ÌâÄ¿ÄÚÈÝ

9£®¼×´¼ÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬ÔÚÉú²úÖÐÓÐ×ÅÖØÒªµÄÓ¦Ó㮹¤ÒµÉÏÓü×ÍéÑõ»¯·¨ºÏ³É¼×´¼µÄ·´Ó¦ÓУº
¢ñCH4£¨g£©+CO2£¨g£©?2CO£¨g£©+2H2£¨g£©¡÷H1=+247.3kJ•mol-1
¢òCO£¨g£©+2H2£¨g£©?CH3OH£¨g£©¡÷H2=-90.1kJ•mol-1
¢ò2CO£¨g£©+O2£¨g£©?2CO2£¨g£©¡÷H3=-566.0kJ•mol-1
£¨1£©ÓÃCH4ºÍO2Ö±½ÓÖƱ¸¼×´¼ÕôÆøµÄÈÈ»¯Ñ§·½³ÌʽΪ2CH4£¨g£©+O2£¨g£©?2CH3OH£¨g£©¡÷H=-251.6kJ•mol-1£®
£¨2£©Ä³Î¶ÈÏ£¬Ïò4LºãÈÝÃܱÕÈÝÆ÷ÖÐͨÈë6mol CO2ºÍ6mol CH4£¬·¢Éú·´Ó¦£¨i£©£¬Æ½ºâÌåϵÖи÷×é·ÖµÄÌå»ý·ÖÊý¾ùΪ$\frac{1}{4}$£¬Ôò´ËζÈϸ÷´Ó¦µÄƽºâ³£ÊýK=1£¬CH4µÄת»¯ÂÊΪ33.3%£®
£¨3£©¹¤ÒµÉÏ¿Éͨ¹ý¼×´¼ôÊ»ù»¯·¨ÖÆÈ¡¼×Ëá¼×õ¥£¬Æä·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCH3OH£¨g£©+CO£¨g£©?HCOOCH3£¨g£©
  ¿ÆÑÐÈËÔ±¶Ô¸Ã·´Ó¦½øÐÐÁËÑо¿£¬²¿·ÖÑо¿½á¹ûÈçÏ£º

¢Ù´Ó·´Ó¦Ñ¹Ç¿¶Ô¼×´¼×ª»¯ÂʵÄÓ°Ï조ЧÂÊ¡±¿´£¬¹¤ÒµÖÆÈ¡¼×Ëá¼×õ¥Ó¦Ñ¡ÔñµÄѹǿÊÇ4.0¡Á106Pa£¨Ìî¡°3.5¡Á106Pa¡°¡°4£®O¡Á106Pa¡°»ò¡°5.0¡Á106Pa¡±£©£®
¢Úʵ¼Ê¹¤ÒµÉú²úÖвÉÓõÄζÈÊÇ80¡æ£¬ÆäÀíÓÉÊǸßÓÚ80¡æʱ£¬Î¶ȶԷ´Ó¦ËÙÂÊÓ°Ïì½ÏС£¬ÇÒ·´Ó¦·ÅÈÈ£¬Éý¸ßζÈʱƽºâÄæÏòÒƶ¯£¬×ª»¯ÂʽµµÍ£®
£¨4£©Ã¾È¼Áϵç³ØÔÚ¿ÉÒƶ¯µç×ÓÉ豸µçÔ´ºÍ±¸ÓõçÔ´µÈ·½ÃæÓ¦ÓÃÇ°¾°¹ãÀ«£®Í¼3Ϊ¡°Ã¾-´ÎÂÈËáÑΡ±È¼Áϵç³ØÔ­ÀíʾÒâͼ£¬µç¼«ÎªÃ¾ºÏ½ðºÍ²¬ºÏ½ð£®
¢ÙEΪ¸ÃȼÁϵç³ØµÄ¸º¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£®Fµç¼«Éϵĵ缫·´Ó¦Ê½ÎªClO-+2e-+H2O¨TCl-+2OH-£®
¢ÚþȼÁϵç³Ø¸º¼«ÈÝÒ×·¢Éú×Ô¸¯Ê´²úÉúÇâÆø£¬Ê¹¸º¼«ÀûÓÃÂʽµµÍ£¬Óû¯Ñ§ÓÃÓï½âÊÍÆäÔ­ÒòMg+2H2O¨TMg£¨OH£©2+H2¡ü£®
