2SO2(g)+O2(g) 2SO3(g)·´Ó¦¹ý³ÌµÄÄÜÁ¿±ä»¯ÈçͼËùʾ¡£

ÒÑÖª1mol SO2(g)Ñõ»¯Îª1mol SO3µÄ¦¤H=-99kJ¡¤mol-1.Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©Í¼ÖÐC±íʾ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ ;¸Ã·´Ó¦Í¨³£ÓÃV2O5×÷´ß»¯¼Á£¬¼ÓV2O5»áʹͼÖÐBµã½µµÍ£¬ÀíÓÉÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ £»

£¨2£©Í¼ÖС÷H= ¡¡¡¡¡¡¡¡kJ¡¤mol-1£»

£¨3£©Èç¹û·´Ó¦ËÙÂʦԣ¨SO2£©Îª0.05 mol¡¤L-1¡¤min-1,Ôò¦Ô£¨O2£©=¡¡ mol¡¤L-1¡¤min-1£»

£¨4£©ÒÑÖªµ¥ÖÊÁòµÄȼÉÕÈÈΪ296 KJ¡¤mol-1£¬¼ÆËãÓÉS(s)Éú³É3 molSO3(g)µÄ¡÷H=¡¡¡¡¡¡¡¡¡¡¡¡ ¡£¡¡¡¡¡¡¡¡¡¡

 (1)Éú³ÉÎï×ÜÄÜÁ¿£»´ß»¯¼ÁÄܽµµÍ·´Ó¦ËùÐèÒªµÄÄÜÁ¿£¬Ìá¸ß»î»¯·Ö×Ó°Ù·ÖÊý

£¨2£©-198    £¨3£©0.025   £¨4£©-1185kJ/mol

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖª723 Kʱ£¬2SO2(g)+O2(g)====2SO3(g)£»¦¤H=-Q kJ¡¤mol-1£¬ÔÚÏàͬÌõ¼þÏ£¬ÏòÒ»ÃܱÕÈÝÆ÷ÖÐͨÈë2 mol SO2ºÍ1 O2 mol£¬´ïƽºâʱ·Å³öÈÈÁ¿ÎªQ1£»ÏòÁíÒ»Ìå»ýÏàͬµÄÃܱÕÈÝÆ÷ÖÐͨÈë1 mol SO2ºÍ0.5 mol O2´ïµ½Æ½ºâʱ·Å³öµÄÈÈÁ¿ÎªQ2¡£ÔòQ1¡¢Q2¡¢QÂú×ãµÄ¹ØϵÊÇ(    )

A.Q=Q1/2                B.Q2£¼Q1/2               C.Q2£¼Q1£¼Q               D.Q=Q1£¾Q2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

Ò»¶¨Ìõ¼þÏ£¬ÏòÒ»´ø»îÈûµÄÃܱÕÈÝÆ÷ÖгäÈë2 mol SO2ºÍ1 mol O2£¬·¢ÉúÏÂÁз´Ó¦£º2SO2 (g)+O2 (g) 2SO3(g)¡£´ïµ½Æ½ºâºó¸Ä±äÏÂÁÐÌõ¼þ£¬SO3ÆøÌåƽºâŨ¶È²»¸Ä±äµÄÊÇ£¨    £©

A.±£³ÖζȺÍÈÝÆ÷Ìå»ý²»±ä£¬³äÈë1 mol SO3 (g)

B.±£³ÖζȺÍÈÝÆ÷ÄÚѹǿ²»±ä£¬³äÈë1 mol SO3 (g)

C.±£³ÖζȺÍÈÝÆ÷ÄÚѹǿ²»±ä£¬³äÈë1 mol O2(g)

D.±£³ÖζȺÍÈÝÆ÷ÄÚѹǿ²»±ä£¬³äÈë1 mol Ar(g)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

