Homogeneous Partial Oxidation of Methane to Methanol-a Simulation by CHEMKIN-III

oxidation

Abstract A simulation of homogeneous partial oxidation of methane to methanol by CHEMKIN-III Gas-Phase Mechanism Interpreter was carried out. The content of different species as functions of time and the effect of pressure, temperature and CH 4 /O 2 ratio were specially discussed HCHO was found the initial product, but it is easy to be oxidized to CO and its selectivity decreased very quickly when the conversion of CH 4 or O 2 was increasing. When O 2 was completely consumed, the conversion of CH 4 And the selectivity of products reached a stable state. CO was predominate product while the amount of required production of methanol was, unexpectedly, very small. In order to improve methanol selectivity, it is necessary to control the oxidation of the intermediate CH 3O· In additional, high pressure and CH 4 /O 2 ratio were favorable to the production of methanol, which was consistent with the experimental results.
Keyword Methane , Methanol , Homogeneous oxidation , Simulation , CHEMKIN-III

Methane gas phase homogeneous selective oxidation to methanol- CHEMKIN-III simulation calculation

Zhang Qijian A * , Zhou Ying-chun A , Dehua b , Zhu Qiming b
A Liaoning Institute of Technology School of Materials and Chemical Engineering, Liaoning, Jinzhou, 121001, China; b Department of Chemistry, Tsinghua University, Beijing 100084, China)

Abstract The gas phase partial oxidation of methane was simulated by CHEMKIN-III Gas-Phase Mechanism Interpreter. The effects of reaction time, reaction pressure, alkoxygen ratio and reaction temperature on partial oxidation of methane were investigated. In the case of a complete reaction of O 2 , the product distribution reaches a steady state, but CO is the main product in the product, and the amount of the target product methanol is small. In order to increase the yield of methanol, it is necessary to control the reaction of the intermediate species CH 3 OO· to reduce it to methanol, instead of oxidizing to form HCHO to form CO . In addition, the high pressure, high alkoxy ratio favors the selectivity of the partial oxidation product methanol, and this conclusion is the same as the experimental results.
Keywords methane, methanol, gas phase oxidation, simulation calculation, CHEMKIN-III

Methane selective oxidation to methanol is the best way to convert and utilize natural gas. However, since the product methanol is more unstable than the reactant methane and is more easily activated on the catalyst, there has been no major breakthrough in the catalytic oxidation of methane to methanol. The study of gas phase homogeneous oxidation of methane to methanol has been reported from time to time with higher yields [1 ,
2 , 3] , so there is a view that “no catalyst is the best catalyst for this reaction”. Methane gas phase homogeneous oxidation to methanol reaction follows the mechanism of free radical chain reaction. A lot of research work has been done on this reaction, but different researchers often propose different mechanism models to match the results of the simulation calculation with their own experimental results. Or, to correct the relevant reactions in the mechanism model, it is hoped that the mechanism model will be in line with its own experimental facts [4, 5] . Since the product distribution is sensitive to the reaction conditions, especially the presence of oxides on the surface of the reactor wall, there are often large differences between the experimental results of different researchers. In the hundreds of primitive steps proposed so far, The identification of the main reaction pathways is different and further exploration is needed.
In order to more clearly understand the homogeneous oxidation reaction of methane gas phase, the effects of operable conditions (temperature, pressure, alkoxygen ratio, etc.) on the reaction are investigated to guide the study of homogeneous oxidation and provide for the development of catalytic oxidation. For reference, the CHEMKIN-III Gas-Phase Mechanism Interpreter was used to simulate the gas phase oxidation of methane.

