ÌâÄ¿ÄÚÈÝ

ÄÜÔ´ÎÊÌâÊǵ±Ç°ÊÀ½ç¸÷¹úËùÃæÁÙµÄÑÏÖØÎÊÌ⣬ͬʱȫÇòÆøºò±äů£¬Éú̬»·¾³ÎÊÌâÈÕÒæÍ»³ö£¬¿ª·¢ÇâÄÜ¡¢ÑÐÖÆȼÁϵç³Ø¡¢·¢Õ¹µÍ̼¾­¼ÃÊÇ»¯Ñ§¹¤×÷ÕßµÄÑо¿·½Ïò¡£
I£®ÇâÆøͨ³£ÓÃÉú²úˮúÆøµÄ·½·¨ÖƵá£ÆäÖÐCO(g)+H2O(g)  CO2(g)+H2(g)
¡÷H<0¡£ÔÚ850¡æʱ£¬Æ½ºâ³£ÊýK=1¡£
£¨1£©Èô½µµÍζȵ½750¡æʱ£¬´ïµ½Æ½ºâʱK    1£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©
£¨2£©850¡æʱ£¬ÈôÏòÒ»ÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖÐͬʱ³äÈë1.0molCO¡¢3molH2O¡¢1.0molCO2
ºÍx molH2£¬Ôò£º
¢Ùµ±x=5.0ʱ£¬ÉÏÊö·´Ó¦Ïò       £¨Ìî¡°Õý·´Ó¦¡±»ò¡°Äæ·´Ó¦¡±£©·½Ïò½øÐС£
¢ÚÈôҪʹÉÏÊö·´Ó¦¿ªÊ¼Ê±ÏòÕý·´Ó¦·½Ïò½øÐУ¬ÔòxÓ¦Âú×ãµÄÌõ¼þÊÇ          ¡£
¢ÛÔÚ850¡æʱ£¬ÈôÉèx=5.0ºÍx=6.0£¬ÆäËüÎïÖʵÄͶÁϲ»±ä£¬µ±ÉÏÊö·´Ó¦´ïµ½Æ½ºâºó£¬
²âµÃH2µÄÌå»ý·ÖÊý·Ö±ðΪa%¡¢b%£¬Ôòa      b£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©
II£®ÒÑÖª4.6gҺ̬ÒÒ´¼ÍêȫȼÉÕÉú³É¶þÑõ»¯Ì¼ºÍҺ̬ˮ·Å³öÈÈÁ¿136kJ¡¤molҺ̬ˮת
»¯ÎªÆøÌåË®ÎüÊÕ44kJµÄÈÈÁ¿¡£
£¨3£©Çëд³öÒÒ´¼È¼ÉÕÉú³ÉÆø̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ
                               ¡£
£¨4£©½«0.1molÒÒ´¼ÔÚ×ãÁ¿ÑõÆøÖÐȼÉÕ£¬µÃµ½µÄÆøÌåÈ«²¿Í¨Èëµ½100mL3mol/LNaOHÈÜÒºÖУ¬ºöÂÔHCO-3µÄµçÀ룬ÔòËùµÃÈÜÒºÖÐc(CO2-3)      c(HCO-3)£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£¬£©Ô­ÒòÊÇ                                                 £¨ÓÃÎÄ×ÖÐðÊö£©¡£
£¨1£©´óÓÚ£¨2·Ö£©
£¨2£©¢ÙÄæ·´Ó¦£¨2·Ö£©  ¢Úx<3£¨2·Ö£©  ¢ÛСÓÚ£¨2·Ö£©
£¨3£©C2H5OH£¨l£©+3C2£¨g£©=2CO2£¨g£©+3H2O£¨g£©  ¡÷H=-1228kJ/mol£¨2·Ö£©
£¨4£©Ð¡ÓÚ£¨2·Ö£©    CO32- Ë®½â³Ì¶È´óÓÚHCO3-Ë®½â³Ì¶È£¨2·Ö£©
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
°±ÓÐ׏㷺µÄÓÃ;£¬Èç¿ÉÓÃÓÚ»¯·Ê¡¢ÏõËá¡¢ºÏ³ÉÏËάµÈ¹¤ÒµÉú²ú¡£°±µÄË®ÈÜÒºÖдæÔÚµçÀëƽºâ£¬³£ÓõçÀë³£ÊýKbºÍµçÀë¶È¦ÁÀ´¶¨Á¿±íʾÆäµçÀë³Ì¶È¡£KbºÍ¦Á³£ÓõIJⶨ·½·¨£ºÔÚÒ»¶¨Î¶ÈʱÓÃËá¶È¼Æ²â¶¨Ò»ÏµÁÐÒÑ֪Ũ¶È°±Ë®µÄpH£¬¿ÉµÃ¸÷Ũ¶È°±Ë®¶ÔÓ¦µÄc(OH-)£¬È»ºóͨ¹ý»»ËãÇóµÃ¸÷¶ÔÓ¦µÄ¦ÁÖµºÍKbÖµ¡£ÏÂÃæÊÇijÖÐѧ»¯Ñ§ÐËȤС×éÔÚ25¡æʱ²â¶¨Ò»ÏµÁÐŨ¶È°±Ë®µÄpHËù¶ÔÓ¦µÄc(OH-)£º
¡¾ÒÇÆ÷ÓëÊÔ¼Á¡¿Ëá¶È¼Æ¡¢50 mL¼îʽµÎ¶¨¹Ü¡¢100mLÉÕ±­¡¢ 0.10 mol¡¤L-1°±Ë®
¡¾ÊµÑéÊý¾Ý¡¿(²»±ØÌî±í¸ñ)
ÉÕ±­ºÅ
V°±Ë® (mL)
VË®(mL)
c (NH3¡¤H2O)£¨mol¡¤L-1£©
c(OH-)
Kb
¦Á
1
50.00
0.00
 
