ÌâÄ¿ÄÚÈÝ
20£®Èô²âµÃ0.5mol•L-1Pb£¨NO3£©2ÈÜÒºµ¼µçÄÜÁ¦Ô¶Ô¶±ÈͬŨ¶ÈµÄ£¨CH3COO£©2PbÈÜÒºµ¼µçÄÜÁ¦Ç¿£¬Ôò£º¢Ùд³ö´×ËáǦµÄµçÀë·½³Ìʽ£¨CH3COO£©2Pb?2CH3COO-+Pb2+£»
¢ÚPbSO4²»ÈÜÓÚË®ºÍÏ¡ÏõËᣬµ«ÄÜÈÜÓÚ½ÏŨµÄ´×ËᣬµÃµ½ÎÞɫ͸Ã÷µÄÈÜÒº£¬µ¼µçÄÜÁ¦ÏÔÖøÔöÇ¿£¬ÏòÎÞÉ«ÈÜÒºÖÐͨÈëH2SÆøÌ壬ÓÖÉú³ÉºÚÉ«³Áµí£¬ÈÜÒºµÄµ¼µçÄÜÁ¦ÉÔÓÐÔöÇ¿£¬Çëд³öÓйط´Ó¦Àë×Ó·½³ÌʽPbSO4+2CH3COOH¨T£¨CH3COO£©2Pb+2H++SO42-£¬²¢½âÊÍÉÏÊöʵÑéÏÖÏóPbSO4ÄÑÈÜ£¬ÓëCH3COOHÈõµç½âÖÊ·´Ó¦ºóÉú³ÉµÄH2SO4ÊÇÇ¿µç½âÖÊ£¬ÍêÈ«µçÀ룬Àë×ÓŨ¶ÈÏÔÖøÔö´ó£»£¨CH3COO£©2Pb+H2S¨TPbS¡ý+2CH3COOH£¬CH3COOH±È£¨CH3COO£©2PbÒ×µçÀ룬Àë×ÓŨ¶ÈÔö´ó£®
£¨3£©ÉÏÊöʵÑéÖÐÉæ¼°µÄ¼¸ÖÖº¬Ç¦»¯ºÏÎïÖУ¬ÊôÓÚÈõµç½âÖʵÄÎïÖÊ»¯Ñ§Ê½ÊÇ£¨CH3COO£©2PbÈܽâ¶È×îСµÄÎïÖÊ»¯Ñ§Ê½ÊÇPbSO4£®
·ÖÎö ¢ÙÈô²âµÃ0.5mol•L-1Pb£¨NO3£©2ÈÜÒºµ¼µçÄÜÁ¦Ô¶Ô¶±ÈͬŨ¶ÈµÄ£¨CH3COO£©2PbÈÜÒºµ¼µçÄÜÁ¦Ç¿£¬ËµÃ÷£¨CH3COO£©2PbÔÚÈÜÒºÖв¿·ÖµçÀ룬ΪÈõµç½âÖÊ£»
¢ÚPbSO4ÈÜÓÚŨ´×Ëᣬ»áÉú³ÉÈõµç½âÖÊ£¨CH3COO£©2Pb£¬Í¬Ê±»¹ÓÐH2SO4Éú³É£¬Àë×ÓŨ¶ÈÔö´ó£¬µ¼µçÄÜÁ¦ÔöÇ¿£¬µ±¼ÓÈëH2SÆøÌåʱ£¬Éú³ÉºÚÉ«³ÁµíPbS£¬Í¬Ê±»¹ÓÐCH3COOH Éú³É£¬ËùÒÔÈÜÒºµ¼µçÄÜÁ¦ÉÔÓÐÔöÇ¿£»
£¨3£©£¨CH3COO£©2PbÔÚÈÜÒºÖдæÔÚµçÀëƽºâ£»ÁòËáǦ²»ÈÜÓÚË®£®
½â´ð ½â£º¢ÙÈô²âµÃ0.5mol•L-1Pb£¨NO3£©2ÈÜÒºµ¼µçÄÜÁ¦Ô¶Ô¶±ÈͬŨ¶ÈµÄ£¨CH3COO£©2PbÈÜÒºµ¼µçÄÜÁ¦Ç¿£¬ËµÃ÷£¨CH3COO£©2PbÔÚÈÜÒºÖв¿·ÖµçÀ룬ΪÈõµç½âÖÊ£¬ÔòÆäµçÀë·½³ÌʽΪ£º£¨CH3COO£©2Pb?2CH3COO-+Pb2+£»
¹Ê´ð°¸Îª£º£¨CH3COO£©2Pb?2CH3COO-+Pb2+£»
¢ÚPbSO4ÈÜÓÚŨ´×Ëᣬ·¢Éú·´Ó¦£ºPbSO4+2CH3COOH¨T£¨CH3COO£©2Pb+2H++SO42-£¬PbSO4ÄÑÈÜ£¬ÓëCH3COOHÈõµç½âÖÊ·´Ó¦ºóÉú³ÉµÄH2SO4ÊÇÇ¿µç½âÖÊ£¬ÍêÈ«µçÀ룬Àë×ÓŨ¶ÈÏÔÖøÔö´ó£¬ËùÒÔµ¼µçÐÔÔöÇ¿£»µ±¼ÓÈëH2SÆøÌåʱ£¬Éú³ÉºÚÉ«³ÁµíPbS£¬·¢Éú·´Ó¦£º£¨CH3COO£©2Pb+H2S¨TPbS¡ý+2CH3COOH£¬CH3COOH±È£¨CH3COO£©2PbÒ×µçÀ룬Àë×ÓŨ¶ÈÔö´ó£¬ËùÒÔµ¼µçÐÔÔöÇ¿£»
¹Ê´ð°¸Îª£ºPbSO4+2CH3COOH¨T£¨CH3COO£©2Pb+2H++SO42-£»PbSO4ÄÑÈÜ£¬ÓëCH3COOHÈõµç½âÖÊ·´Ó¦ºóÉú³ÉµÄH2SO4ÊÇÇ¿µç½âÖÊ£¬ÍêÈ«µçÀ룬Àë×ÓŨ¶ÈÏÔÖøÔö´ó£»£¨CH3COO£©2Pb+H2S¨TPbS¡ý+2CH3COOH£¬CH3COOH±È£¨CH3COO£©2PbÒ×µçÀ룬Àë×ÓŨ¶ÈÔö´ó£»