  1. Calculation Methods
    CHEMKIN is a software package designed to solve the problem of chemical reaction dynamics. It provides a flexible and powerful tool to turn complex chemical dynamics into simple and clear mathematical problems. The mechanism model used in this paper involves 5 types of atoms ( O , H , C , N , Ar ), 48 species, of which CH 4 , × CH 3 , CH 3 O × , CH 3 OO × , CH 3 OOH , CH 3 OH , HCHO , × CH 2 OH , HCO × ,CO , CO 2 , C 2 H 6 , O 2 , × O , × OH , HO 2 × , H 2 O 2 , H 2 O , H × , H 2 and the like. And 273 elementary reactions, the main reactions are as follows:
    ( 1 ) The initiation of the reaction:
    CH 4 + M ® × CH 3 + H × + M * (R-1)
    where MRepresents inert molecules (such as CH , N 2 that do not participate in the reaction ) or reactor walls.
    CH 4 + O ® × CH 3 + HO 2 ×  (R-2)
    ( 2 ) Reaction progress
    × CH 3 + O ® CH 3 OO ×  (R-3)
    CH 3 + O ® HCHO + OH ×  (R-4)
    · CH 3 + O ® CH 3 O × + O ×  (R-5)
    CH3 OO × ® × CH 3 + O 2 (R-6) CH 3 OO × ® CH 3 O × + O × (R-7) CH 3 OO × ® HCHO + OH × (R-8) CH 3 OO × + CH ® CH 3 OOH + × CH 3 (R-9) CH 3 OOH ® CH 3 O × + OH × (R-10)CH 3 O × + CH ® CH 3 OH + × CH  (R-11)
    CH 3 O × + O ® HCHO + HO 2 ×  (R-12)
    CH 3 O × + M ® HCHO + H × + M  (R-13)
    CH 3 OH + O ® × CH 2 OH + HO 2 ×  (R-14)
    × CH 2 OH + O ® HCHO + HO2 ×  (R-15)
    × CH 2 OH + M ® HCHO + H × + M (R-16) HCHO + O ® HCO × + HO × (R-17) HCHO + HO × ® H 2 O 2 + HCO × (R-18) HCHO + × OH ® HCO × + H 2 O (R-19) HCO × + O ® CO + HO × (R-20) HCO + ×OH ® CO × + H 2 O  (R-21)
    HCO × + M ® CO + H × + M  (R-22)
    CO + O ® CO 2 + O ×  (R-23)
    CH 4 + HO 2 × ® × CH 3 + H 2 O 2 (R-24) H 2 O 2 + M ® 2 × OH + M (R-25) CH 4 + × OH ®× CH 3 + H 2 O  (R-26)
    CH 4 + × OH ® × CH 3 + H 2 O  (R-26)
    CH 4 + H × ® × CH 3 + H 2 (R-27) CH 4 + O ® × CH 3 + × OH (R-28) CH 3 OO × + H 2 O ® CH 3 OOH + HO 2

    ×  (R-29)
    CH 3 OO × + HO 2 × ® CH 3 OOH + O 2 (R-30) CH 3 OO × + H 2 ® CH 3 OOH + H × (R-31) CH 3 OO × + CH 3 O × ® CH 3 OOH + HCHO(R-32) CH 3 OO × + CH 3 OH ® CH 3 OOH + × CH 2 OH

    (R-33)
    CH 3 OO × + × CH 2 OH ® CH 3 OOH + HCHO  (R-34)
    CH 3 OO × + HCHO ® CH 3 OOH + CHO ×  (R-35)
    HO 2 × + CH 3 OH ® H 2 O 2 + × CH 2 OH  (R-36)
    HO 2 × + × CH 2 OH ® H 2 O2 + HCHO  (R-37)
    HO 2 × + HO 2 × ® H 2 O 2 + O 2 (R-38) HCHO + O ® HCO × + × OH (R-39) 2CH 3 OO × ® 2CH 3 O × + O 2 (R-40) CH 3 OO × + × CH ® 2CH 3 O × (R-41) CH 3 OO ×

    + H × ® CH 3 O × + × OH  (R-42)
    CH 3 OOH + H × ® CH 3 O × + H 2 O (R-43) CH 3 O × + H ® CH 3 OH + H × (R-44) CH 3 O × + HO × ® CH 3 OH +O 2 (R-45) CH 3 O × + H 2 O

