gaussian基本概念和用法(4)

D19=-111.54855725 D20=-163.81739212 D21=-177.01784202 D22=116.24854025 D23=84.73759842 D24=117.02628819 D25=-54.44419854 D26=120.14740904 D27=-120.73901029 D28=-120.36008831 D29=64.32343577 D30=119.79274556 D31=-121.08069298 D32=-179.41915959 D33=0.20082332 D34=-179.92188311 D35=-0.19458212

优化正常结束，频率计算时出错. .out文件末尾显示： Error #1 in AlGdDF.

Error termination via Lnk1e in d:\\chemistry\\gaussian98w\\l1002.exe. Job cpu time: 0 days 2 hours 34 minutes 4.0 seconds.

File lengths (MBytes): RWF= 129 Int= 0 D2E= 0 Chk= 13 Scr= 1

1。上面我贴的那个问题经过我将内存设定从256M改为192M后频率计算正常结束了。为什么你会想到把内存变小？ 2。你试试不设定内存看看。我觉得你的分子较大，算频率需要内存应该很多，所以不如不设。

输入文件： %chk=t_d %mem=250mb

# rb3lyp/gen opt=(calcfc,ts,noeigentest) freq scf=maxcyc=500 pseudo=read

C60O C2v Epoxide PM3 Optimization 0,1

74 0 0.248370 0.014156 0.033227 6 0 0.403673 -0.936350 1.733392 6 0 2.200917 0.171752 -0.336405 6 0 -0.270860 2.061381 -0.203209 6 0 -0.180546 -1.298567 -1.594701 1 0 0.913675 -1.906338 1.762663 1 0 -0.056469 -0.642812 2.681098 1 0 2.947154 -0.605352 -0.128354 1 0 2.704061 1.060569 -0.742512

11

1 0 -0.333335 2.571891 0.768423 1 0 -1.257830 2.123101 -0.686918 1 0 0.480105 2.561522 -0.830148 1 0 0.004133 -0.784037 -2.549068 1 0 -1.248033 -1.562489 -1.535515 1 0 0.435959 -2.206120 -1.587137 8 0 -2.716252 -0.092883 -0.267346 8 0 -1.769735 -0.112964 0.616623 C H O 0 6-31+G(d) **** W 0 sddall **** W 0 sddall

--Link1-- %NoSave %Chk=t_d %mem=250mb

#T b3lyp/gen irc=(rcfc,stepsize=20) geom=check guess=read pseudo=read

C60O Cs Open Isomer PM3 Optimization 0,1 C H O 0 6-31+G(d,p] **** W 0 sddall **** W 0 sddall

错误信息：过渡态已经找到，在计算IRC中，

Rotational constants (GHZ): 2.3655799 1.4105816 1.3561379

Isotopes: W-184,C-12,C-12,C-12,C-12,H-1,H-1,H-1,H-1,H-1,H-1,H-1,H-1,H-1,H-1,O-16,O-16 QPERR --- A SYNTAX ERROR WAS DETECTED IN THE INPUT LINE. 6-31+G(d,p]

12

Last state=\

TCursr= 1032 LCursr= 10

Error reading general basis specification. Error termination via Lnk1e in F:\\G98W\\l301.exe. Job cpu time: 0 days 0 hours 0 minutes 4.0 seconds.

File lengths (MBytes): RWF= 16 Int= 0 D2E= 0 Chk= 12 Scr= 1 请帮忙！！ 基组错误：6-31+G(d,p]，改成6-31+G(d,p)

个人感觉，做计算还是linux下比较好。windows下的计算我只做小的test，所以没有大体系的经验。不过看别人的帖子，好像主要是内存和硬盘文件分割的问题，你要是能试出解决办法就好了，这样也为大家总结了经验。 输入文件为：%chk=qwq %mem=200mb

# b3lyp/sto-3g opt scf(maxcyc=200)

calculate 0 2

Cu 0-0.6710.9510.000 O -1-0.6712.7810.000 O -1-0.671-1.0090.000 O -1-0.6711.1391.951 O -1-0.6710.763-1.951 C 01.2690.9510.000 C 02.039-0.3810.064 C 01.269-1.7130.128 H 01.2691.0021.069 H 01.8041.876-0.044 H 01.2690.899-1.069 H 02.671-0.3400.926 H 02.671-0.423-0.798 H 02.302-1.9900.141 H 00.270-2.0970.146 H 01.325-2.7810.179 O -1-2.6710.9510.000 输出文件为：

Standard orientation:

--------------------------------------------------------------------- Center Atomic Atomic Coordinates (Angstroms) Number Number Type X Y Z --------------------------------------------------------------------- 1 29 0 0.589550 -0.004850 0.014248 2 8 0 1.785028 1.345274 0.325533 3 8 0 -0.690853 -1.450884 -0.319149 4 8 0 0.702649 0.580333 -1.852973 …… 此处隐藏：1048字，全部文档内容请下载后查看。喜欢就下载吧 ……