乙烷-Hyperchem程序应用.doc
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1、Hyperchem程序应用-乙烷1、 画乙烷分子模型:build-default element加氢:模型化:2、用半经验方法CNDO进行优化3、选用从头算计算方法进行单点计算:4、显示键长:5、显示原子电荷:6、显示键角、二面角:7、分子性质:8、乙烷分子2D、3D静电势图:9、等值面图:10、乙烷分子总电荷密度图(2D、3D):11、分子轨道图-最高占据轨道2D、3D图:12、分子轨道图-最低空轨道2D、3D图:13、分子结构模型表示:14、计算输出结果:HyperChem log start - Sun Dec 12 13:20:41 2010.Geometry optimization
2、, SemiEmpirical, molecule = (untitled).CNDOFletcherReeves optimizerConvergence limit = 0.0001000 Iteration limit = 50Accelerate convergence = YESOptimization algorithm = Fletcher-ReevesCriterion of RMS gradient = 0.1000 kcal/(A mol) Maximum cycles = 120RHF Calculation:Singlet state calculationNumber
3、 of electrons = 14Number of Double Occupied Levels = 7Charge on the System = 0Total Orbitals = 14Starting CNDO calculation with 14 orbitalsE=0.0000 Grad=0.000 Conv=NO(0 cycles 0 points) Iter=1 Diff=4413.54714E=0.0000 Grad=0.000 Conv=NO(0 cycles 0 points) Iter=2 Diff=2.12522E=0.0000 Grad=0.000 Conv=N
4、O(0 cycles 0 points) Iter=3 Diff=0.13330E=0.0000 Grad=0.000 Conv=NO(0 cycles 0 points) Iter=4 Diff=0.00917E=0.0000 Grad=0.000 Conv=NO(0 cycles 0 points) Iter=5 Diff=0.00002E=-1659.8192 Grad=64.946 Conv=NO(0 cycles 1 points) Iter=1 Diff=27.89455E=-1659.8192 Grad=64.946 Conv=NO(0 cycles 1 points) Iter
5、=2 Diff=2.23060E=-1659.8192 Grad=64.946 Conv=NO(0 cycles 1 points) Iter=3 Diff=0.18582E=-1659.8192 Grad=64.946 Conv=NO(0 cycles 1 points) Iter=4 Diff=0.01805E=-1659.8192 Grad=64.946 Conv=NO(0 cycles 1 points) Iter=5 Diff=0.00010E=-1659.8192 Grad=64.946 Conv=NO(0 cycles 1 points) Iter=6 Diff=0.00000E
6、=-1634.7422 Grad=146.122 Conv=NO(0 cycles 2 points) Iter=1 Diff=11.78436E=-1634.7422 Grad=146.122 Conv=NO(0 cycles 2 points) Iter=2 Diff=0.88985E=-1634.7422 Grad=146.122 Conv=NO(0 cycles 2 points) Iter=3 Diff=0.06927E=-1634.7422 Grad=146.122 Conv=NO(0 cycles 2 points) Iter=4 Diff=0.00631E=-1634.7422
7、 Grad=146.122 Conv=NO(0 cycles 2 points) Iter=5 Diff=0.00005E=-1671.0494 Grad=3.886 Conv=NO(1 cycles 3 points) Iter=1 Diff=0.00236E=-1671.0494 Grad=3.886 Conv=NO(1 cycles 3 points) Iter=2 Diff=0.00018E=-1671.0494 Grad=3.886 Conv=NO(1 cycles 3 points) Iter=3 Diff=0.00001E=-1671.1287 Grad=2.282 Conv=N
8、O(1 cycles 4 points) Iter=1 Diff=0.00235E=-1671.1287 Grad=2.282 Conv=NO(1 cycles 4 points) Iter=2 Diff=0.00018E=-1671.1287 Grad=2.282 Conv=NO(1 cycles 4 points) Iter=3 Diff=0.00001E=-1671.1643 Grad=1.471 Conv=NO(1 cycles 5 points) Iter=1 Diff=0.00931E=-1671.1643 Grad=1.471 Conv=NO(1 cycles 5 points)
9、 Iter=2 Diff=0.00070E=-1671.1643 Grad=1.471 Conv=NO(1 cycles 5 points) Iter=3 Diff=0.00006E=-1671.1063 Grad=3.987 Conv=NO(1 cycles 6 points) Iter=1 Diff=0.00655E=-1671.1063 Grad=3.987 Conv=NO(1 cycles 6 points) Iter=2 Diff=0.00049E=-1671.1063 Grad=3.987 Conv=NO(1 cycles 6 points) Iter=3 Diff=0.00004
10、E=-1671.1665 Grad=1.618 Conv=NO(2 cycles 7 points) Iter=1 Diff=0.03525E=-1671.1665 Grad=1.618 Conv=NO(2 cycles 7 points) Iter=2 Diff=0.00283E=-1671.1665 Grad=1.618 Conv=NO(2 cycles 7 points) Iter=3 Diff=0.00025E=-1671.1665 Grad=1.618 Conv=NO(2 cycles 7 points) Iter=4 Diff=0.00003E=-1671.1250 Grad=5.