1.34¡Á10-3
 
 
2
25.00
25.00
 
9.48¡Á10-4
 
 
3
5.00
45.00
 
4.24¡Á10-4
 
 
Çë¸ù¾ÝÒÔÉÏÐÅÏ¢»Ø´ðÏÂÊöÎÊÌ⣺                     
£¨1£©25¡æʱ£¬°±Ë®µÄµçÀë³£Êý£ºKb   ¡ø   £¬Í¨¹ý¼ÆËãËùµÃµÄÊý¾ÝºÍ¼ò½àµÄÎÄ×Ö˵Ã÷µçÀë³£Êý¡¢µçÀë¶ÈÓëÈõµç½âÖʵijõʼŨ¶ÈµÄ¹Øϵ   ¡ø   ¡£
£¨2£©ÓÃ0.10mol¡¤L¡ª1ÑÎËá·Ö±ðµÎ¶¨20.00mL0.10mol¡¤L¡ª1µÄNaOHÈÜÒººÍ20.00mL0.10mol¡¤L¡ª1
°±Ë®ËùµÃµÄµÎ¶¨ÇúÏßÈçÏ£º

ÇëÖ¸³öÑÎËáµÎ¶¨°±Ë®µÄÇúÏßΪ    ¡ø   £¨ÌîA»òB£©£¬Çëд³öÇúÏßaµãËù¶ÔÓ¦µÄÈÜÒºÖи÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄÅÅÁÐ˳Ðò   ¡ø   ¡£
£¨3£©Òº°±×÷ΪһÖÖDZÔÚµÄÇå½àÆû³µÈ¼ÁÏÒÑÔ½À´Ô½±»Ñо¿ÈËÔ±ÖØÊÓ¡£ËüÔÚ°²È«ÐÔ¡¢¼Û¸ñµÈ·½Ãæ½Ï»¯Ê¯È¼ÁϺÍÇâȼÁÏÓÐ׎ϴóµÄÓÅÊÆ¡£°±ÔÚȼÉÕÊÔÑé»úÖÐÏà¹ØµÄ·´Ó¦ÓУº
4NH3£¨g£©+3O2£¨g£©= 2N2£¨g£©+6H2O£¨l£©   ¡÷H1                              ¢Ù
4NH3£¨g£©+5O2£¨g£©= 4NO£¨g£©+6H2O£¨l£©  ¡÷H2               ¢Ú
4NH3£¨g£©+6NO£¨g£©= 5N2£¨g£©+6H2O£¨l£©  ¡÷H3                               ¢Û
Çëд³öÉÏÊöÈý¸ö·´Ó¦ÖС÷H1¡¢¡÷H2¡¢¡÷H3ÈýÕßÖ®¼ä¹ØϵµÄ±í´ïʽ£¬¡÷H1=   ¡ø   ¡£
£¨4£©Allis-ChalmersÖÆÔ칫˾·¢ÏÖ¿ÉÒÔÓð±×÷ΪȼÁϵç³ØµÄȼÁÏ¡£Æä×Ü·´Ó¦Ê½Îª4NH3+3O2= 2N2+6H2O£¬Õý¼«Éϵĵ缫·´Ó¦Ê½ÎªO2+2H2O+4e¡ª=4OH¡ª£¬Ôò¸º¼«Éϵĵ缫·´Ó¦Ê½Îª    ¡ø   ¡£

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

¾«Ó¢¼Ò½ÌÍø