    ® CH 3 OH +HO 2 ×  (R-46)
    CH 3 O × + HCHO ® CH 3 OH + HCO ×  (R-47)
    CH 3 O × + CH 3 O × ® CH 3 OH + HCHO  (R- 48)
    CH 3 O × + CH 3 OH ® CH 3 OH + × CH 2 OH (R-49) CH 3 O × + H × ®
    HCHO+H (R-51)
    CH 3 O × + × OH ® HCHO+H 2 O (R-52) CH 3 OH + × OH ® × CH 2 OH + H 2 O (R-53) CH 3 OH + H × ® × CH 2 OH + H 2 (R-54) CH 3 OH + × CH ® × CH 2 OH + CH 4 (R-55) × CH

    2 OH + × CH ® HCHO + CH 4 (R-56) HCHO + H × ® HCO × + H 2 (R-57) HCHO + × CH ® HCO × + CH 4 (R-58) HCHO + × CH 2 OH ® HCO × + CH 3 OH (R-59) HCO × + H × ® CO + H 2 (R-60) HCO × + × CH 3

    ® CO + CH 4 (R-61) HCO × + × OH ® CO + H 2 O (R-62) HCO × + O ® CO + × OH (R-63) CO + CH 3 O × ® CO 2 + CH × (R-64) CO + × OH ® CO 2 + H × (R-65) CO + HO × ® CO 2 + × OH (R-66) CO + CH

    3 OO × ® CO 2 + CH 3 O ×  (R-67)
    CO + O + M ® CO 2 + M  (R-68)
    HCO × + O ® CO 2 + H ×  (R-69)
    HCO × + HO 2 × ® CO 2 + H × + OH ×  (R-70)
    · CH 3 + × CH ® C 2 H 6 (R-71) HO ×
    + M ® O 2 + H × + M  (R-72)
    HO 2 × + O ® O 2 + × OH  (R-73)
    HO 2 × + H × ® 2 × OH  (R-74)
    HO 2 × + H ® H 2 O 2 + H ×  (R-75)
    2 O + H × ® H 2 + × OH  (R-76)
    2O + O ® 2 × OH  (R-77)
    2 + × OH ® H 2 O + H ×  (R-78)
    2 + O ® × OH + H ×  (R-79)
    2 + H × ® × OH + O  (R-80)
    CH 3 OH + × CH ® CH 3 O × + CH  (R-81)

    * Unless otherwise specified, the rate constant expression k = Aexp ( -E/RT ), Pre-finger factorThe unit of A is s -1 and cm 3 ·mo l -1 ·s – 1 for single molecule and bimolecular reaction respectively ; the unit of Eis kJ · mol -1 .

    FIG . 1 O 2 and CH . 4 concentration in function of the reaction time a case
    Fig.1 Concentrations are an of O 2 and CH . 4 AS Functions of Time
    P = 6.0MPa, T = 770K (497 deg.] C ), CH . 4 / O 2 = 10 /. 1

  2. Results and discussion
    of various species concentrations of 2.1 over time
    to take a reaction temperature of 770K , a reaction pressure of 6.0MPa , using CHEMKIN-III calculate different pressures are simulated, each of the species over time under different system conditions alkoxycarbonyl ratio Variety. The molar fraction (i.e., concentration) of O 2 and CH 4 in the entire system as a function of reaction time under the conditions of starting alkoxygen ratio CH 4 /O 2 = 10 is shown in Fig. 1 . There was a significant lead time for the reaction , and there was no significant change in the concentrations of O 2 and CH 4 in the first 2 seconds . After 2 seconds, the two began to decrease rapidly, especially after 3 seconds until O 2 was completely depleted, and the magnitude of the decrease in the concentration of O 2 and CH 4 was a sharp linear relationship with time, indicating that the reaction proceeded drastically to about 5.3.At the second, O 2 is completely consumed, and the concentration of CH 4 is also stabilized. The existence of the guiding period indicates that the concentration of free radicals in the system needs to be accumulated to a certain extent to trigger a strong chain-locking reaction. In the flow system (under the usual experimental conditions), when the exhaust gas reflects the complete reaction of O 2 , there will always be a certain spatial region inside the reactor, in which the concentration of free radicals is maintained at a relatively high level. When the new reaction gas enters the region, the reaction can be carried out quickly without having to go through the guiding period.FIG 2 changes of concentration of the product over the reaction time of the case Fig.2 Concentrations are an AS Functions of Time Products of P = 6.0MPa, T = 770K (497 deg.] C ), CH . 4 / O 2 = 10 /. 1

