ÌâÄ¿ÄÚÈÝ

£¨22·Ö£©2SO2(g)+O2(g) 2SO3(g)·´Ó¦¹ý³ÌµÄÄÜÁ¿±ä»¯ÈçͼËùʾ¡£ÒÑÖª1mol SO2(g)Ñõ»¯Îª1mol SO3µÄ¦¤H=¡ª99kJ¡¤mol¡ª1£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Í¼ÖÐA¡¢C·Ö±ð±íʾ    ¡¢    £¬EµÄ´óС¶Ô¸Ã·´Ó¦µÄ·´Ó¦ÈÈÓÐÎÞÓ°Ï죿   ¡£¸Ã·´Ó¦Í¨³£ÓÃV2O5×÷´ß»¯¼Á£¬¼ÓV2O5»áʹͼÖÐBµãÉý¸ß»¹ÊǽµµÍ£¿   £¬ÀíÓÉÊÇ       £»

£¨2£©Í¼ÖС÷H=   KJ¡¤mol¡ª1£»

£¨3£©V2O5µÄ´ß»¯Ñ­»·»úÀí¿ÉÄÜΪ£ºV2O5Ñõ»¯SO2ʱ£¬×ÔÉí±»»¹Ô­ÎªËļ۷°»¯ºÏÎËļ۷°»¯ºÏÎïÔÙ±»ÑõÆøÑõ»¯¡£Ð´³ö¸Ã´ß»¯Ñ­»·»úÀíµÄ»¯Ñ§·½³Ìʽ         £»

£¨4£©£¨2·Ö£©Èç¹û·´Ó¦ËÙÂʦԣ¨SO2£©Îª0£®05 mol¡¤L¡ª1¡¤min¡ª1,Ôò¦Ô£¨O2£©=   mol¡¤L¡ª1¡¤min¡ª1¡¢

¦Ô(SO3)=       mol¡¤L¡ª1¡¤min¡ª1£»

£¨5£©£¨2·Ö£©ÒÑÖªµ¥ÖÊÁòµÄȼÉÕÈÈΪ296 KJ¡¤mol¡ª1£¬¼ÆËãÓÉS(s)Éú³É3 molSO3(g)µÄ¡÷H   £¨ÒªÇó¼ÆËã¹ý³Ì£©¡£

£¨6£©¼×ÍéȼÁÏµç³Ø £¨KOH×÷µç½âÖÊÈÜÒº£©

¸º¼«·´Ó¦·½³ÌʽÊÇ£º                                                  

Õý¼«·´Ó¦·½³ÌʽÊÇ£º                                                     

×Ü·´Ó¦·½³ÌʽÊÇ£º                                                     

£¨7£©£¨2·Ö£©³£ÎÂÏ£¬ÉèpH ¾ùΪ5µÄH2SO4ºÍA12(SO4)3ÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)·Ö±ðΪc1 ¡¢c2£¬Ôòc1£ºc2=                   ¡£

£¨8£©£¨2·Ö£©Å¨¶ÈΪ0.5 mol/LµÄÑÎËáÓëµÈŨ¶ÈµÄ°±Ë®ÈÜÒº·´Ó¦£¬Ê¹ÈÜÒº³ÊÖÐÐÔ£¬Ôò»ìºÏǰÌå»ýVËá________V¼î(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)

 (9)£¨2·Ö£©È¡10 mLÈÜÒº0.5 mol/LµÄÑÎËᣬ¼ÓˮϡÊ͵½500 mL£¬Ôò´ËʱÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H£«)£½________¡£

 

£¨22·Ö£©£¨1£©·´Ó¦ÎïÄÜÁ¿  Éú³ÉÎïÄÜÁ¿  ÎÞ  ½µµÍ  ÒòΪ´ß»¯¼Á¸Ä±äÁË·´Ó¦µÄÀú³Ìʹ»î»¯ÄÜE½µµÍ  £¨2£©£¨2·Ö£©¡ª198 