(8·Ö) ½Ó´¥·¨ÖÆÁòËṤÒÕÖУ¬ÆäÖ÷·´Ó¦ÔÚ450¡æ²¢Óд߻¯¼Á´æÔÚϽøÐУº

2SO2(g)+O2(g) 2SO3(g)  ¡÷H=£­190 kJ¡¤mo1£­1

£¨1£©ÔÚÒ»¹Ì¶¨ÈÝÆ÷ÖгäÈë2mol SO2ºÍ1molO2 £¬ÔÚÒ»¶¨µÄÌõ¼þÏ´ﵽƽºâ£¬·´Ó¦·Å³öµÄÈÈÁ¿__________(Ìî´óÓÚ¡¢Ð¡ÓÚ»òµÈÓÚ) 190 kJ

£¨2£©ÔÚÒ»¸ö¹Ì¶¨ÈÝ»ýΪ5LµÄÃܱÕÈÝÆ÷ÖгäÈë0.20mol SO2ºÍ0.10molO2£¬°ë·ÖÖÓºó´ïµ½Æ½ºâ£¬²âµÃÈÝÆ÷Öк¬SO30.18mol£¬Ôòv(O2)=______mol¡¤L-1¡¤min-1

£¨3£©ÏÂÁÐÌõ¼þµÄ¸Ä±ä¶ÔÆä·´Ó¦ËÙÂʼ¸ºõÎÞÓ°ÏìµÄÊÇ         (Ñ¡ÌîÐòºÅ)

¢ÙÉý¸ßζȠ   ¢Ú±£³ÖÌå»ý²»±ä£¬Ö»Ôö¼ÓÑõÆøµÄÖÊÁ¿   ¢Û±£³ÖÌå»ý²»±ä£¬³äÈëNeʹÌåϵѹǿÔö´ó  ¢Ü±£³Öѹǿ²»±ä£¬³äÈëNeʹÈÝÆ÷µÄÌå»ýÔö´ó 

£¨4£©ÏÂÁÐÃèÊöÖÐÄÜ˵Ã÷ÉÏÊö(1)·´Ó¦ÒÑ´ïƽºâµÄÊÇ           (Ñ¡ÌîÐòºÅ)

¡¡¢Ùv(O2)Õý£½2v(SO3)Äæ¡¡¢ÚSO2¡¢O2 ¡¢SO3µÄŨ¶ÈÖ®±ÈΪ2£º1£º2

¢Ûµ¥Î»Ê±¼äÄÚÉú³É2n molSO2µÄͬʱÉú³É2n mol SO3

¢ÜÈÝÆ÷ÖÐÆøÌåµÄƽ¾ù·Ö×ÓÁ¿²»Ëæʱ¼ä¶ø±ä»¯

¡¡¡¡¢ÝÈÝÆ÷ÖÐÆøÌåµÄÃܶȲ»Ëæʱ¼ä¶ø±ä»¯  ¢ÞÈÝÆ÷ÖÐÆøÌåѹǿ²»Ëæʱ¼ä¶ø±ä»¯

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêÕã½­Ê¡º¼ÖÝÊÐÎ÷ºþ¸ßÖи߶þ3ÔÂÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨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×ÍéȼÁϵç³Ø£¨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»ìºÏÇ°Ìå»ýVËá________V¼î(Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±)
(9)£¨2·Ö£©È¡10 mLÈÜÒº0.5 mol/LµÄÑÎËᣬ¼ÓˮϡÊ͵½500 mL£¬Ôò´ËʱÈÜÒºÖÐÓÉË®µçÀë³öµÄc(H£«)£½________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêËÄ´¨Ê¡³É¶¼ÊÐÁùУЭ×÷Ìå¸ß¶þÏÂѧÆÚÆÚÖÐÁª¿¼»¯Ñ§ÊÔ¾í£¨´ø½âÎö£© ÌâÐÍ£ºÌî¿ÕÌâ

(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¨¶ÈË淴Ӧʱ¼äµÄ±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»¯ÂÊËæʱ¼ä±ä»¯µÄͼÏñÈçͼ3Ëùʾ£¬ÇëÔÚͼ3Öл­³öÒÒÈÝÆ÷ÖеÄת»¯ÂÊËæʱ¼ä±ä»¯µÄͼÏñ¡£

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

´Ëʱ·´Ó¦¢ÛÖÐÕý¡¢Äæ·´Ó¦ËÙÂʵĹØϵʽÊÇ            (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¡°=¡±£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