£¨5£©ÒÒÈ©ËᣨHOOC-CHO£©ÊÇÓлúºÏ³ÉµÄÖØÒªÖмäÌ壮¹¤ÒµÉÏÓá°Ë«¼«ÊҳɶԵç½â·¨¡±Éú²úÒÒÈ©ËᣬԭÀíÈçͼ4Ëùʾ£¬¸Ã×°ÖÃÖÐÒõ¡¢ÑôÁ½¼«Îª¶èÐԵ缫£¬Á½¼¶ÊÒ¾ù¿É²úÉúÒÒÈ©ËᣬÆäÖÐÒÒ¶þÈ©ÓëMµç¼«µÄ²úÎï·´Ó¦Éú³ÉÒÒÈ©Ëᣮ
¢ÙNµç¼«Éϵĵ缫·´Ó¦Ê½ÎªHOOC-COOH+2e-+2H+¨THOOC-CHO+H2O£®
¢ÚÈôÓÐ2molH+ͨ¹ýÖÊ×Ó½»»»Ä¤£¬²¢ÍêÈ«²ÎÓëÁË·´Ó¦£¬Ôò¸Ã×°ÖÃÖÐÉú³ÉµÄÒÒÈ©ËáΪ2mol£®
£¨6£©³£ÎÂÏÂÓöèÐԵ缫µç½â200mLNaCl¡¢CuSO4µÄ»ìºÏÈÜÒº£¬ËùµÃÆøÌåµÄÌå»ýËæʱ¼ä±ä»¯Èçͼ5Ëùʾ£¬¸ù¾Ýͼ5ÖÐÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣮£¨ ÆøÌåÌå»ýÒÑ»»Ëã³É±ê×¼×´¿öϵÄÌå»ý£¬ÇÒºöÂÔÆøÌåÔÚË®ÖеÄÈܽâºÍÈÜÒºÌå»ýµÄ±ä»¯£©
¢ÙÇúÏßII £¨Ìî¢ñ»ò¢ò£©±íʾÑô¼«²úÉúÆøÌåµÄ±ä»¯£»
¢ÚNaClµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.1mol/L£®CuSO4µÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.1mol/L£®
¢Ût2ʱËùµÃÈÜÒºµÄpHΪ1£®

·ÖÎö £¨1£©¸ù¾Ý¸Ç˹¶¨Âɽ«£¨¢ñ£©¡Á2+£¨¢ò£©¡Á2+£¨¢ó£©¼ÆËãÊéдÈÈ»¯Ñ§·½³Ìʽ£»
£¨2£©¸ù¾ÝƽºâÌåϵÖи÷×é·ÖµÄÌå»ý·ÖÊý¾ùΪ$\frac{1}{4}$£¬ÀûÓÃÈý¶Îʽ¼ÆËã³öƽºâʱ¸÷×é·ÖµÄº¬Á¿£¬¼ÆËãƽºâ³£ÊýºÍת»¯ÂÊ£»
£¨3£©¢ÙÒÀ¾Ýת»¯ÂÊÇúÏß·ÖÎöÅжϣ»¢ÚͼÏó·ÖÎöËÙÂÊËæζȱ仯µÄÇ÷ÊÆ·ÖÎö»Ø´ð£»
£¨4£©¢ÙÔ­µç³ØÖÐʧµç×ÓµÄΪ¸º¼«£»Õý¼«ÉÏClO-µÃµç×ÓÉú³ÉÂÈÀë×Ó£»¢ÚMgµÄ»îÆÃÐÔ½ÏÇ¿ÄÜÓëË®·´Ó¦Éú³ÉÇâÆø£»
£¨5£©¢ÙNµç¼«ÉÏHOOC-COOHµÃµç×ÓÉú³ÉHOOC-CHO£»¢Ú2mol H+ͨ¹ýÖÊ×Ó½»»»Ä¤£¬Ôòµç³ØÖÐתÒÆ2molµç×Ó£¬¸ù¾Ýµç¼«·½³Ìʽ¼ÆË㣻
£¨6£©±¾ÌâµÄ¹Ø½¡ÊǶÔͼÏóµÄ½â¶Á£®¸Õ¿ªÊ¼Ê±Òõ¼«Cu2+µÃµç×Ó£¬ÎÞÆøÌå·Å³ö£¬Cu2+·´Ó¦ÍêÈÜÒºÖеÄH+·Åµç£¬¢ñÊÇH2£¬Ñô¼«ÏÈÊÇÈÜÒºÖеÄCl-·Åµç£¬·´Ó¦ÍêºóÈÜÒºÖеÄOH-·Åµç£¬µç½â200mLÒ»¶¨Å¨¶ÈµÄNaClÓëCuSO4»ìºÏÈÜÒº£¬Ñô¼«·¢Éú2Cl--2e-=Cl2¡ü¡¢4OH--4e-=O2¡ü+2H2O£¬Òõ¼«·¢ÉúCu2++2e-=Cu¡¢2H++2e-=H2¡ü£¬½áºÏͼ¿ÉÖª£¬¢ñΪÒõ¼«ÆøÌåÌå»ýÓëʱ¼äµÄ¹Øϵ£¬¢òΪÑô¼«ÆøÌåÌå»ýÓëʱ¼äµÄ¹Øϵ£¬¼ÆËãʱץסµç×ÓÊغ㣮