(3) £¨2·Ö£©SO2+V2O5=SO3+2VO2  4VO2+O2=2V2O5  

(4)£¨2·Ö£©0£®025  0£®05

(5) £¨2·Ö£©S(s)+O2(g)=2SO2(g)  ¡÷H1=¡ª296KJ¡¤mol¡ª1

SO2(g)+1/2O2(g) SO3(g)¡÷H2=¡ª99 KJ¡¤mol¡ª1 

3 S(s)+9/2O2(g)=3SO3(g)  ¡÷H=3(¡÷H1+¡÷H2)=¡ª1185 KJ¡¤mol¡ª1 

£¨6£©¸º¼«£ºCH4+10OH- £­8e-=CO32-+7H2O  Õý¼«£ºO2£«2H2O+4e-=4OH-

   ×Ü·´Ó¦£ºCH4£«2 O2+2OH-=CO32-+3H2O       

£¨7£©£¨2·Ö£©1£º10£¨8£©£¨2·Ö£©Ð¡ÓÚ£¨9£©£¨2·Ö£©10-12

½âÎö:

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨22·Ö£©2SO2(g)+O2(g) 2SO3(g)·´Ó¦¹ý³ÌµÄÄÜÁ¿±ä»¯ÈçͼËùʾ¡£ÒÑÖª1mol SO2(g)Ñõ»¯Îª1molµÄ¦¤H=¡ª99kJ¡¤mol¡ª1£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Í¼ÖÐA¡¢C·Ö±ð±íʾ   ¡¢    £¬EµÄ´óС¶Ô¸Ã·´Ó¦µÄ·´Ó¦ÈÈÓÐÎÞÓ°Ï죿  ¡£¸Ã·´Ó¦Í¨³£ÓÃV2O5×÷´ß»¯¼Á£¬¼ÓV2O5»áʹͼÖÐBµãÉý¸ß»¹ÊǽµµÍ£¿  £¬ÀíÓÉÊÇ      £»
£¨2£©Í¼ÖС÷H=  KJ¡¤mol¡ª1£»
£¨3£©V2O5µÄ´ß»¯Ñ­»·»úÀí¿ÉÄÜΪ£ºV2O5Ñõ»¯SO2ʱ£¬×ÔÉí±»»¹Ô­ÎªËļ۷°»¯ºÏÎËļ۷°»¯ºÏÎïÔÙ±»ÑõÆøÑõ»¯¡£Ð´³ö¸Ã´ß»¯Ñ­»·»úÀíµÄ»¯Ñ§·½³Ìʽ         £»
£¨4£©£¨2·Ö£©Èç¹û·´Ó¦ËÙÂʦԣ¨SO2£©Îª0£®05 mol¡¤L¡ª1¡¤min¡ª1,Ôò¦Ô£¨O2£©=  mol¡¤L¡ª1¡¤min¡ª1¡¢
¦Ô(SO3)=       mol¡¤L¡ª1¡¤min¡ª1£»
£¨5£©£¨2·Ö£©ÒÑÖªµ¥ÖÊÁòµÄȼÉÕÈÈΪ296 KJ¡¤mol¡ª1£¬¼ÆËãÓÉS(s)Éú³É3 molSO3(g)µÄ¡÷H  £¨ÒªÇó¼ÆËã¹ý³Ì£©¡£
£¨6£©¼×ÍéȼÁÏµç³Ø£¨KOH×÷µç½âÖÊÈÜÒº£©
¸º¼«·´Ó¦·½³ÌʽÊÇ£º                                                  
Õý¼«·´Ó¦·½³ÌʽÊÇ£º                                                     
×Ü·´Ó¦·½³ÌʽÊÇ£º                                                     
£¨7£©£¨2·Ö£©³£ÎÂÏ£¬ÉèpH ¾ùΪ5µÄH2SO4ºÍA12(SO4)3ÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H+)·Ö±ðΪc1¡¢c2£¬Ôòc1£ºc2=                   ¡£
£¨8£©£¨2·Ö£©Å¨¶ÈΪ0.5 mol/LµÄÑÎËáÓëµÈŨ¶ÈµÄ°±Ë®ÈÜÒº·´Ó¦£¬Ê¹ÈÜÒº³ÊÖÐÐÔ£¬Ôò»ìºÏǰÌå»ýVËá________V¼î(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)
(9)£¨2·Ö£©È¡10 mLÈÜÒº0.5 mol/LµÄÑÎËᣬ¼ÓˮϡÊ͵½500 mL£¬Ôò´ËʱÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H£«)£½________¡£

