Question

10£®Ä³ÓлúÎïA£¨C₄H₆O₅£©¹ã·º´æÔÚÓÚÐí¶àË®¹ûÄÚ£¬ÓÈÒÔÆ»¹û¡¢ÆÏÌÑ¡¢Î÷¹Ï¡¢É½é«ÄÚΪ¶à£¬ÊÇÒ»ÖÖ³£ÓõÄʳƷÌí¼Ó¼Á£®¸Ã»¯ºÏÎï¾ßÓÐÈçÏÂÐÔÖÊ£º
¢ÙÔÚ25¡æʱ£¬AµÄµçÀëƽºâ³£ÊýK₁=3.9¡Á10^-4£¬K₂=5.5¡Á10^-6
¢ÚA+RCOOH£¨»òROH£©$\stackrel{ŨÁòËá}{¡ú}$ÓÐÏãζµÄ²úÎï
¢Û1mol A$\stackrel{×ãÁ¿µÄÄÆ}{¡ú}$ÂýÂý²úÉú1.5molÆøÌå
¢ÜºË´Å¹²ÕñÇâÆ×±íÃ÷A·Ö×ÓÖÐÓÐ5ÖÖ²»Í¬»¯Ñ§»·¾³µÄÇâÔ­×Ó£¬ÓëAÏà¹ØµÄ·´Ó¦¿òͼÈçÏ£º

£¨1£©¸ù¾Ý»¯ºÏÎïAµÄÐÔÖÊ£¬¶ÔAµÄ½á¹¹¿É×÷³öµÄÅжÏÊÇ£¨b£©£¨c£©£®
£¨a£©¿Ï¶¨ÓÐ̼̼˫¼ü             £¨b£©ÓÐÁ½¸öôÈ»ù
£¨c£©¿Ï¶¨ÓÐôÇ»ù                 £¨d£©ÓÐ-COOR¹ÙÄÜÍÅ
£¨2£©Ð´³öA¡¢D¡¢F¡¢MµÄ½á¹¹¼òʽ£ºA£ºHOOCCH£¨OH£©CH₂COOH£»D£ºNaOOCCH=CHCOONa£»F£ºHOOCC¡ÔCCOOH£»M£ºHOOCCH£¨Cl£©CH₂COOH£®
£¨3£©Ð´³öA¡úB¡¢B¡úEµÄ·´Ó¦ÀàÐÍ£ºA¡úBÏûÈ¥·´Ó¦£» B¡úE¼Ó³É·´Ó¦£®
£¨4£©Ð´³öM¡úAµÄµÚ¢Ù²½·´Ó¦µÄ·´Ó¦Ìõ¼þ£ºÇâÑõ»¯ÄÆË®ÈÜÒº¡¢¼ÓÈÈ£»
д³öE¡úFµÄµÚ¢Ù²½·´Ó¦µÄ·´Ó¦·½³Ìʽ£ºHOOCCH£¨Br£©CH£¨Br£©COOH+4NaOH$¡ú_{¡÷}^{´¼}$NaOOCC¡ÔCCOONa+2NaBr+4H₂O£®
£¨5£©ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬BÓëÒÒ¶þ´¼¿É·¢ÉúËõ¾Û·´Ó¦£¬Éú³ÉµÄ¸ß·Ö×Ó»¯ºÏÎïÓÃÓÚÖÆÔì²£Á§¸Ö£®Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ£ºn+n HOCH₂CH₂OH$\stackrel{Ò»¶¨Ìõ¼þ}{¡ú}$+£¨2n-1£©H₂O£®
£¨6£©Ð´³öÒ»ÖÖÓëA¾ßÓÐÏàͬ¹ÙÄÜÍŵÄAµÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£®