The concentration of each product changes with time as shown in Figure 2. HCHO starts to form at the beginning of the reaction. As the reaction progresses, CO , CH 3 OH , C 2 H 6 , CO 2 , C 2 H 4 , H 2 As the products appear one after the other, their concentration in the system also increases with the consumption of O 2 . At the time when O 2 is about to be completely converted, the concentration of methanol reaches a maximum value, and then begins to fall and gradually stabilizes. The concentration of C 2 H 4 is similar to that of methanol, but the case of C 2 H 6 is different. Before the complete reaction of O 2 , the concentration gradually increases with the progress of the reaction. When O 2 completely disappears, CThe concentration of 2 H 6 suddenly increased sharply and then stabilized in a short period of time. This is · CH . 3 and O 2 Reaction ( Rl ) and · CH . 3 and · CH . 3 coupling reaction between (R2) with the result of competition in the O 2 when molecules disappear, ? CH . 3 inevitably linked with each other generating C 2 H . 6 , such that C 2 H . 6 generates a greatly increased rate.
· CH 3 +O 2 → · CH 3 OO · (R1) · CH 3 +
· CH 3 → C 2 H . 6 (R2) in FIG. 3 system, various species conversion or selectivity with reaction time changes Fig.3 Conversion of Molecules or the SELECTIVITY Time AS Functions of P = 6.0MPa, T = 770K (497 deg.] C ), CH 4 /O 2 =10/1

Seen by the change in product selectivity (FIG. 3 ), the initial product of the reaction of formaldehyde, described HCHO directly via ? CH 3 generated, but this time the reaction conversion rate is very low, HCHO amount generating particularly low. As the reaction progressed, the selectivity of HCHO decreased sharply, and the selectivity of CO increased rapidly and gradually became dominant, indicating that CO was formed by oxidation of HCHO .
When the O 2 conversion rate rose to around 10% , the selectivity of HCHO has dropped to a very low level ( about 10% ). When O 2 is completely consumed, the selectivity of HCHO is reduced to negligible, so HCHO can hardly be detected in actual experiments . However, in some reports (such as Chun and Anthony et al [6,7] ), HCHO exists in the product, in order to match the results of the simulation calculation with the experimental results, ChunIn the mechanism model, et al. multiplied the “enhancement factor” of the pre-exponential factor of the relevant primitive reaction. In this paper, it is believed that the presence of HCHO in the product is likely to be due to the presence of certain metal oxides in the reaction system leading to the decomposition of methanol in the later stage of the reaction, rather than the gas phase reaction. Because in our experiments, the effects of metals were virtually eliminated, HCHO was not detected in the reaction products [1] .
In the results of the simulation calculation, in the initial initiation stage of the reaction, there is no methanol formation of the target product, methanol gradually appears as the reaction progresses, and reaches a maximum at the moment of complete consumption of O 2 , taking into account the concentration of CO in At this time, the decrease is small, indicating that methanol and HCHO are separately produced by parallel reaction of the same intermediate. The intermediate is generally considered to be CH 3 O?, so in the partial oxidation of methane, the formation and reaction of CH 3 O? The key issues to be considered, we believe that in order to improve the selectivity of the target product methanol, it is necessary to control the CH 3 O to the direction of reduction while avoiding or limiting its oxidation reaction. Therefore, the addition of a reducing gas to the reaction can increase the selectivity of methanol, as pointed out by Chellappa AS et al . [2] .
2.2 Influence of reaction pressure
The reaction pressure has a very large effect on the oxidation of methane. Figure 4A shows the simulated O 2 conversion and methanol selectivity as a function of reaction time at different pressures. The pressure is reduced and the O 2 is required for complete reaction. The time will increase, that is, the reaction rate will decrease. Since a fixed residence time is generally employed in the experiment, the reaction temperature required for the complete reaction of O 2 is correspondingly increased. After the complete consumption of O 2 , the selectivity of methanol also decreases with the decrease of pressure, indicating that the low pressure is obviously not conducive to the formation of methanol, which is consistent with the general experimental results.     Looking at the changes in CO and CO 2 (Fig. 4B ), the selectivity of CO increases with the decrease of pressure, which is consistent with the experimental results, but the selectivity of CO 2 is special, with the decrease of pressure. On the contrary, it has been reduced, which is quite different from the experimental results, indicating that there are still some influencing factors that require further research, and further improvement is needed for the mechanism model. 4 FIG reaction pressure of O 2 conversion and methanol and CO.’s, CO.’s 2 selectivity of Fig.4 Effect of Conversion ON pressure of O 2 and CH of the SELECTIVITY . 3

