ÌâÄ¿ÄÚÈÝ
3£®£¨1£©¼ÓÈÈÌõ¼þÏÂÓÃCO»¹ÔPbO¿ÉµÃµ½µ¥ÖÊǦ£®¼ºÖª£º
¢Ù2Pb£¨s£©+O2£¨g£©=2PbO£¨s£©¡÷H=-438kJ/mol
¢Ú2CO£¨g£©+O2£¨g£©=2CO2£¨g£©¡÷H=-566kJ/mol
ÔòCO»¹ÔPbOµÄÈÈ»¯Ñ§·½³ÌʽΪCO£¨g£©+PbO£¨s£©=Pb£¨s£©+CO£¨g£©¡÷H=-64kJ/mol£®
£¨2£©ÏõËáǦÓëÎý¿É·¢ÉúÖû»·´Ó¦£ºPb2++Sn?Pb+Sn2+£¬³£ÎÂÏ£¬·´Ó¦¹ý³ÌÖнðÊôÀë×Ó£¨R2+±íʾPb2+¡¢Sn2+£©Å¨¶ÈËæÊ±¼äµÄ±ä»¯Çé¿öÈçͼËùʾ£º
¢Ùt1¡¢t2ʱ¿ÌÉú³ÉSn2+ËÙÂʽϴóµÄÊÇt1£®
¢ÚÈô½«500ml 1.2mol/LPb£¨NO3£©2Óë2.4mol/LSn£¨NO3£©2µÈÌå»ý»ìºÏºó£¬ÔÙÏòÈÜÒºÖмÓÈë0.2molPb¡¢0.2molSn£¬´Ëʱv£¨Õý£©£¾v£¨Ä棩£¨Ìî¡°£¾¡±¡¢¡°£¼¡±£©£®
£¨3£©Ïû³ý·ÏË®ÖÐPb2+Ôì³ÉÎÛȾµÄ·½·¨ÓжàÖÖ£¬ÆäÖÐÖ®Ò»Êǽ«Æäת»¯Îª³Áµí£¬¼ºÖª³£ÎÂϼ¸ÖÖǦµÄÄÑÈÜ»¯ºÏÎïµÄKsp£ºKsp[Pb£¨OH£©2]=1.2¡Á10-16£¬Ksp£¨PbS£©=8¡Á10-28£¬´Ó¸üÓÐÀûÓÚ³ýÈ¥ÎÛË®ÖÐPb2+µÄ½Ç¶È¿´£¬´¦ÀíЧ¹û×î²îµÄÊǽ«Pb2+ת»¯ÎªPb£¨OH£©2³Áµí£®ÈôÓÃFeS½«Pb2+ת»¯ÎªPbS£¬ÔòÏàӦת»¯·´Ó¦µÄƽºâ³£ÊýK=7.5¡Á109£¨¼ºÖªKsp£¨FeS£©=6¡Á10-18£©£®
£¨4£©Ç¦ÔªËØÔÚÉú»îÖг£ÓÃ×öǦÐîµç³Ø£¬Çëд³öǦÐîµç³Ø³äµçʱµÄ×Ü·´Ó¦·½³Ìʽ£º2PbSO4£¨s£©+2H2O£¨l£©=Pb£¨s£©+PbO2£¨s£©+2H2SO4£¨aq£©£®
·ÖÎö £¨1£©¢Ù2Pb£¨s£©+O2£¨g£©=2PbO£¨s£©¡÷H=-438kJ/mol
¢Ú2CO£¨g£©+O2£¨g£©=2CO2£¨g£©¡÷H=-566kJ/mol
$\frac{¢Ú-¢Ù}{2}$µÃ£¬CO£¨g£©+PbO£¨s£©=Pb£¨s£©+CO£¨g£©¡÷H=-64kJ/mol¾Ý´Ë½øÐзÖÎö£»
£¨2£©¢Ùt1ʱ¿ÌÉú³ÉSn2+Ũ¶È±ä»¯±Èt2ʱ¿Ì¿ì£»
¢Ú¼ÓÈë0.2molSn£¬Ê¹Æ½ºâÕýÏòÒÆ¶¯£¬¹Êv£¨Õý£©£¾v£¨Ä棩£»
£¨3£©½áºÏÄÑÈÜ»¯ºÏÎïµÄKsp½øÐзÖÎö£»K=$\frac{c£¨F{e}^{2+}£©}{c£¨P{b}^{2+}£©}$¾Ý´Ë½øÐмÆË㣻
£¨4£©Ç¦ÔªËØÔÚÉú»îÖг£ÓÃ×öǦÐîµç³Ø£¬Çëд³öǦÐîµç³Ø³äµçʱµÄ×Ü·´Ó¦·½³Ìʽ2PbSO4£¨s£©+2H2O£¨l£©=Pb£¨s£©+PbO2£¨s£©+2H2SO4£¨aq£©£®
½â´ð ½â£º£¨1£©¢Ù2Pb£¨s£©+O2£¨g£©=2PbO£¨s£©¡÷H=-438kJ/mol
¢Ú2CO£¨g£©+O2£¨g£©=2CO2£¨g£©¡÷H=-566kJ/mol
$\frac{¢Ú-¢Ù}{2}$µÃ£¬CO£¨g£©+PbO£¨s£©=Pb£¨s£©+CO£¨g£©¡÷H=-64kJ/mol£¬
¹Ê´ð°¸Îª£ºCO£¨g£©+PbO£¨s£©=Pb£¨s£©+CO£¨g£©¡÷H=-64kJ/mol£»
£¨2£©¢Ùt1ʱ¿ÌÉú³ÉSn2+Ũ¶È±ä»¯±Èt2ʱ¿Ì¿ì£¬¹Êt1¡¢t2ʱ¿ÌÉú³ÉSn2+ËÙÂʽϴóµÄÊÇt1£¬
¹Ê´ð°¸Îª£ºt1£»
¢Ú¼ÓÈë0.2molSn£¬Ê¹Æ½ºâÕýÏòÒÆ¶¯£¬¹Êv£¨Õý£©£¾v£¨Ä棩£¬
¹Ê´ð°¸Îª£º£¾£»
£¨3£©½áºÏÄÑÈÜ»¯ºÏÎïµÄKsp£¬Ksp¸üС£¬¸üÄÑÈÜ£¬¹Ê´Ó¸üÓÐÀûÓÚ³ýÈ¥ÎÛË®ÖÐPb2+µÄ½Ç¶È¿´£¬´¦ÀíЧ¹û×î²îµÄÊǽ«Pb2+ת»¯ÎªPb£¨OH£©2³Áµí£»
FeS+Pb2+=PbS+Fe2+£¬ÔòK=$\frac{c£¨F{e}^{2+}£©}{c£¨P{b}^{2+}£©}$=$\frac{c£¨F{e}^{2+}£©c£¨{S}^{2-}£©}{c£¨P{b}^{2+}£©c£¨{S}^{2-}£©}$=$\frac{KspFeS}{KspPbS}$=$\frac{6¡Á1{0}^{-18}}{8¡Á1{0}^{-28}}$=7.5¡Á109£»
¹Ê´ð°¸Îª£ºPb£¨OH£©2£»7.5¡Á109£»
£¨4£©Ç¦ÔªËØÔÚÉú»îÖг£ÓÃ×öǦÐîµç³Ø£¬Çëд³öǦÐîµç³Ø³äµçʱµÄ×Ü·´Ó¦·½³Ìʽ2PbSO4£¨s£©+2H2O£¨l£©=Pb£¨s£©+PbO2£¨s£©+2H2SO4£¨aq£©£¬
¹Ê´ð°¸Îª£º2PbSO4£¨s£©+2H2O£¨l£©=Pb£¨s£©+PbO2£¨s£©+2H2SO4£¨aq£©£®
µãÆÀ ±¾Ì⿼²éÈÈ»¯Ñ§·½³ÌʽµÄÊéд£¬Æ½ºâµÄÒÆ¶¯£¬Æ½ºâ³£ÊýµÄ¼ÆË㣬µç¼«·½³ÌʽµÄÊéдµÈ£¬±¾ÌâÄѶÈÖеȣ®
| A£® | ±ê×¼×´¿öÏ£¬22.4 L H2OËùº¬µÄË®·Ö×ÓÊýΪNA | |