½â´ð ½â£º£¨1£©¸ù¾Ý£¨¢ñ£©CH4£¨g£©+CO2£¨g£©?2CO£¨g£©+2H2£¨g£©¡÷H1=+247.3kJ•mol-1
£¨¢ò£©CO£¨g£©+2H2£¨g£©?CH3OH£¨g£©¡÷H2=-90.1kJ•mol-1
£¨¢ó£©2CO£¨g£©+O2£¨g£©?2CO2£¨g£©¡÷H3=-566.0kJ•mol-1
ÓÉ£¨¢ñ£©¡Á2+£¨¢ò£©¡Á2+£¨¢ó£©µÃ2CH4£¨g£©+O2£¨g£©?2CH3OH£¨g£©¡÷H£¬¹Ê¡÷H=2¡÷H1+2¡÷H2+¡÷H3=£¨+247.3kJ•mol-1£©¡Á2+£¨-90.1kJ•mol-1£©¡Á2+£¨-566.0kJ•mol-1£©=-251.6kJ•mol-1£¬ËùÒÔÓÃCH4ºÍO2Ö±½ÓÖƱ¸¼×´¼ÕôÆøµÄÈÈ»¯Ñ§·½³ÌʽΪ2CH4£¨g£©+O2£¨g£©?2CH3OH£¨g£©¡÷H=-251.6kJ•mol-1£¬¹Ê´ð°¸Îª£º2CH4£¨g£©+O2£¨g£©?2CH3OH£¨g£©¡÷H=-251.6kJ•mol-1£»
£¨2£©CH4£¨g£©+CO2£¨g£©?2CO£¨g£©+2H2£¨g£©
Æðʼ£¨mol£© 6        6        0         0
±ä»¯£¨mol£© x       x         2x        2x
ƽºâ£¨mol£©6-x      6-x       2x         2x
ƽºâÌåϵÖи÷×é·ÖµÄÌå»ý·ÖÊý¾ùΪ$\frac{1}{4}$£¬ËùÒÔ6-x=2x£¬½âµÃx=2£¬
Ôò´ËζÈϸ÷´Ó¦µÄƽºâ³£ÊýK=$\frac{{c}^{2}£¨CO£©¡Á{c}^{2}£¨{H}_{2}£©}{c£¨C{H}_{4}£©¡Ác£¨C{O}_{2}£©}=\frac{{1}^{2}¡Á{1}^{2}}{1¡Á1}$=1£¬
CH4µÄת»¯ÂÊΪ$\frac{2}{6}$=33.3%£¬¹Ê´ð°¸Îª£º1£»33.3%£»
£¨3£©¢Ù´Ó·´Ó¦Ñ¹Ç¿¶Ô¼×´¼×ª»¯ÂʵÄÓ°Ï조ЧÂÊ¡°¿´£¬Í¼ÏóÖÐת»¯Âʱ仯×î´óµÄÊÇ4.0¡Á106Pa£¬¹Ê´ð°¸Îª£º4.0¡Á106Pa£»
¢ÚÒÀ¾ÝͼÏó·ÖÎöζÈÔÚ¸ßÓÚ80¡ãC¶Ô·´Ó¦ËÙÂÊÓ°Ïì²»´ó£¬·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬Î¶ȹý¸ß£¬Æ½ºâÄæÏò½øÐУ¬²»ÀûÓÚת»¯ÂÊÔö´ó£¬
¹Ê´ð°¸Îª£º¸ßÓÚ80¡æʱ£¬Î¶ȶԷ´Ó¦ËÙÂÊÓ°Ïì½ÏС£¬ÇÒ·´Ó¦·ÅÈÈ£¬Éý¸ßζÈʱƽºâÄæÏòÒƶ¯£¬×ª»¯ÂʽµµÍ£»
£¨4£©¢Ù¡°Ã¾-´ÎÂÈËáÑΡ±È¼Áϵç³ØÖÐʧµç×ÓµÄΪ¸º¼«£¬ÔòMgΪ¸º¼«£»Õý¼«ÉÏClO-µÃµç×ÓÉú³ÉÂÈÀë×Ó£¬ÔòÕý¼«µÄµç¼«·´Ó¦Ê½Îª£ºClO-+2e-+H2O¨TCl-+2OH-£»¹Ê´ð°¸Îª£º¸º£»ClO-+2e-+H2O¨TCl-+2OH-£»
¢ÚMgµÄ»îÆÃÐÔ½ÏÇ¿ÄÜÓëË®·´Ó¦Éú³ÉÇâÆø£¬Æ䷴ӦΪ£ºMg+2H2O¨TMg£¨OH£©2+H2¡ü£¬¹Ê´ð°¸Îª£ºMg+2H2O¨TMg£¨OH£©2+H2¡ü£»
£¨5£©¢ÙNµç¼«ÉÏHOOC-COOHµÃµç×ÓÉú³ÉHOOC-CHO£¬Ôòµç¼«·´Ó¦Ê½ÎªHOOC-COOH+2e-+2H+¨THOOC-CHO+H2O£¬¹Ê´ð°¸Îª£ºHOOC-COOH+2e-+2H+¨THOOC-CHO+H2O£»
¢Ú2mol H+ͨ¹ýÖÊ×Ó½»»»Ä¤£¬Ôòµç³ØÖÐתÒÆ2molµç×Ó£¬¸ù¾Ýµç¼«·½³ÌʽHOOC-COOH+2e-+2H+¨THOOC-CHO+H2O£¬¿ÉÖªÉú³É1molÒÒÈ©ËᣬÓÉÓÚÁ½¼«¾ùÓÐÒÒÈ©ËáÉú³ÉËùÒÔÉú³ÉµÄÒÒÈ©ËáΪ2mol£»¹Ê´ð°¸Îª£º2£»
£¨6£©¢Ùµç½â200mLÒ»¶¨Å¨¶ÈµÄNaClÓëCuSO4»ìºÏÈÜÒº£¬Ñô¼«·¢Éú2Cl--2e-=Cl2¡ü¡¢4OH--4e-=O2¡ü+2H2O£¬Òõ¼«·¢ÉúCu2++2e-=Cu¡¢2H++2e-=H2¡ü£¬ËùÒÔÑô¼«ÉÏÏȲúÉúÆøÌ壬ÇúÏߢò±íʾÑô¼«²úÉúÆøÌåµÄ±ä»¯£¬¹Ê´ð°¸Îª£º¢ò£»
¢Ú½áºÏͼ¿ÉÖª£¬¢ñΪÒõ¼«ÆøÌåÌå»ýÓëʱ¼äµÄ¹Øϵ£¬¢òΪÑô¼«ÆøÌåÌå»ýÓëʱ¼äµÄ¹Øϵ¢ÙÓÉͼ¿ÉÖª£¬²úÉúÂÈÆøΪ224mL£¬ÔòÓÉ2Cl--2e-=Cl2¡ü¿ÉÖª£¬n£¨NaCl£©=$\frac{0.224L}{22.4L/mol}$¡Á2=0.02mol£¬ËùÒÔc£¨NaCl£©=$\frac{0.02mol}{0.2L}$=0.1mol/L£¬
ÓÉt2ʱÉú³ÉÑõÆøΪ112mL£¬n£¨O2£©=$\frac{0.112L}{22.4L/mol}$=0.005mol£¬Ôò¹²×ªÒƵç×ÓΪ0.02mol+0.005mol¡Á4=0.04mol£¬
¸ù¾Ýµç×ÓÊغ㼰Cu2++2e-=Cu¿ÉÖª£¬n£¨CuSO4£©=$\frac{0.04mol}{2}$=0.02mol£¬ËùÒÔc£¨CuSO4£©=$\frac{0.02mol}{0.2L}$=0.1mol/L£¬
¹Ê´ð°¸Îª£º0.1mol/L£»0.1mol/L£»
¢ÛÓÉt2ʱ4OH--4e-=O2¡ü+2H2O¡«4H+£¬n£¨H+£©=0.005mol¡Á4=0.02mol£¬ÔòÈÜÒºµÄc£¨H+£©=$\frac{0.02mol}{0.2L}$=0.1mol/L£¬pH=1£¬¹Ê´ð°¸Îª£º1£®

