ÓÃH2O2ºÍH2SO4µÄ»ìºÏÈÜÒº¿ÉÈܳöÓ¡Ë¢µç·°å½ðÊô·ÛÄ©ÖеÄÍ­¡£ÒÑÖª£º
Cu(s)£«2H£«(aq)=Cu2£«(aq)£«H2(g) ¡÷H=64.39kJ¡¤mol£­1
2H2O2(l)=2H2O(l)£«O2(g) ¡÷H=£­196.46kJ¡¤mol£­1
H2(g)£«O2(g)=H2O(l) ¡÷H=£­285.84kJ¡¤mol£­1 
ÔÚ H2SO4ÈÜÒºÖÐCuÓëH2O2·´Ó¦Éú³ÉCu2£«ºÍH2OµÄÈÈ·½³ÌʽΪ              
Cu(s)+H2O2(l)+2H£«(aq)=Cu2£«(aq)+2H2O(l) ¡÷H=-319.68kJ.mol-1
¿¼²é¸Ç˹¶¨ÂɵÄÓ¦Ó᣸ù¾Ý¸Ç˹¶¨ÂÉ¿ÉÖª£¬¢Ù£«¢Ú¡Â2£«¢Û¼´µÃµ½Cu(s)+H2O2(l)+2H£«(aq)=Cu2£«(aq)+2H2O(l)£¬ËùÒԸõķ´Ó¦ÈÈ¡÷H=64.39kJ¡¤mol£­1£­196.46kJ¡¤mol£­1¡Â2£­285.84kJ¡¤mol£­1£½£­319.68kJ.mol-1¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨4·Ö£©ÒÑÖª25¡æ¡¢101kpaʱ£¬CH3OH(l)µÄȼÉÕÈÈΪ726.5 KJ/mol¡£
¢Åд³ö¸ÃÌõ¼þÏÂCH3OH(l)ÍêȫȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º                           ¡£
¢ÆÒÑÖªÓйط´Ó¦µÄÄÜÁ¿±ä»¯ÈçÏÂͼ£¬Ôò·´Ó¦CH4+H2O(g)CO+3H2µÄìʱä¡÷H=_______¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£º¼ÆËãÌâ

ÒÑÖªÏÂÁз´Ó¦µÄ·´Ó¦ÈÈ£º
£¨1£©CH3COOH(l)+2O2(g)=2CO2(g)+2H2O(l)     ¦¤H1=£­870.3kJ/mol
£¨2£©C(s)+O2(g)=CO2(g)      ¦¤H2=£­393.5kJ/mol
£¨3£©H2(g)+1/2O2(g)£½H2O(l)       ¦¤H3=£­285.8kJ/mol
ÊÔ¼ÆËãÏÂÊö·´Ó¦µÄ·´Ó¦ÈÈ2C(s)+O2(g)+2 H2(g)£½CH3COOH(l)

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨19·Ö£©ÎÒ¹ú¡¶³µÓÃȼÁϼ״¼¹ú¼Ò±ê×¼¡·µÄʵʩÀ­¿ªÁ˳µÓÃȼÁϵÍ̼¸ïÃüµÄ´óÄ»£¬Ò»Ð©Ê¡ÊÐÕýÔÚ½ÐøÊÔµãÓëÍƹãʹÓü״¼ÆûÓÍ¡£¼×´¼¿Éͨ¹ý½«ÃºµÄÆø»¯¹ý³ÌÖÐÉú³ÉµÄCOºÍH2ÔÚÒ»¶¨Ìõ¼þÏ·¢ÉúÈçÏ·´Ó¦ÖƵãºCO(g) + 2H2(g) CH3OH(g)¡£Í¼I¡¢Í¼¢òÊǹØÓڸ÷´Ó¦½øÐÐÇé¿öµÄͼʾ¡£

Çë¸ù¾Ýͼʾ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Í¼IÊÇ·´Ó¦Ê±COºÍCH3OHµÄŨ¶ÈËæʱ¼äµÄ±ä»¯Çé¿ö£¬´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬ÓÃCOŨ¶È±ä»¯±íʾƽ¾ù·´Ó¦ËÙÂÊv(CO)=_______________¡£
£¨2£©Í¼¢ò±íʾ¸Ã·´Ó¦½øÐйý³ÌÖÐÄÜÁ¿µÄ±ä»¯£¬ÇúÏßa±íʾ²»Ê¹Óô߻¯¼Áʱ·´Ó¦µÄÄÜÁ¿±ä»¯£¬ÔÚͼ¢òÖл­³öʹÓô߻¯¼ÁºóµÄÄÜÁ¿±ä»¯ÇúÏßb¡£
£¨3£©Ð´³ö¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ                                  ¡£
£¨4£©¸Ã·´Ó¦µÄƽºâ³£ÊýKµÄ±í´ïʽΪ                 £»µ±Î¶ÈÉý¸ßʱ£¬¸Ãƽºâ³£ÊýK½«________£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©¡£
£¨5£©ºãÈÝÌõ¼þÏ£¬ÏÂÁдëÊ©ÖÐÄÜʹÔö´óµÄÓР        ¡£
A£®Éý¸ßζȠ                             B£®³äÈËHeÆø   
C£®ÔÙ³äÈë1molCOºÍ2molH2               D£®Ê¹Óô߻¯¼Á
£¨6£©ÔÚºãÎÂÌõ¼þÏ£¬±£³ÖCOŨ¶È²»±ä£¬À©´óÈÝÆ÷Ìå»ý£¬Ôòƽºâ      £¨Ìî¡°ÄæÏòÒƶ¯¡±¡¢¡°ÕýÏòÒƶ¯¡±¡¢¡°²»Òƶ¯¡±£©
£¨7£©ÔÚζȡ¢ÈÝ»ýÏàͬµÄÈý¸öÃܱÕÈÝÆ÷ÖУ¬°´²»Í¬·½Ê½Í¶ÁÏ£¬±£³ÖºãΡ¢ºãÈÝ£¬²âµÃ·´Ó¦´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏÂ
ÈÝÆ÷
¼×
ÒÒ
±û
ͶÁÏÁ¿
1mol CO ¡¢2mol H2
1molCH3OH
2molCH3OH
CH3OHµÄŨ¶È£¨mol¡¤L-1£©
c1
c2
c3
·´Ó¦µÄÄÜÁ¿±ä»¯
·Å³öQ1kJ
ÎüÊÕQ2kJ
ÎüÊÕQ3kJ
Ìåϵѹǿ£¨Pa£©
P1
P2
P3
·´Ó¦Îïת»¯ÂÊ
¦Á1
¦Á2
¦Á3
ÔòÏÂÁйØϵÕýÈ·µÄÊÇ          
A£®c1= c2           B£®Q3= 2Q2        C£®2 P1£¼P3 
D£®¦Á1+¦Á2=1      E£®2¦Á2=¦Á3
F£®¸Ã·´Ó¦ÈôÉú³É1molCH3OH·Å³öµÄÈÈÁ¿Îª£¨Q1+ Q2£©kJ
£¨8£©ÈôÔÚÒ»Ìå»ý¿É±äµÄÃܱÕÈÝÆ÷ÖгäÈë1mol CO¡¢2mol H2ºÍ1molCH3OH£¬´ïµ½Æ½ºâʱ²âµÃ»ìºÏÆøÌåµÄÃܶÈÊÇͬÎÂͬѹÏÂÆðʼµÄ1.6±¶£¬Ôò¸Ã·´Ó¦Ïò      £¨Ìî¡°Õý¡±¡¢¡°Ä桱£©·´Ó¦·½ÏòÒƶ¯£¬ÀíÓÉÊÇ                                        

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

25¡æ¡¢101 kPaÏ£¬2gÇâÆøȼÉÕÉú³ÉҺ̬ˮ£¬·Å³ö285.8kJÈÈÁ¿£¬±íʾ¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ
A£®2H2(g)+O2(g)£½2H2O(1)¡÷H= ¨D285.8kJ£¯mol
B£®2H2(g)+ O2(g)£½2H2O(1)¡÷H=" +571.6" kJ£¯mol
C£®2H2(g)+O2(g)£½2H2O(g)¡÷H=" ¨D571.6" kJ£¯mol
D£®H2(g)+O2(g)£½H2O(1)¡÷H = ¨D285.8kJ£¯mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÂÈÆø¡¢äåÕôÆø·Ö±ð¸úÇâÆø·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÈçÏÂ(Q 1¡¢Q2¾ùΪÕýÖµ)£º
H2(g)£«Cl2(g) = 2HCl(g)¡¡¡÷H1£½£­Q1 kJ¡¤mol-1
H2(g)£«Br2(g) = 2HBr(g)¡¡¡÷H2£½£­Q2 kJ¡¤mol-1
ÓйØÉÏÊö·´Ó¦µÄÐðÊöÕýÈ·µÄÊÇ(¡¡¡¡)
A£®Q1< Q2
B£®Éú³ÉÎï×ÜÄÜÁ¿¾ù¸ßÓÚ·´Ó¦Îï×ÜÄÜÁ¿
C£®1 mol HBr(g)¾ßÓеÄÄÜÁ¿¸ßÓÚ1 mol HBr(l)¾ßÓеÄÄÜÁ¿
D£®Éú³É1 mol HClÆøÌåʱ·Å³öQ1ÈÈÁ¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÔÚ101kPaºÍ25¡ãCʱ£¬Óйط´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÈçÏ£º
C(s)+1/2 O2(g)=CO(g) ¡÷H1= -110.5KJ/mol
  =
    =
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨     £©
A£®£»
B£®È¼ÉÕÈȵĻ¯Ñ§·½³ÌʽΪ£º
C£®
D£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙCH4(g)+2O2(g)==CO2(g)+2H2O(l)    ¡÷H=-889.5kJ/mol
¢Ú2C2H6(g)+7O2(g)==4CO2(g)+6H2O(l) ¡÷H=-3116.7kJ/mol
¢ÛC2H4(g)+3O2(g)==2CO2(g)+2H2O(l)  ¡÷H=-1409.6kJ/mol
¢Ü2C2H2(g)+5O2(g)==4CO2(g)+2H2O(l) ¡÷H=-2596.7kJ/mol
¢ÝC3H8(g)+5O2(g)==3CO2(g)+4H2O(l)  ¡÷H=-2217.8kJ/mol
ÏÖÓÐÓÉÉÏÊöÎåÖÖÌþÖеÄÁ½ÖÖ×éºÏ³ÉµÄ»ìºÏÆøÌå2mol£¬¾­³ä·ÖȼÉÕºó·Å³ö3037.6kJÈÈÁ¿£¬ÔòÏÂÁÐÄÄЩ×éºÏÊDz»¿ÉÄܵĠ   £¨ £©
A£®C2H4ºÍC2H6B£®C2H2ºÍC3H8C£®C2H6ºÍC3H8D£®C2H6ºÍCH4

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁи÷×éÈÈ»¯Ñ§·½³ÌʽÖУ¬»¯Ñ§·´Ó¦µÄ¦¤HÇ°Õß´óÓÚºóÕßµÄÊÇ(¡¡¡¡)
¢ÙC(s)£«O2(g)===CO2(g)¡¡¦¤H1¡¡ C(s)£«1/2O2(g)===CO(g)¡¡¦¤H2
¢ÚNaOH(aq)£«HCl(aq)===NaCl(aq)£«H2O(l)¡¡¦¤H3¡¡
NaOH(aq)£«HNO3(aq)===NaNO3(aq)£«H2O(l)¡¡¦¤H4
¢ÛH2(g)£«1/2O2(g)===H2O(l)¡¡¦¤H5¡¡2H2(g)£«O2(g)===2H2O(l)¡¡¦¤H6
¢ÜCaCO3(s)===CaO(s)£«CO2(g)¡¡¦¤H7¡¡CaO(s)£«H2O(l)===Ca(OH)2(s)¡¡¦¤H8
A£®¢ÙB£®¢Ú¢ÜC£®¢Û¢ÜD£®¢Ù¢Ú¢Û

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