OH, CO and CO 2
T=770K(497 °C ), CH 4 /O 2 =10/1

2.3 Effect of
alkoxygen ratio The alkoxy ratio is also a factor that has a large influence on the selectivity of methanol. The experimental results show that the increase of alkoxygen ratio is beneficial to increase the selectivity of methanol. From the results of the simulation calculation, the alkoxy ratio has little effect on the initiation period of the reaction initiation, and the time required for the complete reaction of O 2 is substantially the same for the alkoxy ratio from 4 to 10 (Fig. 5A ). However, the selectivity of methanol does increase with the increase of the alkoxygen ratio. The selectivity of methanol increases from 3.0 % at an alkoxy ratio of 4 to 6.6% at a ratio of alkoxy to 10 to 3.0% , while CO 2 selectivity is From nearly 20% to 10% , but the selectivity of CO has not changed much, only when the alkoxy ratio is 4 , it is significantly lower than other results (Figure 5B ). Although there are gaps in the specific values, the above results are consistent with the experimental results in the trend of change. For example, as Zhang Qijun et al reported, as the alkoxygen ratio increased from 4 to 10 , the selectivity of CO2 decreased from more than 20 % to 5.    %, while the selectivity of CO does not change much, and the selectivity of methanol is greatly improved [1] .

FIG 5 alkoxycarbonyl than O 2 conversion and methanol, CO.’s X selectivity of
Fig.5 Effect of CH . 4 / O 2 ratio of ON Conversion O 2 and CH of the SELECTIVITY . 3 OH and CO.’s X
P = 6.0MPa, T 770K = (497 deg.] C )

of FIG. 6 reactant conversion and product selectivity at low temperatures with reaction time changes
in Fig.6 conversion of reactants and products of the sELECTIVITY AS Functions of lower Temperature time aT
P = 6.0MPa, T = 670K (397 °C ), CH 4 /O 2 =10/1

