ÌâÄ¿ÄÚÈÝ

ijºãÎÂÃܱÕÈÝÆ÷·¢Éú¿ÉÄæ·´Ó¦£ºZ(£¿£©£«W(£¿£©X(g£©£«Y(£¿£© ¦¤H£¬ÔÚt1ʱ¿Ì·´Ó¦´ïµ½Æ½ºâ£¬ÔÚt2ʱ¿ÌËõСÈÝÆ÷Ìå»ý£¬t3ʱ¿ÌÔٴδﵽƽºâ״̬ºóδÔٸıäÌõ¼þ¡£ÏÂÁÐÓйØËµ·¨ÖÐÕýÈ·µÄÊÇ£¨ £©

A£®ZºÍWÔÚ¸ÃÌõ¼þϾùΪ·ÇÆøÌ¬

B£®t1¡«t2ʱ¼ä¶ÎÓët3ʱ¿Ìºó£¬Á½Ê±¼ä¶Î·´Ó¦ÌåϵÖÐÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿²»¿ÉÄÜÏàµÈ

C£®ÈôÔÚ¸ÃζÈÏ´˷´Ó¦Æ½ºâ³£Êý±í´ïʽΪK£½c(X£©£¬Ôò t1¡«t2ʱ¼ä¶ÎÓët3ʱ¿ÌºóµÄXŨ¶È²»ÏàµÈ

D£®Èô¸Ã·´Ó¦Ö»ÔÚijζÈTÒÔÉÏ×Ô·¢½øÐУ¬Ôò¸Ã·´Ó¦µÄƽºâ³£ÊýKËæÎ¶ÈÉý¸ß¶ø¼õС

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¶þ¼×ÃÑ£¨DME£©±»ÓþΪ¡°21ÊÀ¼ÍµÄÇå½àȼÁÏ¡±¡£ÓÉºÏ³ÉÆøÖƱ¸¶þ¼×ÃѵÄÖ÷ÒªÔ­ÀíÈçÏ£º

¢ÙCO(g£©+2H2(g£©CH3OH(g£©¡÷H1=£­90.7kJ¡¤mol-1

¢Ú2CH3OH(g£©CH3OCH3(g£©+H2O(g£©¡÷H2=£­23.5kJ¡¤mol-1

¢ÛCO(g£©+H2O(g£©CO2(g£©+H2(g£©¡÷H3=£­41.2kJ¡¤mol-1

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Ôò·´Ó¦3H2(g£©£«3CO(g£©CH3OCH3(g£©£«CO2(g£©µÄ¡÷H£½ kJ¡¤mol-1¡£

£¨2£©ÏÂÁдëÊ©ÖУ¬ÄÜÌá¸ßCH3OCH3²úÂʵÄÓÐ £¨Ìî×Öĸ£©¡£

A£®Ê¹ÓùýÁ¿µÄCO B£®Éý¸ßÎÂ¶È C£®Ôö´óѹǿ

£¨3£©·´Ó¦¢ÛÄÜÌá¸ßCH3OCH3µÄ²úÂÊ£¬Ô­ÒòÊÇ ¡£

£¨4£©½«ºÏ³ÉÆøÒÔn(H2£©/n(CO£©=2ͨÈë1 LµÄ·´Ó¦Æ÷ÖУ¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º4H2(g£©+2CO(g£©CH3OCH3(g£©+H2O(g£© ¡÷H£¬ÆäCOµÄƽºâת»¯ÂÊËæÎ¶ȡ¢Ñ¹Ç¿±ä»¯¹ØÏµÈçͼ1Ëùʾ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ £¨Ìî×Öĸ£©¡£

A£®¡÷H <0

B£®P1<P2<P3

C£®ÈôÔÚP3ºÍ316¡æÊ±£¬Æðʼn(H2£©/n(CO£©=3£¬Ôò´ïµ½Æ½ºâʱ£¬COת»¯ÂÊСÓÚ50£¥

£¨5£©²ÉÓÃÒ»ÖÖÐÂÐ͵Ĵ߻¯¼Á£¨Ö÷Òª³É·ÖÊÇCu-MnµÄºÏ½ð£©£¬ÀûÓÃCOºÍH2ÖÆ±¸¶þ¼×ÃÑ¡£¹Û²ìͼ2»Ø´ðÎÊÌâ¡£´ß»¯¼ÁÖÐn(Mn£©/n(Cu£©Ô¼Îª ʱ×îÓÐÀûÓÚ¶þ¼×Ãѵĺϳɡ£

