ! Optimized Parameters ! ! (Angstroms and Degrees) ! -------------------------- -------------------------- ! Name Definition Value Derivative Info. ! -------------------------------------------------------------------------------- ! R1 R(1,2) 0.9578 -DE/DX = 0.0 ! ! R2 R(1,3) 0.9578 -DE/DX = 0.0 ! ! A1 A(2,1,3) 104.9622 -DE/DX = 0.0 ! -------------------------------------------------------------------------------- GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad 1|1|UNPC-UNK|Freq|RmPW1PW91|6-311+G(d,p)|H2O1|PCUSER|14-Sep-2009|0||#N GEOM=ALLCHECK GUESS=READ SCRF=CHECK GENCHK RMPW1PW91/6-311+G(D,P) FRE Q||Title Card Required||0,1|O,-0.0952734031,0.,-0.0673325842|H,-0.0573 541316,0.,0.8897195459|H,0.8195413565,0.,-0.3510588722||Version=x86-Wi n32-G03RevB.03|State=1-A1|HF=-76.4346818|RMSD=1.110e-010|RMSF=9.664e-0 08|Dipole=0.6952636,0.,0.4913637|DipoleDeriv=-0.430751,0.,0.0776326,0. ,-0.7723616,0.,0.0776326,0.,-0.4857332,0.2777636,0.,-0.0837002,0.,0.38 61808,0.,-0.038118,0.,0.1804785,0.1529874,0.,0.0060676,0.,0.3861808,0. ,-0.0395146,0.,0.3052547|Polar=7.0700986,0.,6.5393675,-0.681324,0.,7.5 526376|PG=C02V [C2(O1),SGV(H2)]|NImag=0||0.54534068,0.,-0.00012431,-0. 11480464,0.,0.62664959,-0.04445742,0.,-0.05313401,0.04643966,0.,0.0000 6215,0.,0.,-0.00009869,0.00630978,0.,-0.54153772,0.00406631,0.,0.55722 138,-0.50088326,0.,0.16793865,-0.00198225,0.,-0.01037609,0.50286551,0. ,0.00006215,0.,0.,0.00003654,0.,0.,-0.00009869,0.10849486,0.,-0.085111 87,0.04906770,0.,-0.01568366,-0.15756256,0.,0.10079554||0.00000019,0., 0.00000014,-0.00000011,0.,-0.00000006,-0.00000009,0.,-0.00000008|||@ 进行能量校正我按以下步骤操作，请问对吗？请高手指导！谢谢。 水分子的能量校正步骤： 1.计算结果中，读取能量：HF=-76.4346818 2.读取零点能：Zero-point correction= 0.021609 3.如果不考虑校正因子：校正后的能量E=-76.4346818+0.021609=-76.413073 当考虑ZPE校正因子时：校正后的能量E=-76.4346818+0.021609*0.9580=-76.413980 [ Last edited by fegg7502 on 2009-9-15 at 23:34 ] |
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3楼2009-09-16 08:03:02
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2楼2009-09-15 23:28:24