(12·Ö)ºÏ³ÉÆøµÄÖ÷Òª³É·ÖÊÇÒ»Ñõ»¯Ì¼ºÍÇâÆø£¬¿ÉÓÃÓںϳɶþ¼×ÃѵÈÇå½àȼÁÏ¡£´ÓÌìÈ»Æø»ñµÃºÏ³ÉÆø¹ý³ÌÖпÉÄÜ·¢ÉúµÄ·´Ó¦ÓУº
¢ÙCH4(g)+H2O(g)CO(g)+3H2(g)    ¨SH1="+206.1" kJ/mol
¢ÚCH4(g)+CO2(g)2CO(g)+2H2(g)   ¨SH2="+247.3" kJ/mol
¢ÛCO(g)+H2O(g)CO2(g)+ H2(g)    ¨SH3
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÔÚÒ»ÃܱÕÈÝÆ÷ÖнøÐз´Ó¦¢Ù£¬²âµÃµÄÎïÖʵÄÁ¿Å¨¶ÈËæ·´Ó¦Ê±¼äµÄ±ä»¯Èçͼ1Ëùʾ¡£10minʱ£¬¸Ä±äµÄÍâ½çÌõ¼þ¿ÉÄÜÊÇ                          ¡£

(2)Èçͼ2Ëùʾ£¬Ôڼס¢ÒÒÁ½ÈÝÆ÷Öзֱð³äÈëµÈÎïÖʵÄÁ¿µÄºÍ£¬Ê¹¼×¡¢ÒÒÁ½ÈÝÆ÷³õʼÈÝ»ýÏàµÈ¡£ÔÚÏàͬζÈÏ·¢Éú·´Ó¦¢Ú£¬²¢Î¬³Ö·´Ó¦¹ý³ÌÖÐζȲ»±ä¡£ÒÑÖª¼×ÈÝÆ÷ÖеÄת»¯ÂÊËæÊ±¼ä±ä»¯µÄͼÏñÈçͼ3Ëùʾ£¬ÇëÔÚͼ3Öл­³öÒÒÈÝÆ÷ÖеÄת»¯ÂÊËæÊ±¼ä±ä»¯µÄͼÏñ¡£

(3)·´Ó¦¢ÛÖР         ¡£800¡æÊ±£¬·´Ó¦¢ÛµÄ»¯Ñ§Æ½ºâ³£ÊýK=1.0£¬Ä³Ê±¿Ì²âµÃ¸ÃζÈϵÄÃܱÕÈÝÆ÷Öи÷ÎïÖʵÄÎïÖʵÄÁ¿¼ûÏÂ±í£º

´Ëʱ·´Ó¦¢ÛÖÐÕý¡¢Äæ·´Ó¦ËÙÂʵĹØÏµÊ½ÊÇ            (Ìî´úºÅ)¡£
a£®(Õý)(Äæ)   b£®(Õý)<(Äæ)   c£®(Õý)=(Äæ)  d£®ÎÞ·¨ÅжÏ
£¨4£©800KʱÏÂÁÐÆðʼÌå»ýÏàͬµÄÃܱÕÈÝÆ÷ÖгäÈë2mol SO2¡¢1mol O2£¬Æä·´Ó¦ÊÇ2SO2(g)+O2(g) 2SO3(g)£»¡÷H=£­96.56 kJ?mol-1£¬¼×ÈÝÆ÷ÔÚ·´Ó¦¹ý³ÌÖб£³Öѹǿ²»±ä£¬ÒÒÈÝÆ÷±£³ÖÌå»ý²»±ä£¬±ûÈÝÆ÷ά³Ö¾øÈÈ£¬ÈýÈÝÆ÷¸÷×Ô½¨Á¢»¯Ñ§Æ½ºâ¡£

¡¾1¡¿´ïµ½Æ½ºâʱ£¬Æ½ºâ³£ÊýK (¼×)    K (ÒÒ)    K(±û)£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£
¡¾2¡¿´ïµ½Æ½ºâʱSO2µÄŨ¶ÈC(SO2)(¼×)  C(SO2) (ÒÒ)  C(SO2) (±û)£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£

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

¾«Ó¢¼Ò½ÌÍø