£¨3£©£¨CH3COO£©2PbÔÚÈÜÒºÖдæÔÚµçÀëƽºâ£¬²»ÄÜÍêÈ«µçÀëÊôÓÚÈõµç½âÖÊ£»ÁòËáǦ²»ÈÜÓÚË®£¬ÊôÓÚÄÑÈÜÎïÖÊ£¬ËùÒÔÈܽâ¶È×îСµÄÎïÖÊÊÇPbSO4£»
¹Ê´ð°¸Îª£º£¨CH3COO£©2Pb£»PbSO4£®
µãÆÀ ±¾Ì⿼²éÁËÈõµç½âÖʵĵçÀëƽºâµÄÓ¦Óᢵç½âÖÊÇ¿ÈõÅжϡ¢Àë×Ó·½³ÌʽµÄÊéдµÈ£¬ÌâÄ¿ÄѶÈÖеȣ¬²àÖØÓÚ¿¼²éѧÉúµÄ·ÖÎöÄÜÁ¦ºÍ¶Ô»ù´¡ÖªÊ¶µÄ×ÛºÏÓ¦ÓÃÄÜÁ¦£®
A£® | AlCl3 | B£® | NaAl02 | C£® | Mgcl2 | D£® | BaCl2 |
£¨1£©À×ÓêÌìÆøÖз¢Éú×ÔÈ»¹Ìµªºó£¬µªÔªËØת»¯ÎªÏõËáÑζø´æÔÚÓÚÍÁÈÀÖУ®´¦ÓÚÑо¿½×¶ÎµÄ»¯Ñ§¹ÌµªÐ·½·¨ÊÇN2ÔÚ´ß»¯¼Á±íÃæÓëË®·¢ÉúÈçÏ·´Ó¦£º
2N2£¨g£©+6H2O£¨l£©=4NH3£¨g£©+3O2£¨g£©¡÷H K ¢Ù
ÒÑÖª£ºN2£¨g£©+3H2£¨g£©=2NH3£¨g£©¡÷H1=-92.4kJ•mol-1 K1¢Ú
2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H2=-571.6kJ•mol-1 K2¢Û
Ôò¡÷H=+1530kJ•mol-1£»K=$\frac{{{K}_{1}}^{2}}{{{K}_{2}}^{3}}$£¨ÓÃK1ºÍ K2±íʾ£©£®
£¨2£©ÔÚËĸöÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬·Ö±ð³äÈë1mol N2¡¢3mol H2O£¬ÔÚ´ß»¯¼ÁÌõ¼þϽøÐз´Ó¦¢Ù3Сʱ£¬ÊµÑéÊý¾Ý¼ûÏÂ±í£º
ÐòºÅ | µÚÒ»×é | µÚ¶þ×é | µÚÈý×é | µÚËÄ×é |
t/¡æ | 30 | 40 | 50 | 80 |
NH3Éú³ÉÁ¿/£¨10-6mol£© | 4.8 | 5.9 | 6.0 | 2.0 |
a£®NH3ºÍO2µÄÎïÖʵÄÁ¿Ö®±ÈΪ4£º3
b£®·´Ó¦»ìºÏÎïÖи÷×é·ÝµÄÖÊÁ¿·ÖÊý²»±ä
c£®µ¥Î»Ê±¼äÄÚÿÏûºÄ1molN2µÄͬʱÉú³É2molNH3
d£®ÈÝÆ÷ÄÚÆøÌåÃܶȲ»±ä
ÈôµÚÈý×é·´Ó¦3hºóÒÑ´ïƽºâ£¬µÚÈý×éN2µÄת»¯ÂÊΪ3¡Á10-4%£»ÓëÇ°Èý×éÏà±È£¬µÚËÄ×é·´Ó¦ÖÐNH3Éú³ÉÁ¿×îСµÄÔÒò¿ÉÄÜÊÇ´ß»¯¼ÁÔÚ80¡æ»îÐÔ¼õС£¬·´Ó¦ËÙÂÊ·´¶ø¼õÂý£®
¢ÙÑõ»¯¼ÁÊÇHNO2¡¡
¢Ú»¹ÔÐÔ£ºCl-£¾N2
¢ÛÿÉú³É2.8g N2£¬»¹Ô¼ÁʧȥµÄµç×ÓΪ0.6mol
¢ÜxΪ4£¬yΪ2 ¡¡
¢ÝSnCl${\;}_{x}^{y-}$ÊÇÑõ»¯²úÎ
A£® | ¢Ù¢Û¢Ý | B£® | ¢Ù¢Ú¢Ü¢Ý | C£® | ¢Ù¢Ú¢Û¢Ü | D£® | Ö»ÓÐ¢Ù¢Û |
¢Ù£»¢ÚUr-£¨aq£©+Na+£¨aq£©?NaUr£¨s£©£®
A£® | Õý·´Ó¦ÎªÎüÈÈ·´Ó¦ | B£® | Õý·´Ó¦Îª·ÅÈÈ·´Ó¦ | ||
C£® | Éý¸ßζȣ¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯ | D£® | ½µµÍζȣ¬Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯ |