°Ñú×÷ΪȼÁÏ¿Éͨ¹ýÏÂÁÐÁ½ÖÖ;¾¶£º
;¾¶I£ºC(s) +O2 (g) == CO2(g)     ¡÷H1<0                     ¢Ù
;¾¶II£ºÏÈÖƳÉˮúÆø£ºC(s) +H2O(g) == CO(g)+H2(g)  ¡÷H2>0    ¢Ú
ÔÙȼÉÕˮúÆø£º2CO(g)+O2 (g) == 2CO2(g)  ¡÷H3<0               ¢Û
2H2(g)+O2 (g) == 2H2O(g)  ¡÷H4<0               ¢Ü
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£© ;¾¶I·Å³öµÄÈÈÁ¿          ( Ìî¡°´óÓÚ¡±¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±) ;¾¶II·Å³öµÄÈÈÁ¿¡£
£¨2£© ¡÷H1¡¢¡÷H2¡¢¡÷H3¡¢¡÷H4µÄÊýѧ¹ØϵʽÊÇ                      ¡£
£¨3£©12gÌ¿·ÛÔÚÑõÆøÖв»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼£¬·Å³ö110£®35kJÈÈÁ¿¡£ÆäÈÈ»¯Ñ§·½³ÌʽΪ                                              ¡£
£¨4£©ÃºÌ¿×÷ΪȼÁϲÉÓÃ;¾¶IIµÄÓŵãÓР                             

£¨1£©µÈÓÚ £»   £¨2£©¡÷H1=¡÷H2+£¨¡÷H3+¡÷H4£©
£¨3£©C(s) +O2 (g) = CO(g)   ¡÷H=-110£®35 kJ¡¤mol-1
£¨4£©È¼ÁÏȼÉÕ³ä·Ö£¬ÀûÓÃÂʸߣ¬·ÅÈȶ࣬ÎÛȾС¡£

½âÎöÊÔÌâ·ÖÎö£ºÓɸÇ˹¶¨ÂÉÖª·´Ó¦·ÅÈÈÓë;¾¶Î޹أ¬ËùÒÔ;¾¶¢ñ·Å³öµÄÈÈÁ¿µÈÓÚ;¾¶¢ò·Å³öµÄÈÈÁ¿£»ÇÒ²»ÄÑÈ·¶¨¡÷H1¡¢¡÷H2¡¢¡÷H3¡¢¡÷H4µÄÊýѧ¹ØϵʽÊÇ¡÷H1=¡÷H2+1/2£¨¡÷H3+¡÷H4£©;
¿¼µã£ºÈÈ»¯Ñ§·½³ÌʽµÄÊéд¼°¸Ç˹¶¨ÂÉ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÒÑÖª£º¢ÙCO(g) + 1/2O2(g) = CO2(g)     ¡÷H= £­283.0kJ¡¤mol£­1
¢ÚCH3OH(l) + 3/2O2(g) = CO2(g)+2H2O(l)  ¡÷H= £­726.5kJ¡¤mol£­1  
Çëд³ö¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ£º____________________________________________________________________£»
¢ÚÒÑÖª²ð¿ª1molH£­H¼ü¡¢1molCl£­Cl¼ü¡¢1molH¡ªCl¼ü·Ö±ðÐèÒªµÄÄÜÁ¿ÊÇ436kJ¡¢243kJ¡¢432kJ£¬Ôò·´Ó¦Ôò·´Ó¦£ºH2(g)+ Cl2(g)=2HCl (g) µÄ¡÷H=         ¡£
£¨2£©ÒÑÖª25¡æ¡¢101 kPaÏ£¬Ï¡µÄÇ¿ËáÓëÏ¡µÄÇ¿¼îÈÜÒº·´Ó¦µÄÖкÍÈÈΪ -57.3 kJ/mol¡£
¢ÙÔò±íʾϡÁòËáÓëÏ¡ÉÕ¼îÈÜÒºÖкͷ´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º                       ¡£
¢Ú²â¶¨ÖкÍÈÈʵÑéʱËùÐèµÄ²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢Á¿Í²¡¢       ¡¢               ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¡°½ÚÄܼõÅÅ¡±£¬¼õÉÙÈ«ÇòÎÂÊÒÆøÌåÅÅ·Å£¬ÒâÒåÊ®·ÖÖش󡣶þÑõ»¯Ì¼µÄ²¶×½Óë·â´æÊÇʵÏÖÎÂÊÒÆøÌå¼õÅŵÄÖØҪ;¾¶Ö®Ò»£¬¿Æѧ¼ÒÀûÓÃÈÜÒºÅçÁÜ¡°²¶×½¡±¿ÕÆøÖеġ£
£¨1£©Ê¹ÓùýÁ¿ÈÜÒºÎüÊÕ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ________£»Èôº¬ÓÐ3molNaOHµÄÈÜÒº¡°²¶×½¡±ÁË22£®4LÆøÌ壨±ê×¼×´¿ö£©£¬ÔòËùµÃÈÜÒºÖÐÄÆÓë̼ԪËصÄÎïÁÏÊغã¹ØϵʽΪ__________£¨ÓÃÀë×ÓŨ¶ÈµÄ¹Øϵʽ±íʾ£©¡£
£¨2£©¢ÙÒÔºÍΪԭÁϿɺϳɻ¯·ÊÄòËØ[]¡£ÒÑÖª£º
   ¢Ù
 ¢Ú
  ¢Û
ÊÔд³öºÍºÏ³ÉÄòËغÍҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ__________¡£
¢Úͨ¹ý·´Ó¦¿Éת»¯Îª£¬ÔÚ´ß»¯¼Á×÷ÓÃÏÂCOºÍ·´Ó¦Éú³É¼×´¼£ºÄ³ÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖгäÓÐ10molCOÓë20mol£¬COµÄƽºâת»¯ÂÊ£¨a£©Óëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçÏÂͼËùʾ¡£

A£®ÈôAµã±íʾÔÚijʱ¿Ì´ïµ½µÄƽºâ״̬£¬´ËʱÈÝÆ÷µÄÈÝ»ýΪVL£¬Ôò¸ÃζÈϵÄƽºâ³£ÊýK=__________£»Æ½ºâ״̬BµãʱÈÝÆ÷µÄÈÝ»ý_______VL¡££¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©
B£®ÈôA¡¢CÁ½µã¶¼±íʾ´ïµ½µÄƽºâ״̬£¬Ôò×Ô·´Ó¦¿ªÊ¼µ½´ïƽºâ״̬ËùÐèµÄʱ¼ä_______£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©
C£®ÔÚ²»¸Ä±ä·´Ó¦ÎïÓÃÁ¿µÄÇé¿öÏ£¬ÎªÌá¸ßCOµÄת»¯ÂʿɲÉÈ¡µÄ´ëÊ©ÊÇ________£¨Ð´³öÒ»ÖÖ¼´¿É£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÒÑÖªH2£¨g£©ºÍCO£¨g£©µÄȼÉÕÈÈ¡÷H·Ö±ðΪ-285.8kJ¡¤mol-1¡¢-283.0kJ¡¤mol¡£  
ÔòÒ»Ñõ»¯Ì¼ÓëҺ̬ˮ·´Ó¦£¬Éú³É¶þÑõ»¯Ì¼ºÍÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽΪ   
________________________________________________________£»

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨8·Ö£©°Ñú×÷ΪȼÁÏ¿Éͨ¹ýÏÂÁÐÁ½ÖÖ;¾¶£º
;¾¶I£ºC(s) +O2 (g)=CO2(g)     ¡÷H1<0                    ¢Ù
;¾¶II£ºÏÈÖƳÉˮúÆø£ºC(s) +H2O(g)=CO(g)+H2(g)  ¡÷H2>0   ¢Ú
ÔÙȼÉÕˮúÆø£º2CO(g)+O2 (g)=2CO2(g)  ¡÷H3<0               ¢Û
2H2(g)+O2 (g)=2H2O(g)  ¡÷H4<0                ¢Ü
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Í¾¾¶I·Å³öµÄÈÈÁ¿  ( Ìî¡°´óÓÚ¡±¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±) ;¾¶II·Å³öµÄÈÈÁ¿¡£
£¨2£©¡÷H1¡¢¡÷H2¡¢¡÷H3¡¢¡÷H4µÄÊýѧ¹ØϵʽÊÇ              ¡£
£¨3£©12gÌ¿·ÛÔÚÑõÆøÖв»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼£¬·Å³ö110.35kJÈÈÁ¿¡£ÆäÈÈ»¯Ñ§·½³ÌʽΪ                                             ¡£
£¨4£©ÃºÌ¿×÷ΪȼÁϲÉÓÃ;¾¶IIµÄÓŵãÓР                                             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨1£©Ä³»¯Ñ§ÐËȤС×éÒªÍê³É·´Ó¦ÈȵIJⶨ¡£ÊµÑé×ÀÉϱ¸ÓÐÉÕ±­£¨´ó¡¢Ð¡Á½¸öÉÕ±­£©¡¢ÅÝÄ­ËÜÁÏ¡¢ÅÝÄ­ËÜÁÏ°å¡¢½ºÍ·µÎ¹Ü¡¢»·Ðβ£Á§½Á°èÆ÷¡¢0£®50mol¡¤ L£­1ÑÎËá¡¢0£®50mol¡¤ L£­1NaOHÈÜÒº£¬ÊµÑéÉÐȱÉٵIJ£Á§ÓÃÆ·ÊÇ_____________¡¢_______________¡£
£¨2£©ÒÑÖª2molCOÆøÌåÍêȫȼÉÕÉú³ÉCO2 ÆøÌå·Å³ö566 kJÈÈÁ¿£»1 molÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö286 kJÈÈÁ¿£»1 molCH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍҺ̬ˮ·Å³ö890 kJÈÈÁ¿¡£Ð´³öÓñê׼ȼÉÕÈÈ×÷Ϊ·´Ó¦ÈȵÄCOȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ__________________¡£
Èô1 molCH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2¹ÌÌåºÍҺ̬ˮ£¬·Å³öÈÈÁ¿_____890 kJ£¨ Ìî¡°£¾¡±¡¢ ¡°£¼¡±¡¢¡°=¡±£©¡£Èô½«a molCH4¡¢COºÍH2µÄ»ìºÏÆøÌåÍêȫȼÉÕ£¬Éú³É CO2ÆøÌåºÍҺ̬ˮ£¬ÇÒCO2ºÍË®µÄÎïÖʵÄÁ¿ÏàµÈʱ£¬Ôò·Å³öÈÈÁ¿£¨Q£©µÄµÄÈ¡Öµ·¶Î§ÊÇ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÇâÑõÁ½ÖÖÔªËØÐγɵij£¼ûÎïÖÊÓÐH2OÓëH2O2£¬ÔÚÒ»¶¨Ìõ¼þϾù¿É·Ö½â¡£
£¨1£©ÒÑÖª£º

»¯Ñ§¼ü
¶Ï¿ª1mol»¯Ñ§¼üËùÐèµÄÄÜÁ¿£¨kJ£©
H¡ªH
436
O¡ªH
463
O=O
498
¢ÙH2OµÄµç×ÓʽÊÇ       ¡£
¢ÚH2O£¨g£©·Ö½âµÄÈÈ»¯Ñ§·½³ÌʽÊÇ       ¡£
¢Û11.2 L£¨±ê×¼×´¿ö£©µÄH2ÍêȫȼÉÕ£¬Éú³ÉÆø̬ˮ£¬·Å³ö       kJµÄÈÈÁ¿¡£
£¨2£©Ä³Í¬Ñ§ÒÔH2O2·Ö½âΪÀý£¬Ì½¾¿Å¨¶ÈÓëÈÜÒºËá¼îÐÔ¶Ô·´Ó¦ËÙÂʵÄÓ°Ïì¡£³£ÎÂÏ£¬°´ÕÕÈç±íËùʾµÄ·½°¸Íê³ÉʵÑé¡£
ʵÑé±àºÅ
·´Ó¦Îï
´ß»¯¼Á
a
50 mL 5% H2O2ÈÜÒº
 
1 mL 0.1 mol¡¤L-1 FeCl3ÈÜÒº
b
50 mL 5% H2O2ÈÜÒº
ÉÙÁ¿Å¨ÑÎËá
1 mL 0.1 mol¡¤L-1 FeCl3ÈÜÒº
c
50 mL 5% H2O2ÈÜÒº
ÉÙÁ¿Å¨NaOHÈÜÒº
1 mL 0.1 mol¡¤L-1 FeCl3ÈÜÒº
d
50 mL 5% H2O2ÈÜÒº
 
MnO2
¢Ù²âµÃʵÑéa¡¢b¡¢cÖÐÉú³ÉÑõÆøµÄÌå»ýËæʱ¼ä±ä»¯µÄ¹ØϵÈçͼ1Ëùʾ¡£
 
ͼ1                              Í¼2
ÓɸÃͼÄܹ»µÃ³öµÄʵÑé½áÂÛÊÇ_________¡£
¢Ú²âµÃʵÑédÔÚ±ê×¼×´¿öÏ·ųöÑõÆøµÄÌå»ýËæʱ¼ä±ä»¯µÄ¹ØϵÈçͼ2Ëùʾ¡£½âÊÍ·´Ó¦ËÙÂʱ仯µÄÔ­Òò£º         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

½ðÊôÎÙÓÃ;¹ã·º£¬Ö÷ÒªÓÃÓÚÖÆÔìÓ²ÖÊ»òÄ͸ßεĺϽð£¬ÒÔ¼°µÆÅݵĵÆË¿¡£¸ßÎÂÏ£¬ÔÚÃܱÕÈÝÆ÷ÖÐÓÃH2»¹Ô­WO3¿ÉµÃµ½½ðÊôÎÙ£¬Æä×Ü·´Ó¦Îª£º
WO3 (s) + 3H2 (g)W (s) + 3H2O (g)
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÅÉÏÊö·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪ___________________________¡£
¢ÆijζÈÏ·´Ó¦´ïƽºâʱ£¬H2ÓëË®ÕôÆøµÄÌå»ý±ÈΪ2:3£¬ÔòH2µÄƽºâת»¯ÂÊΪ_____________________£»ËæζȵÄÉý¸ß£¬H2ÓëË®ÕôÆøµÄÌå»ý±È¼õС£¬Ôò¸Ã·´Ó¦Îª·´Ó¦_____________________£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©¡£
¢ÇÉÏÊö×Ü·´Ó¦¹ý³Ì´óÖ·ÖΪÈý¸ö½×¶Î£¬¸÷½×¶ÎÖ÷Òª³É·ÖÓëζȵĹØϵÈçϱíËùʾ£º