Answer 1

·ÖÎö A·Ö×ÓʽΪC₄H₆O₅£¬²»±¥ºÍ¶ÈΪ$\frac{2¡Á4+2-6}{2}$=2£¬AÓë´¼£¨ROH£©»òôÈËᣨRCOOH£©ÔÚŨH₂SO₄ºÍ¼ÓÈÈÌõ¼þϾù¿É·´Ó¦Éú³ÉÓÐÏãζµÄ²úÎÔòAº¬ÓÐôÇ»ù¡¢ôÈ»ù£¬1mol AÓë×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦²úÉú1.5mol H₂£¬½áºÏAµÄµçÀëƽºâ³£Êý£¬¿ÉÖªAΪ¶þÔªÈõËᣬÔòA·Ö×ÓÖк¬ÓÐÒ»¸öôÇ»ùºÍ¶þ¸öôÈ»ù£¬ºË´Å¹²ÕñÇâÆ×±íÃ÷A·Ö×ÓÖÐÓÐ5ÖÖ²»Í¬»¯Ñ§»·¾³µÄÇâÔ­×Ó£¬ÔòAµÄ½á¹¹¼òʽΪ£ºHOOCCH£¨OH£©CH₂COOH£¬MÔÚÇâÑõ»¯ÄÆË®ÈÜÒº¡¢¼ÓÈÈÌõ¼þÏ·¢ÉúË®½â·´Ó¦¡¢ËữµÃµ½A£¬¹ÊMΪHOOCCH£¨Cl£©CH₂COOH£¬A·¢ÉúÏûÈ¥·´Ó¦Éú³ÉBΪHOOCCH=CHCOOH£¬BÓë×ãÁ¿µÄNaOH·´Ó¦Éú³ÉDΪNaOOCCH=CHCOONa£¬BÓëäå·¢Éú¼Ó³É·´Ó¦Éú³ÉEΪHOOCCH£¨Br£©CH£¨Br£©COOH£¬EÔÚÇâÑõ»¯ÄÆ´¼ÈÜÒº¡¢¼ÓÈÈÌõ¼þÏ·¢ÉúÏûÈ¥·´Ó¦Éú³É£¬ÔÙËữµÃµ½FΪHOOCC¡ÔCCOOH£¬¾Ý´Ë½â´ð£®