2.4 Influence of
reaction temperature The reaction temperature has a great influence on the radical reaction. For the methane oxidation reaction, due to the special stability of the CH 4 molecule, a relatively high temperature is required to allow the reaction to proceed. At 770K , 5.0MPa pressure, when CH 4 /O 2 = 10 , the time required for complete reaction of O 2 is less than 6 seconds, but at 670K temperature, other conditions are unchanged, even if the reaction time is extended to 1000 seconds. , O 2 still does not fully react (Figure 6 ). It is indicated that at 670K , the reaction proceeds very slowly, and the guiding period of the reaction is too long, which is not conducive to the progress of the radical chain reaction. From the distribution of the product, HCHO is still the initial product of the reaction, but as the reaction progresses, its selectivity decreases rapidly, and the selectivity of CO increases greatly, which indicates that CO is further oxidized from HCHO . As the reaction time prolongs, the conversion rate of O 2 gradually increases.The selectivity of CO and CH 3 OH is gradually decreasing, while the selectivity of CO 2 is slowly increasing, which indicates that the prolongation of reaction time before O 2 is not completely reacted, which is beneficial to the formation of thermodynamically stable products. The longer the time, the more the reaction will be in equilibrium. At the same time, it was noted that the formation of the C 2 product in the product was suppressed at a very low level, which was mainly due to the fact that the decrease in temperature caused a significant decrease in the CH 3 concentration in the system .
When the reaction temperature is 870K and other conditions are unchanged, the complete reaction of O 2 is completed in less than half a second. Since the reaction occurs too fast, the selectivity of CO is no longer the same as when 770K and 670K are used. It reaches a maximum value and then decreases slightly, but it is always rising. Although some changes occur when O 2 is completely reacted, it still maintains a tendency to increase and eventually tends to a relatively stable value (Fig. 7 ). The trend of other products is similar to that of 770K , but C 2 H 6 andThe selectivity of C 2 H 4 is greatly improved compared with that at 770 K. This is because high temperature favors the stabilization of CH 3 radicals, and the amount of C 2 H 6 is inevitably increased.

FIG 7 where reactant conversion and product selectivity with reaction time at elevated temperature
Fig.7 on Conversions and Functions of the SELECTIVITY Time AS AT IN AREAS OF COMMUNICAITIONS temperature
P = 6.0MPa, T = 870k (597 deg.] C ), CH . 4 / O 2 = 10 /1

Figure 8 Product mole fraction as a function of reaction time Fig.8 Concentrations of molecules as functions of time P=6.0MPa, T=870K(597 °C ), CH 4 /O 2 =10/1

Due to the high temperature of 870K , it is possible to make the reaction more equilibrium. The reaction time was extended to 1000 seconds for analysis (Fig. 8 ). It was found that after the complete reaction of O 2 , the distribution of the product was stable for a certain period of time, but when the reaction time exceeded 10 seconds, the products were in the system. The mole fraction began to change, and the species such as HCHO , CH 3 OH and C 2 H 4 gradually decreased, while CO , H 2 and C 2 H 6 gradually increased, but the mole fraction of CO 2 did not change. This may be because at this temperature, when the reaction time is long, HCHO and CH 3 OH are unstable, and decomposition reaction may occur to form more stable CO and H 2 .H 2 in turn undergoes an addition reaction with C 2 H 4 , which also leads to a decrease in C 2 H 4 and an increase in C 2 H 6 . This is consistent with the usual experimental results. The methane selective oxidation to methanol reaction temperature is too low, the activity is poor, and too high, the selectivity is poor. Also at this time since the system has no O 2 is present, there would be no strong oxidizing species capable CO is oxidized to CO 2 , so that CO 2 concentration does not change.

  1. Conclusions It
    can be seen from the simulation calculation of the selective oxidation reaction of methane by CHEMKIN-III that the oxidation reaction of methane first undergoes the initiation of the reaction during the lead time. After the boot time, the speed of the oxidation reaction of methane is very fast, since O 2 is greatly insufficient, O 2 will soon be completely consumed, the reaction product is distributed quickly reach a plateau.
    HCHO is formed at the initiation stage of the reaction, but as the reaction progresses, since it is easily oxidized to form CO , the concentration in the product is maintained at a very low level. CH 3 OHand HCHO are formed separately by parallel reaction, and the concentration of CH 3 OH in the product gradually increases, but when O 2 is completely consumed, the selectivity is lowered. In general, the selectivity of CH 3 OH is very low. The increase in pressure reduces the time required for the complete reaction of O 2 . In the flow reactor, the temperature of the complete reaction of O 2 can be lowered , while increasing the pressure also increases the selectivity of the target product CH 3 OH .
    Decreasing the alkoxy ratio, ie increasing the concentration of O 2 in the feed gas , reduces the selectivity of methanol and increases the selectivity of CO 2 .

Acknowledgments thank Dr. Japanese Institute of Earth Environment Industrial Technology Research catalytic Yao Liang in water CHEMKIN help give the simulation aspects.

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