£¨6£©Í¼3ΪÂÌÉ«µçÔ´¡°¶þ¼×ÃÑȼÁÏµç³Ø¡±µÄ¹¤×÷Ô­ÀíʾÒâͼ£¬aµç¼«µÄµç¼«·´Ó¦Ê½Îª ¡£

¹¤ÒµÉÏÀûÓÃп±ºÉ°£¨Ö÷Òªº¬ZnO¡¢ZnFe2O4£¬»¹º¬ÓÐÉÙÁ¿FeO¡¢CuOµÈÔÓÖÊ£©ÖÆÈ¡½ðÊôпµÄ¹¤ÒÕÁ÷³ÌÈçÏ£º

»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©ZnFe2O4ÊÇÒ»ÖÖÐÔÄÜÓÅÁ¼µÄÈí´Å²ÄÁÏ£¬Ò²ÊÇÒ»ÖÖ´ß»¯¼Á£¬ÄÜ´ß»¯Ï©ÀàÓлúÎïÑõ»¯ÍÑÇâµÈ·´Ó¦¡£

¢Ù ZnFe2O4ÖÐFeµÄ»¯ºÏ¼ÛÊÇ___________£¬´ÓÎïÖÊ·ÖÀà½Ç¶È˵£¬ZnFe2O4ÊôÓÚ__________£¨Ìî¡°Ëᡱ¡¢¡°¼î¡±»ò¡°ÑΡ±£©¡£

¢Ú ¹¤ÒµÉÏÀûÓ÷´Ó¦ZnFe2£¨C2O4£©3¡¤6H2OZnFe2O4+2CO2¡ü+4CO¡ü+6H2OÖÆ±¸ZnFe2O4¡£¸Ã·´Ó¦ÖÐÑõ»¯²úÎïÊÇ______________£¨Ìѧʽ£©£¬Ã¿Éú³É1mol ZnFe2O4£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿ÊÇ___________¡£

£¨2£©Ëá½þʱҪ½«Ð¿±ºÉ°·ÛË飬ÆäÄ¿µÄÊÇÌá¸ßËá½þЧÂÊ¡£Îª´ïµ½ÕâһĿµÄ£¬»¹¿É²ÉÓõĴëÊ©ÊÇ___________£¨ÈδðÒ»Ìõ£©£»ÒÑÖªZnFe2O4ÄÜÈÜÓÚËᣬÔòËá½þºóÈÜÒºÖдæÔڵĽðÊôÀë×ÓÓÐ______________¡£

£¨3£©¾»»¯¢ñÖÐH2O2²ÎÓë·´Ó¦µÄÀë×Ó·½³ÌʽΪ_______________£»ÊÔ¼ÁXµÄ×÷ÓÃÊÇ___________¡£

£¨4£©¸ÖÌú¶ÆÐ¿ÊǸÖÌú·À»¤µÄÒ»ÖÖÓÐЧ·½·¨¡£°´Í¼¼××°ÖýøÐÐÄ£ÄâÌúÉ϶ÆÐ¿µÄʵÑ飬ʵÑé½á¹ûÈçͼÒÒËùʾ¡£ÒÒÖкá×ø±êx±íʾµç·ÖÐͨ¹ýµç×ÓµÄÎïÖʵÄÁ¿£¬×Ý×ø±êy±íʾ·´Ó¦Îï»òÉú³ÉÎïµÄÎïÖʵÄÁ¿¡£

¢Ù Cµç¼«µÄµç¼«·´Ó¦Ê½Îª________________________¡£

¢Ú E¿ÉÒÔ±íʾµÄÁ¿ÊÇ_________________________£¨ÈÎдһÖÖ£©¡£

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø