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[求助]
過渡態(tài)
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用qst2計算過渡態(tài)時反應(yīng)物有兩個,輸入文件中反應(yīng)物的坐標(biāo)不知道應(yīng)該怎么處理,這種情況是不是需要插入虛原子?我計算的是NH3+H2O=NH3.H2O的過渡態(tài),以前沒接觸過過渡態(tài),輸入文件這里都耗費了好久了,請各位大俠出手相助,將不勝感激。 [ Last edited by B612 on 2012-12-5 at 11:02 ] |
催化與分子模擬 |
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金蟲 (正式寫手)

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嘿嘿..謝謝,我發(fā)現(xiàn)是202錯誤,結(jié)構(gòu)不合理,我一直認(rèn)為是離得太遠了,這是我的輸入文件,麻煩你給我看看吧 #p rb3lyp/6-311G**opt=qst2 test Gaussian: reactant NH3 +H2O -> NH3.H2O ts with constrained transition vector 0 1 N H 1 1.01361086 H 1 1.01326522 2 107.35196282 H 1 1.01316406 3 107.52252921 2 -115.23991561 0 O 1 2.93782288 4 120.25735978 2 -114.44004352 0 H 5 0.97655421 1 6.08698782 4 -40.51648096 0 H 5 0.96051900 1 111.77963430 4 -40.22358808 0 product NH3 +H2O -> NH3.H2O ts with constrained transition vector 0 1 O N 1 10.36481339 H 1 0.96187209 2 90.11621611 H 1 0.96188189 2 164.82057944 3 -179.42505302 0 H 2 1.01315154 1 95.78919154 3 18.42061254 0 H 2 1.01313153 1 67.04873005 3 -87.88685457 0 H 2 1.01315170 1 47.48992514 3 126.56007817 0 |
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不好意思,剛剛我把輸入文件給錯了,是這個 #p rb3lyp/6-311G** opt=qst2 test Gaussian: reactant NH3 +H2O -> NH3.H2O ts with constrained transition vector 0 1 N 0.00000000 0.00000000 0.00000000 H 0.00000000 0.00000000 1.01361086 H 0.96715224 0.00000000 -0.30219687 H -0.41402999 -0.87391355 -0.30225104 O -1.34873353 2.54903428 -0.56048674 H -0.92828188 1.67232201 -0.46963449 H -1.73459256 2.56222882 -1.43999539 product NH3 +H2O -> NH3.H2O ts with constrained transition vector 0 1 N 0.00000000 0.00000000 10.36481339 H 0.95633752 0.31851332 10.46700859 H 0.03439987 -0.93229441 9.96974468 H -0.44487470 0.59989710 9.68020669 O 0.00000000 0.00000000 0.00000000 H 0.96187011 0.00000000 -0.00195102 H -0.25184892 0.00252732 -0.92832241 |
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