½â´ð ½â£ºA·Ö×ÓʽΪC₄H₆O₅£¬²»±¥ºÍ¶ÈΪ$\frac{2¡Á4+2-6}{2}$=2£¬AÓë´¼£¨ROH£©»òôÈËᣨRCOOH£©ÔÚŨH₂SO₄ºÍ¼ÓÈÈÌõ¼þϾù¿É·´Ó¦Éú³ÉÓÐÏãζµÄ²úÎÔòAº¬ÓÐôÇ»ù¡¢ôÈ»ù£¬1mol AÓë×ãÁ¿µÄ½ðÊôÄÆ·´Ó¦²úÉú1.5mol H₂£¬½áºÏAµÄµçÀëƽºâ³£Êý£¬¿ÉÖªAΪ¶þÔªÈõËᣬÔòA·Ö×ÓÖк¬ÓÐÒ»¸öôÇ»ùºÍ¶þ¸öôÈ»ù£¬ºË´Å¹²ÕñÇâÆ×±íÃ÷A·Ö×ÓÖÐÓÐ5ÖÖ²»Í¬»¯Ñ§»·¾³µÄÇâÔ­×Ó£¬ÔòAµÄ½á¹¹¼òʽΪ£ºHOOCCH£¨OH£©CH₂COOH£¬MÔÚÇâÑõ»¯ÄÆË®ÈÜÒº¡¢¼ÓÈÈÌõ¼þÏ·¢ÉúË®½â·´Ó¦¡¢ËữµÃµ½A£¬¹ÊMΪHOOCCH£¨Cl£©CH₂COOH£¬A·¢ÉúÏûÈ¥·´Ó¦Éú³ÉBΪHOOCCH=CHCOOH£¬BÓë×ãÁ¿µÄNaOH·´Ó¦Éú³ÉDΪNaOOCCH=CHCOONa£¬BÓëäå·¢Éú¼Ó³É·´Ó¦Éú³ÉEΪHOOCCH£¨Br£©CH£¨Br£©COOH£¬EÔÚÇâÑõ»¯ÄÆ´¼ÈÜÒº¡¢¼ÓÈÈÌõ¼þÏ·¢ÉúÏûÈ¥·´Ó¦Éú³É£¬ÔÙËữµÃµ½FΪHOOCC¡ÔCCOOH£®
£¨1£©ÓÉÉÏÊö·ÖÎö¿ÉÖª£¬¸ù¾Ý»¯ºÏÎïAµÄÐÔÖÊ£¬A·Ö×ÓÖк¬ÓÐÒ»¸öôÇ»ùºÍ¶þ¸öôÈ»ù£¬¹ÊÑ¡£º£¨b£©£¨c£©£»
£¨2£©ÓÉÓÚÉÏÊö·ÖÎö¿ÉÖª£¬AΪHOOCCH£¨OH£©CH₂COOH£¬DΪNaOOCCH=CHCOONa£¬FΪHOOCC¡ÔCCOOH£¬MΪHOOCCH£¨Cl£©CH₂COOH£¬
¹Ê´ð°¸Îª£ºHOOCCH£¨OH£©CH₂COOH£»NaOOCCH=CHCOONa£»HOOCC¡ÔCCOOH£»HOOCCH£¨Cl£©CH₂COOH£»
£¨3£©ÓÉÉÏÊö·ÖÎö¿ÉÖª£¬A¡úBÊôÓÚÏûÈ¥·´Ó¦£» B¡úEÊôÓڼӳɷ´Ó¦£¬¹Ê´ð°¸Îª£ºÏûÈ¥·´Ó¦£»¼Ó³É·´Ó¦£»
£¨4£©M¡úAµÚ¢Ù²½Â±´úÌþµÄË®½â·´Ó¦·´Ó¦£¬·´Ó¦Ìõ¼þΪ£ºÇâÑõ»¯ÄÆË®ÈÜÒº¡¢¼ÓÈÈ£¬
E¡úFµÚ¢Ù²½·´Ó¦·½³ÌʽΪ£ºHOOCCH£¨Br£©CH£¨Br£©COOH+4NaOH$¡ú_{¡÷}^{´¼}$NaOOCC¡ÔCCOONa+2NaBr+4H₂O£¬
¹Ê´ð°¸Îª£ºÇâÑõ»¯ÄÆË®ÈÜÒº¡¢¼ÓÈÈ£»HOOCCH£¨Br£©CH£¨Br£©COOH+4NaOH$¡ú_{¡÷}^{´¼}$NaOOCC¡ÔCCOONa+2NaBr+4H₂O£»
£¨5£©ÔÚ´ß»¯¼Á×÷ÓÃÏ£¬HOOCCH=CHCOOHÓëÒÒ¶þ´¼¿É·¢ÉúËõ¾Û·´Ó¦£¬Éú³ÉµÄ¸ß·Ö×Ó»¯ºÏÎïÓÃÓÚÖÆÔì²£Á§¸Ö£®¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºn+n HOCH₂CH₂OH$\stackrel{Ò»¶¨Ìõ¼þ}{¡ú}$+£¨2n-1£©H₂O£¬
¹Ê´ð°¸Îª£ºn+n HOCH₂CH₂OH$\stackrel{Ò»¶¨Ìõ¼þ}{¡ú}$+£¨2n-1£©H₂O£»
£¨6£©ÓëA¾ßÓÐÏàͬ¹ÙÄÜÍŵÄAµÄͬ·ÖÒì¹¹ÌåÓУº£¬
¹Ê´ð°¸Îª£º£®

µãÆÀ ±¾Ì⿼²éÓлúÎïµÄÍƶϣ¬È·¶¨AµÄ½á¹¹¼òʽÊǹؼü£¬ÔÙ½áºÏ·´Ó¦Ìõ¼þÓëÓлúÎï·Ö×Óʽ½øÐÐÍƶϣ¬²àÖØÓÚѧÉúµÄ·ÖÎöÄÜÁ¦µÄ¿¼²é£¬ÐèҪѧÉúÊìÁ·ÕÆÎÕ¹ÙÄÜÍŵÄÐÔÖÊÓëת»¯£¬ÄѶÈÖеȣ®