ÌâÄ¿ÄÚÈÝ

11£®ÄÜÔ´ÊÇÈËÀàÉú´æºÍ·¢Õ¹µÄÖØÒªÖ§Öù£®Ñо¿²¢ÓÐЧµØ¿ª·¢ÐÂÄÜÔ´ÔÚÄÜÔ´½ôȱµÄ½ñÌì¾ßÓÐÖØÒªµÄÀíÂÛÒâÒ壮ÒÑÖªH2ÓëCO·´Ó¦Éú³ÉCH3OHµÄ¹ý³ÌÈçͼËùʾ£ºCOµÄȼÉÕÈÈ¡÷H2=-b kJ•mol-1£¬CH3OHµÄȼÉÕÈÈ
¡÷H3=-c kJ•mol-1£®Çë¸ù¾ÝÒÔÉÏÐÅÏ¢£¬»Ø´ðÏÂÁÐÓйØÎÊÌ⣺
£¨1£©¼×´¼ÊÇÒ»ÖÖ¿É£¨Ìî¡°¿É¡±»ò¡°²»¿É¡±£©ÔÙÉúÄÜÔ´£¬¼×´¼µÄ¹ÙÄÜÍŵĵç×ÓʽÊÇ£®
£¨2£©CH3OHȼÉÕµÄÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪ2CH3OH£¨l£©+3O2£¨g£©¡ú2CO2£¨g£©+4H2O£¨l£©¡÷H=-2ckJ•mol-1£®
£¨3£©H2µÄȼÉÕÈÈ¡÷H=$\frac{a+c-b}{2}$KJ/mol£®
£¨4£©Ò»Ð©»¯Ñ§¼ü¼üÄÜÈçÏ£ºC=OΪd kJ/mol£»O=OΪekJ/mol£»C=OΪfkJ/mol£®ÔòÓÉÒÔÉÏÊý¾ÝÓÐb=2f-d-$\frac{e}{2}$£¨ÓÃd¡¢e¡¢fµÄ´úÊýʽ±íʾ£©£®
£¨5£©H2ºÍCOµÄ»ìºÏÆøÌån mol£¬³ä·ÖȼÉÕ¹²·Å³öQ kJÈÈÁ¿£¬Ôò»ìºÏÆøÌåÖÐH2ºÍCOµÄÎïÖʵÄÁ¿Ö®±ÈΪ£¨2bn-2Q£©£º£¨bn-an-cn+2Q£©£®
£¨6£©COÓëO2¿ÉÒÔ×é³ÉÐÂÐÍȼÁÏµç³Ø£¬Èô¸Ãµç³ØÒÔPtΪµç¼«£¬ÒÔKOHΪµç½âÖÊÈÜÒº£¬Ð´³ö¸ÃȼÁÏµç³ØµÄÕý¼«·´Ó¦Ê½O2+4e-+2H2O=4OH-£®

·ÖÎö £¨1£©¼×´¼ÊÇÒ»ÖÖ¿ÉÔÙÉúÄÜÔ´£¬¼×´¼µÄ¹ÙÄÜÍÅÊÇôÇ»ù£¬³ÊµçÖÐÐÔ£»
£¨2£©ÒÀ¾Ý¼×´¼µÄȼÉÕÈȸÅÄîÊéдÈÈ»¯Ñ§·½³Ìʽ£»
£¨3£©·ÖÎöͼÏó½áºÏ¼×´¼ºÍÒ»Ñõ»¯Ì¼µÄȼÉÕÈÈ£¬ÒÀ¾Ý¸Ç˹¶¨ÂɼÆËãÇâÆøµÄȼÉÕÈÈ£»
£¨4£©ÒÀ¾Ý·´Ó¦µÄìʱä=·´Ó¦ÎïµÄ¼üÄÜÖ®ºÍ-Éú³ÉÎïµÄ¼üÄÜÖ®ºÍ£»
£¨5£©ÒÀ¾ÝÇâÆøºÍÒ»Ñõ»¯Ì¼È¼ÉյĻ¯Ñ§·½³Ìʽ¼ÆË㣻
£¨6£©Õý¼«ÊÇÑõÆøµÃµç×ÓÉú³ÉÇâÑõ¸ùÀë×Ó£¬µç¼«·´Ó¦Ê½Îª£ºO2+4e-+2H2O=4OH-£»

