ÌâÄ¿ÄÚÈÝ
£¨14·Ö£©¼×´¼±»³ÆΪ21ÊÀ¼ÍµÄÐÂÐÍȼÁÏ£¬¹¤ÒµÉÏͨ¹ýÏÂÁз´Ó¦¢ÙºÍ¢Ú£¬ÓÃCH4ºÍH2OΪÔÁÏÀ´ÖƱ¸¼×´¼¡£¢ÙCH4£¨g£©£«H2O £¨g£©CO£¨g£©£«3H2£¨g£© ¡÷H1 ¢ÚCO£¨g£©£«2H2£¨g£©CH3OH£¨g£©¡÷H2 ½«0£®2mol CH4ºÍ0£®3mol H2O£¨g£©Í¨ÈëÈÝ»ýΪ10LµÄÃܱÕÈÝÆ÷ÖУ¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦¢Ù£¬´ïµ½Æ½ºâʱ£¬CH4µÄת»¯ÂÊÓëζȣ®Ñ¹Ç¿µÄ¹ØϵÈçͼËùʾ£º
£¨1£©Î¶Ȳ»±ä£¬ËõСÌå»ý£¬Ôö´óѹǿ£¬¢ÙµÄ·´Ó¦ËÙÂÊ £¨Ìî¡°Ôö´ó¡±£®¡°¼õС¡±£®¡°²»±ä¡±£©£¬Æ½ºâÏò ·½ÏòÒƶ¯¡£
£¨2£©·´Ó¦¢ÙµÄ¡÷H1 0£¬£¨Ìî¡°©‚¡±¡°©„¡±¡°©ƒ¡±£©£¬Æäƽºâ³£Êý±í´ïʽΪK= £¬100¡æ£¬Ñ¹Ç¿Îªp1ʱƽºâ³£ÊýµÄÖµÊÇ ¡£
£¨3£©ÔÚѹǿΪ0£®1MpaÌõ¼þÏ£¬½«a mol COÓë3a molH2µÄ»ìºÏÆøÌåÔÚ´ß»¯¼Á×÷ÓÃϽøÐз´Ó¦¢Ú¡£ÎªÁËÑ°ÕҺϳɼ״¼µÄζȺÍѹǿµÄÊÊÒËÌõ¼þ£¬Ä³Í¬Ñ§Éè¼ÆÁËÈý×éʵÑ飬²¿·ÖʵÑéÌõ¼þÒѾÌîÔÚÏÂÃæʵÑéÉè¼Æ±íÖС£ÇëÔÚÏ¿ոñÖÐÌîÈëÊ£ÓàµÄʵÑéÌõ¼þÊý¾Ý¡£
£¨4£©½ü¼¸Ä꿪·¢µÄ¼×´¼È¼Áϵç³ØÊDzÉÓò¬µç¼«£¬µç³ØÖеÄÖÊ×Ó½»»»Ä¤Ö»ÔÊÐíÖÊ×ÓºÍË®·Ö×Óͨ¹ý£¬Æ乤×÷ÔÀíʾÒâͼÈçÏ£º
Çë»Ø´ð
¢ÙPt£¨a£©µç¼«µÄµç¼«·´Ó¦Ê½Îª
¢ÚÈç¹û¸Ãµç³Ø¹¤×÷ʱµç·ÖÐͨ¹ý2molµç×ÓÔòÏûºÄµÄCH3OHÓÐ mol¡£
£¨1£©Î¶Ȳ»±ä£¬ËõСÌå»ý£¬Ôö´óѹǿ£¬¢ÙµÄ·´Ó¦ËÙÂÊ £¨Ìî¡°Ôö´ó¡±£®¡°¼õС¡±£®¡°²»±ä¡±£©£¬Æ½ºâÏò ·½ÏòÒƶ¯¡£
£¨2£©·´Ó¦¢ÙµÄ¡÷H1 0£¬£¨Ìî¡°©‚¡±¡°©„¡±¡°©ƒ¡±£©£¬Æäƽºâ³£Êý±í´ïʽΪK= £¬100¡æ£¬Ñ¹Ç¿Îªp1ʱƽºâ³£ÊýµÄÖµÊÇ ¡£
£¨3£©ÔÚѹǿΪ0£®1MpaÌõ¼þÏ£¬½«a mol COÓë3a molH2µÄ»ìºÏÆøÌåÔÚ´ß»¯¼Á×÷ÓÃϽøÐз´Ó¦¢Ú¡£ÎªÁËÑ°ÕҺϳɼ״¼µÄζȺÍѹǿµÄÊÊÒËÌõ¼þ£¬Ä³Í¬Ñ§Éè¼ÆÁËÈý×éʵÑ飬²¿·ÖʵÑéÌõ¼þÒѾÌîÔÚÏÂÃæʵÑéÉè¼Æ±íÖС£ÇëÔÚÏ¿ոñÖÐÌîÈëÊ£ÓàµÄʵÑéÌõ¼þÊý¾Ý¡£
ʵÑé±àºÅ | T¡æ | n£¨CO£©/n£¨H2£© | p £¨Mpa£© |
I | 150 | 1/3 | 0£®1 |
¢ò | | | 5 |
¢ó | 350 | | 5 |
Çë»Ø´ð
¢ÙPt£¨a£©µç¼«µÄµç¼«·´Ó¦Ê½Îª
¢ÚÈç¹û¸Ãµç³Ø¹¤×÷ʱµç·ÖÐͨ¹ý2molµç×ÓÔòÏûºÄµÄCH3OHÓÐ mol¡£
£¨1£©Ôö´ó£¨1·Ö£© Äæ·´Ó¦£¨»òÏò×󣩣¨1·Ö£©
£¨2£©©ƒ£¨1·Ö£© c £¨CO£©¡¤c3£¨H2£©/c £¨CH4£©¡¤c£¨H2O£©£¨2·Ö£© 1£®35¡Á10-3£¨2·Ö£©
£¨3£©II 150 1/3 III 1/3 £¨Ã¿¿Õ1·Ö¹²3·Ö£©
£¨4£©¢ÙCH3OH + H2O -6e- = CO2¡ü+ 6 H+ £¨2·Ö£©
¢Ú1/3 £¨2·Ö£©
£¨2£©©ƒ£¨1·Ö£© c £¨CO£©¡¤c3£¨H2£©/c £¨CH4£©¡¤c£¨H2O£©£¨2·Ö£© 1£®35¡Á10-3£¨2·Ö£©
£¨3£©II 150 1/3 III 1/3 £¨Ã¿¿Õ1·Ö¹²3·Ö£©
£¨4£©¢ÙCH3OH + H2O -6e- = CO2¡ü+ 6 H+ £¨2·Ö£©
¢Ú1/3 £¨2·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