我們利用密度泛涵理論來計算在氧化石墨烯上吸附兩顆金屬鉑原子，並在兩個金屬原子上各別吸附兩個甲烷氣體，利用催化表面的氧化能力將其轉換成甲醇的過程。根據早期學術上的研究發現鉑金屬對於碳氫化合物的吸附能比起大部分之金屬較大，並且能加速碳氫鍵的斷裂，有利於將甲烷分子氧化成甲醇。所以在計算中，兩顆鉑原子分別吸附兩個甲烷氣體之後(其中鉑原子已和氧化石墨烯有了良好鍵結)，我們預測了多條不同氧化路徑，主要分為兩大部分。第一部分總括有三個步驟 (1) 斷其中一顆甲烷的碳氫鍵，使氫原子與氧化石墨烯上的氧原子鍵結形成氫氧根，這時所需之能量為0.34 eV。(2) 斷另一個甲烷上的碳氫鍵，使其吸附在鉑原子上形成甲基，這時所需0.51 eV。(3) 最後我們將甲基和氫氧根結合產生甲醇，此時所需能量為0.2 eV。最後甲醇脫附就完成前半部反應。而後半部反應則是再通入氧氣分子，總括有兩個步驟(1)透過O-O鍵的斷裂和氫氧基的形成，須越過0.34eV 的能障。(2) 將甲基和氫氧根結合產生第二個甲醇，此時所需能量為1.25eV，並同時填補催化表面的氧原子，使其恢復一開始的結構。
關鍵字: 鉑原子、氧化石墨烯、甲烷、理論計算

Based on density functional theory (DFT) calculation, the synthesis of methanol from the methane oxidation has been investigated on the graphene oxide nanosheets supporting two platinum atoms (Pt2-graphene oxide). According to previous literature, the platinum atoms could increase the adsorption energies of hydrocarbon species, and facilitate C-H bond scission to form the methanol. In our calculation, two CH4 molecule could be adsorbed on each Pt atom of the Pt2-graphene oxide surface, and the calculated possible mechanism of the methane oxidation reaction includes two parts; before and after the oxygen molecule adding to the system. First section involves the following three sequential steps: (I) dehydrogenation of one methane to the surface oxide to form methyl-Pt(1) and hydroxyl (Ea = 0.34 eV), (II) dehydrogenation of the other methane to Pt atom to form methyl and H-Pt(2) (Ea = 0.51 eV), and then (III) the coupling of the hydroxyl and the methyl to produce the methanol (Ea = 0.20 eV); Then,methanol could be desorbed from Pt2-graphene oxide catalyst. The second section contains two successive steps after adding oxygen molecule: the initial step is the scission of O-O bond and formation of hydroxyl group with the H atom on Pt(2) atom (Ea = 0.34 eV), and then, the hydroxyl would couple with methyl-Pt(1) to produce the other methanol (Ea = 1.25 eV). At last, we could gain two methanol and return to the original graphene oxide .
Key point: Platinum atom 、Graphene-oxide 、methane 、 DFT

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