µãÆÀ ±¾Ì⿼²éÁ˸Ç˹¶¨ÂɵÄÓ¦Óá¢ÈÈ»¯Ñ§·½³ÌʽµÄ¼ÆËãÊéд¡¢»¯Ñ§Æ½ºâµÄ¼ÆË㡢ͼÏó·ÖÎöÅжÏÒÔ¼°µç»¯Ñ§µÄ֪ʶµÈ£¬×¢ÒâÕÆÎÕ»ù´¡ÖªÊ¶µÄÕÆÎÕ£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
17£®£¨1£©ÔÚÒ»¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷ÖнøÐÐ×ÅÈçÏ·´Ó¦£º
CO2£¨g£©+H2£¨g£©?CO£¨g£©+H2O£¨g£© Æäƽºâ³£ÊýKºÍζÈtµÄ¹ØϵÈçÏ£º
t¡æ70080085010001200
K2.61.71.00.90.6
£¨a£©¸Ã·´Ó¦Æ½ºâ³£ÊýKµÄ±í´ïʽΪ£º$\frac{c£¨CO£©•c£¨{H}_{2}O£©}{c£¨C{O}_{2}£©•c£¨{H}_{2}£©}$£»
£¨b£©µ±Î¶ÈΪ850¡æ£¬ÔÚ2LÃܱÕÈÝÆ÷ÖÐͨÈë1.0mol CO2ºÍ1.0mol H2£¬Ôòƽºâºó£¬CO2µÄת»¯ÂÊΪ50%£®
£¨2£©ÔÚ΢ÉúÎï×÷ÓõÄÌõ¼þÏ£¬NH4+¾­¹ýÁ½²½·´Ó¦±»Ñõ»¯³ÉNO3-£®ÕâÁ½²½µÄÄÜÁ¿±ä»¯Ê¾ÒâͼÈçÏ£º

£¨a£©µÚ¶þ²½·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¨Ñ¡Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±£©
£¨b£©1molNH4+£¨aq£©È«²¿Ñõ»¯³ÉNO3-£¨aq£©µÄÈÈ»¯Ñ§·½³ÌʽÊÇNH4+ £¨aq£©+2O2£¨g£©¨TNO3- £¨aq£©+2H+£¨aq£©+H2O£¨l£©¡÷H=-346 kJ/mol£®
£¨3£©ÒÑÖª£º
»¯Ñ§Ê½µçÀëƽºâ³£Êý
HCNK=4.9¡Á10-10
H2CO3K1=4.3¡Á10-7¡¢K2=5.6¡Á10-11
¢Ù25¡æʱ£¬²âµÃHCNºÍNaCNµÄ»ìºÏÈÜÒºµÄpH=11£¬Ôòc£¨HCN£©/c£¨CN-£©=2.0¡Á10-2£®
¢ÚÏòNaCNÈÜÒºÖÐͨÈëÉÙÁ¿CO2£¬Ôò·¢Éú·´Ó¦µÄÀë×Ó·½³ÌʽΪ£ºCN-+CO2+H2O=HCN+HCO3-£®
£¨4£©ÒÑÖª£ºKsp£¨CaCO3£©=4.96¡Á10-9£¬²»¿¼ÂÇÆäËüÒòËØÓ°Ï죬ÏÖ½«0.40mol/LµÄNa2CO3ÈÜÒººÍ0.20mol/LµÄCaCl2ÈÜÒºµÈÌå»ý»ìºÏ£¬Ôò»ìºÏºóÈÜÒºÖÐCa2+Ũ¶ÈΪ4.96¡Á10-8mol/L£®

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

¾«Ó¢¼Ò½ÌÍø