11、375 Conv=NO(2 cycles 8 points) Iter=1 Diff=0.01699E=-1671.1250 Grad=5.375 Conv=NO(2 cycles 8 points) Iter=2 Diff=0.00136E=-1671.1250 Grad=5.375 Conv=NO(2 cycles 8 points) Iter=3 Diff=0.00012E=-1671.1250 Grad=5.375 Conv=NO(2 cycles 8 points) Iter=4 Diff=0.00001E=-1671.1765 Grad=1.170 Conv=NO(3 cycles
12、 9 points) Iter=1 Diff=0.00024E=-1671.1765 Grad=1.170 Conv=NO(3 cycles 9 points) Iter=2 Diff=0.00003E=-1671.1899 Grad=0.985 Conv=NO(3 cycles 10 points) Iter=1 Diff=0.00024E=-1671.1899 Grad=0.985 Conv=NO(3 cycles 10 points) Iter=2 Diff=0.00003E=-1671.2013 Grad=0.804 Conv=NO(3 cycles 11 points) Iter=1
13、 Diff=0.00096E=-1671.2013 Grad=0.804 Conv=NO(3 cycles 11 points) Iter=2 Diff=0.00013E=-1671.2013 Grad=0.804 Conv=NO(3 cycles 11 points) Iter=3 Diff=0.00002E=-1671.2177 Grad=0.462 Conv=NO(3 cycles 12 points) Iter=1 Diff=0.00381E=-1671.2177 Grad=0.462 Conv=NO(3 cycles 12 points) Iter=2 Diff=0.00053E=-
14、1671.2177 Grad=0.462 Conv=NO(3 cycles 12 points) Iter=3 Diff=0.00008E=-1671.2252 Grad=0.440 Conv=NO(3 cycles 13 points) Iter=1 Diff=0.00028E=-1671.2252 Grad=0.440 Conv=NO(3 cycles 13 points) Iter=2 Diff=0.00004E=-1671.2266 Grad=0.298 Conv=NO(4 cycles 14 points) Iter=1 Diff=0.03095E=-1671.2266 Grad=0
15、.298 Conv=NO(4 cycles 14 points) Iter=2 Diff=0.00240E=-1671.2266 Grad=0.298 Conv=NO(4 cycles 14 points) Iter=3 Diff=0.00020E=-1671.2266 Grad=0.298 Conv=NO(4 cycles 14 points) Iter=4 Diff=0.00002E=-1671.1543 Grad=5.510 Conv=NO(4 cycles 15 points) Iter=1 Diff=0.02754E=-1671.1543 Grad=5.510 Conv=NO(4 c
16、ycles 15 points) Iter=2 Diff=0.00214E=-1671.1543 Grad=5.510 Conv=NO(4 cycles 15 points) Iter=3 Diff=0.00018E=-1671.1543 Grad=5.510 Conv=NO(4 cycles 15 points) Iter=4 Diff=0.00002E=-1671.2268 Grad=0.117 Conv=NO(5 cycles 16 points) Iter=1 Diff=0.00000E=-1671.2268 Grad=0.058 Conv=NO(5 cycles 17 points)
17、 Iter=1 Diff=0.00000E=-1671.2268 Grad=0.005 Conv=NO(5 cycles 18 points) Iter=1 Diff=0.00000E=-1671.2268 Grad=0.120 Conv=NO(5 cycles 19 points) Iter=1 Diff=0.00000E=-1671.2268 Grad=0.005 Conv=YES(6 cycles 20 points) Iter=1 Diff=0.00000ENERGIES AND GRADIENTTotal Energy = -11813.3264338 (kcal/mol)Total
18、 Energy = -18.825358113 (a.u.)Binding Energy = -1671.2268529 (kcal/mol)Isolated Atomic Energy = -10142.0995808 (kcal/mol)Electronic Energy = -28506.1734829 (kcal/mol)Core-Core Interaction = 16692.8470491 (kcal/mol)Heat of Formation = -1016.8348529 (kcal/mol)Gradient = 0.0047345 (kcal/mol/Ang)MOLECUL
19、AR POINT GROUP D3DEIGENVALUES(eV)Symmetry: 1 A1G 1 A2U 1 EU 1 EU 2 A1G Eigenvalue: -40.735771 -29.080317 -23.689325 -23.689322 -18.800663Symmetry: 1 EG 1 EG 2 EU 2 EU 2 A2U Eigenvalue: -15.713766 -15.713758 7.124663 7.124665 7.752848Symmetry: 3 A1G 2 EG 2 EG 3 A2U Eigenvalue: 8.382067 10.037918 10.0
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