½â´ð ½â£º£¨1£©¼×´¼ÊÇÒ»ÖÖ¿ÉÔÙÉúÄÜÔ´£¬¼×´¼µÄ¹ÙÄÜÍÅÊÇôÇ»ù£¬³ÊµçÖÐÐÔ£¬µç×ÓʽΪ£º£¬¹Ê´ð°¸Îª£º¿É£»£»
£¨2£©CH3OHµÄȼÉÕÈÈ¡÷H3=-ckJ•mol-1£»È¼ÉÕÈÈÊÇ¿ÉȼÎïÍêȫȼÉÕÉú³ÉÎȶ¨Ñõ»¯Îï·Å³öµÄÈÈÁ¿£¬
ÈÈ»¯Ñ§·½³ÌʽΪ£º2CH3OH£¨l£©+3O2£¨g£©¡ú2CO2£¨g£©+4H2O£¨l£©¡÷H=-2ckJ•mol-1£¬¹Ê´ð°¸Îª£º2CH3OH£¨l£©+3O2£¨g£©¡ú2CO2£¨g£©+4H2O£¨l£©¡÷H=-2ckJ•mol-1£»
£¨3£©ÒÀ¾Ý¼×´¼ºÍÒ»Ñõ»¯Ì¼µÄȼÉÕÈȼÆËãÇâÆøµÄȼÉÕÈÈ£º
¢Ù2CH3OH£¨l£©+3O2£¨g£©¡ú2CO2£¨g£©+4H2O£¨l£©¡÷H=-2ckJ•mol-1£»
¢Ú2CO£¨g£©+O2£¨g£©=2CO2£¨g£©¡÷H=-2bkJ•mol-1£»
¢Û2H2£¨g£©+CO£¨g£©=CH3OH£¨l£©¡÷H=-aKJ/mol£»
ÒÀ¾Ý¸Ç˹¶¨ÂÉ£¬¢Û¡Á2-¢Ú+¢ÙµÃµ½£º4H2£¨g£©+2O2£¨g£©=4H2O£¨l£©¡÷H=-£¨2a-2b+2c£©KJ/mol£»
ÒÀ¾ÝÇâÆøµÄȼÉÕÈÈÊÇ1molÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³öµÄÈÈÁ¿£¬ÇâÆøµÄȼÉÕÈÈ=$\frac{a+c-b}{2}$KJ/mol£»¹Ê´ð°¸Îª£º$\frac{a+c-b}{2}$KJ/mol
£¨4£©Ò»Ð©»¯Ñ§¼ü¼üÄÜÈçÏ£ºC¡ÔOΪd kJ/mol£»O=OΪekJ/mol£»C=OΪfkJ/mol£®
·´Ó¦Îª£º2CO£¨g£©+O2£¨g£©=2CO2£¨g£©¡÷H=-2bkJ•mol-1£»
·´Ó¦µÄìʱä=·´Ó¦ÎïµÄ¼üÄÜÖ®ºÍ-Éú³ÉÎïµÄ¼üÄÜÖ®ºÍ=2d+e-2¡Áf¡Á2=¡÷H=-2b£¬b=2f-d-$\frac{e}{2}$£»¹Ê´ð°¸Îª£º2f-d-$\frac{e}{2}$£»
£¨5£©H2ºÍCOµÄ»ìºÏÆøÌånmol£¬ÉèÇâÆøÎªÎïÖʵÄÁ¿ÎªX£¬Ò»Ñõ»¯Ì¼ÎïÖʵÄÁ¿Îªn-X£»³ä·ÖȼÉÕ¹²·Å³öQkJÈÈÁ¿
2CO£¨g£©+O2£¨g£©=2CO2£¨g£©¡÷H=-2bkJ•mol-1£»
2                          2b
n-X                        b£¨n-X£©
2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H=-£¨a-b+c£©KJ/mol 
2                                                            a+c-b
X                                                       $\frac{X}{2}$£¨a+c-b£©         
µÃµ½£ºb£¨n-X£©+$\frac{X}{2}$£¨a+c-b£©=Q£»
¼ÆËãµÃµ½£ºX=$\frac{2bn-2Q}{3b-a-c}$
 »ìºÏÆøÌåÖÐH2ºÍCOµÄÎïÖʵÄÁ¿Ö®±È=$\frac{2bn-2Q}{3b-a-c}$£º$\frac{bn-an-cn+2Q}{3b-a-c}$=£¨2bn-2Q£©£º£¨bn-an-cn+2Q£©
¹Ê´ð°¸Îª£º£¨2bn-2Q£©£º£¨bn-an-cn+2Q£©£»
£¨6£©Õý¼«ÊÇÑõÆøµÃµç×ÓÉú³ÉÇâÑõ¸ùÀë×Ó£¬µç¼«·´Ó¦Ê½Îª£ºO2+4e-+2H2O=4OH-£¬¹Ê´ð°¸Îª£ºO2+4e-+2H2O=4OH-£®

µãÆÀ ±¾Ì⿼²éÁËÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨£¬È¼ÉÕÈȵĸÅÄîÀí½âºÍÓ¦Ó㬸Ç˹¶¨ÂɵļÆËãÓ¦Ó㬼üÄܼÆËãìʱäµÄÓ¦Ó㬻ìºÏÎïµÄ¼ÆËã·ÖÎö£¬Ô­µç³ØµÄµç¼«·´Ó¦µÈ£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø