我国某科研团队通过计算机模拟技术提出在12—磷钨酸(固体强酸)催化作用下甲醇气相脱水合成二甲醚的反应机理:CH
3OH+H
+→IM1(中间体1)→TS1(过渡态1)→IM3(中间体3)→CH
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsubsup%3E%3Cmrow%3E%3Cmn%3E3%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmo%3E%2B%3C%2Fmo%3E%3C%2Fmrow%3E%3C%2Fmsubsup%3E%3C%2Fmath%3E)
+H
2O,CH
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmsubsup%3E%3Cmrow%3E%3Cmn%3E3%3C%2Fmn%3E%3C%2Fmrow%3E%3Cmrow%3E%3Cmo%3E%2B%3C%2Fmo%3E%3C%2Fmrow%3E%3C%2Fmsubsup%3E%3C%2Fmath%3E)
+CH
3OH→IM2(中间体2)→TS2(过渡态2)→IM4(中间体4)→CH
3OCH
3+H
+。反应过程和相对能量E(kJ•mol
-1)如图所示。下列说法不正确的是( )
![](http://tikupic.21cnjy.com/2022/02/21/18/e6/18e654cf1dcc57176c74281cd9103b45_709x305.png)
- A、该反应机理中的最大能垒(活化能)E正=513.4kJ•mol-1
- B、2CH2OH(g)
CH3OCH3(g)+H2O(g) △H=-18.2kJ•mol-1 - C、已知IM1为CH2=OH++H2 , CH2=OH+H+→IM1是一个无垒(活化能为0)过程,可推测该过程从断键开始,断键时吸收的能量大于成键释放的能量
- D、类推IM2应为CH2=OH++CH4