| B£® | ³£Î³£Ñ¹Ï£¬32 g³ôÑõËùº¬Ô×ÓÊýΪNA | |
| C£® | 1molFeÓë1molCl2ÍêÈ«·´Ó¦×ªÒƵç×ÓÊý3NA | |
| D£® | ±ê×¼×´¿öÏ£¬2NA¸ö¶þÑõ»¯Ì¼·Ö×ÓËùÕ¼µÄÌå»ýΪ44.8 L |
| A£® | ÔÚÌú´ß»¯×÷ÓÃÏ£¬±½ÓëäåË®ÄÜ·¢ÉúÈ¡´ú·´Ó¦ | |
| B£® | ¾ÛÒÒÏ©ËÜÁÏÖк¬ÓдóÁ¿Ì¼Ì¼Ë«¼ü£¬ÈÝÒ×ÀÏ»¯ | |
| C£® | ÓлúÎï | |
| D£® | ÒÒËáÒÒõ¥¡¢ÓÍÖ¬·Ö±ðÓëÈÈNaOHÈÜÒº·´Ó¦¾ùÓд¼Éú³É |
¢ÙÏòÈÜÒºÖмÓÈë¹ýÁ¿£¨NH4£©2CO3£¬µÃµ½ÆøÌå¼×¡¢ÈÜÒº¼×ºÍ°×É«³Áµí¼×£»
¢ÚÏòÈÜÒº¼×ÖмÓÈë¹ýÁ¿Ba£¨OH£©2£¬¼ÓÈȵõ½ÆøÌåÒÒ¡¢ÈÜÒºÒҺͰ×É«³ÁµíÒÒ£»
¢ÛÏòÈÜÒºÒÒÖмÓÈëCu·ÛºÍ¹ýÁ¿Ï¡ÁòËᣬµÃµ½ÆøÌå±û¡¢ÈÜÒº±ûºÍ°×É«³Áµí±û£®
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
| A£® | ¸Ã·ÛÄ©ÖпÉÄÜÊÇFe2£¨SO4£©3ºÍAl£¨NO3£©3µÄ»ìºÏÎï | |
| B£® | ³ÁµíÒҺͱû¶¼ÊÇBaSO4 | |
| C£® | ÆøÌåÒҺͱû·Ö±ðÊÇNH3ºÍH2 | |
| D£® | ÆøÌå¼×¿ÉÄÜÓÉA13+ÓëCO32-Ï໥´Ù½øË®½âµÃµ½ |
| A£® | äåÒÒÍé¼ÓÇâÑõ»¯ÄÆÈÜÒº¼ÓÈȺó£¬ÔÙ¼ÓAgNO3ÈÜÒºÓÐdz»ÆÉ«³ÁµíÉú³É | |
| B£® | ³ýÈ¥±½ÖеÄÉÙÁ¿±½·Ó£¬¿ÉÓÃÏȼÓŨäåË®£¬³ä·Ö³Áµíºó¹ýÂ˼´¿É | |
| C£® | ÅäÖÆÒø°±ÈÜҺʱ£¬½«Ï¡°±Ë®ÂýÂýµÎ¼Óµ½ÏõËáÒøÈÜÒºÖУ¬Ö±ÖÁ²úÉúµÄ³ÁµíÇ¡ºÃÈܽâΪֹ | |
| D£® | Óû¼ìÑéÕáÌÇË®½â²úÎïÊÇ·ñ¾ßÓл¹ÔÐÔ£¬¿ÉÏòË®½âºóµÄÈÜÒºÖÐÖ±½Ó¼ÓÈëÐÂÖÆµÄCu£¨OH£©2Ðü×ÇÒº²¢¼ÓÈÈ |
| A£® | v£¨NO2£©=v£¨O2£© | |
| B£® | ·´Ó¦Æ÷ÖеÄѹǿ²»ËæÊ±¼ä±ä»¯¶ø±ä»¯ | |
| C£® | »ìºÏÆøÌåµÄÑÕÉ«±£³Ö²»±ä | |
| D£® | »ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿±£³Ö²»±ä |