ζÈ
25¡æ  ~  550¡æ  ~  600¡æ  ~  700¡æ
Ö÷Òª³É·Ý
WO3      W2O5      WO2        W
 
µÚÒ»½×¶Î·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________________£»580¡æʱ£¬¹ÌÌåÎïÖʵÄÖ÷Òª³É·ÖΪ________£»¼ÙÉèWO3Íêȫת»¯ÎªW£¬ÔòÈý¸ö½×¶ÎÏûºÄH2ÎïÖʵÄÁ¿Ö®±ÈΪ____________________________________¡£
¢È ÒÑÖª£ºÎ¶ȹý¸ßʱ£¬WO2 (s)ת±äΪWO2 (g)£»
WO2 (s) + 2H2 (g)  W (s) + 2H2O (g)£»¦¤H £½ +66.0 kJ¡¤mol£­1
WO2 (g) + 2H2(g)  W (s) + 2H2O (g)£»¦¤H £½ £­137.9 kJ¡¤mol£­1
ÔòWO2 (s)  WO2 (g) µÄ¦¤H £½ ______________________¡£
¢ÉÎÙË¿µÆ¹ÜÖеÄWÔÚʹÓùý³ÌÖлºÂý»Ó·¢£¬Ê¹µÆË¿±äϸ£¬¼ÓÈëI2¿ÉÑÓ³¤µÆ¹ÜµÄʹÓÃÊÙÃü£¬Æ乤×÷Ô­ÀíΪ£ºW (s) +2I2 (g)WI4 (g)¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÓÐ____________¡£
a£®µÆ¹ÜÄÚµÄI2¿ÉÑ­»·Ê¹ÓÃ
b£®WI4ÔÚµÆË¿ÉϷֽ⣬²úÉúµÄWÓÖ³Á»ýÔÚµÆË¿ÉÏ
c£®WI4ÔڵƹܱÚÉϷֽ⣬ʹµÆ¹ÜµÄÊÙÃüÑÓ³¤   
d£®Î¶ÈÉý¸ßʱ£¬WI4µÄ·Ö½âËÙÂʼӿ죬WºÍI2µÄ»¯ºÏËÙÂʼõÂý

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Íê³ÉÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£¨»¯Ñ§·½³Ìʽ¡¢µç¼«·´Ó¦Ê½¡¢±í´ïʽµÈ£©µÄÊéд£º
£¨1£©ÒÑÖª£º2Cu(s)£«1/2O2(g)=Cu2O(s)£»¡÷H = -169kJ¡¤mol-1£¬
C(s)£«1/2O2(g)=CO(g)£»¡÷H = -110.5kJ¡¤mol-1£¬
Cu(s)£«1/2O2(g)=CuO(s)£»¡÷H = -157kJ¡¤mol-1
ÓÃÌ¿·ÛÔÚ¸ßÎÂÌõ¼þÏ»¹Ô­CuOÉú³ÉCu2OµÄÈÈ»¯Ñ§·½³ÌʽÊÇ£º                      
£¨2£©ÔÚÒ»¶¨Ìõ¼þÏ£¬¶þÑõ»¯ÁòºÍÑõÆø·¢ÉúÈçÏ·´Ó¦£º2SO2(g)+O2(g)2SO3(g)£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽ£º                      
£¨3£©ÒÔ¼×Íé¡¢¿ÕÆøΪ·´Ó¦ÎKOHÈÜÒº×÷µç½âÖÊÈÜÒº¹¹³ÉȼÁϵç³Ø£¬Ôò¸º¼«·´Ó¦Ê½Îª£º                 ¡£
£¨4£©ÎÞË®AlCl3Æ¿¸Ç´ò¿ªÓа×Îí£¬Æä·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                 ¡£
£¨5£©¡°Ã¾¡ª´ÎÂÈËáÑΡ±È¼Áϵç³Ø£¬Æä×°ÖÃʾÒâͼÈçͼ£¬¸Ãµç³Ø·´Ó¦µÄ×Ü·´Ó¦·½³ÌʽΪ_____________________¡£

£¨6£©¹¤ÒµÉϵç½â±¥ºÍʳÑÎË®µÄÀë×Ó·½³ÌʽΪ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