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
C
̼ÍêȫȼÉշųöµÄÈÈÁ¿¶à£¬µ«·ÅÈÈÔ½¶à£¬¡÷HԽС£¬ËùÒÔ¢ÙÖЦ¤HÇ°ÕßСÓÚºóÕߣ»¢ÚÖж¼ÊÇÖкÍÈÈ£¬¡÷HÏàµÈ£»ÏûºÄµÄÇâÆøÔ½¶à£¬·Å³öµÄÈÈÁ¿Ô½¶à£¬Í¬Ñù·ÅÈÈÔ½¶à£¬¡÷HԽС£¬ËùÒÔ¢ÛÖЦ¤HÇ°Õß´óÓÚºóÕߣ»¢ÜÖÐÇ°ÕßÊÇÎüÈÈ·´Ó¦£¬ºóÕßÊÇ·ÅÈÈ·´Ó¦£¬¦¤HÇ°Õß´óÓÚºóÕߣ¬´ð°¸Ñ¡C¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

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              

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨10·Ö£©ÒÑÖª£ºa¡¢H+(aq) + OH-(aq) = H2O(l) ¡÷H=-57.3 kJ?mol-1£»
b¡¢1.6gCH4ÍêȫȼÉÕÉú³ÉË®ÕôÆøʱ·ÅÈÈ80.2kJ£¬1gË®ÕôÆøת»¯³ÉҺ̬ˮ·ÅÈÈ2.444kJ¡£
£¨1£©ÇâÑõ»¯ÄÆÓëÁòËáÁ½Ï¡ÈÜÒº·¢Éú·´Ó¦£¬Ð´³ö±íÕ÷ÆäÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ£º
                                                                      
£¨2£©Ð´³ö±íÕ÷¼×ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ£º
                                                                      
£¨3£©ÒÑÖª2SO2(g)+O2(g) 2SO3(g)·´Ó¦¹ý³ÌµÄÄÜÁ¿±ä»¯ÈçͼËùʾ¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
    
¢ÙͼÖÐC±íʾ              EµÄ´óС¶Ô¸Ã·´Ó¦µÄ·´Ó¦ÈÈÓÐÎÞÓ°Ï죿         ¡£
¢Ú¸Ã·´Ó¦Í¨³£ÓÃV2O5×÷´ß»¯¼Á£¬¼ÓV2O5»áʹͼÖÐBµãÉý¸ß»¹ÊǽµµÍ£¿     £¬ÀíÓÉÊÇ                                         £»
£¨4£©¸Ç˹¶¨ÂÉÔÚÉú²úºÍ¿ÆѧÑо¿ÖÐÓкÜÖØÒªµÄÒâÒå¡£ÓÐЩ·´Ó¦µÄ·´Ó¦ÈÈËäÈ»ÎÞ·¨Ö±½Ó²âµÃ£¬µ«¿Éͨ¹ý¼ä½ÓµÄ·½·¨²â¶¨¡£ÏÖ¸ù¾ÝÏÂÁÐ3¸öÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º
¢ÙFe2O3(s)+3CO(g)=2Fe(s)+3CO2(g)       ¡÷H£½¨D24.8 kJ?mol-1
¢Ú 3Fe2O3(s)+ CO(g)=2Fe3O4(s)+ CO2(g)     ¡÷H£½¨D47.2 kJ?mol-1
¢ÛFe3O4(s)+CO(g)=3FeO(s)+CO2(g)        ¡÷H£½ +640.5 kJ?mol-1
д³öCOÆøÌ廹ԭFeO¹ÌÌåµÃµ½Fe¹ÌÌåºÍCO2ÆøÌåµÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º
_________________________________________________¡£ 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
(1)CH3COOH(l)+2O2(g)==2CO2(g)+2H2O(l)    ¡÷H1= £­870 kJ¡¤mol-1
(2)C(s)+O2(g) ==CO2(g)                     ¡÷H2= £­393.5kJ¡¤mol-1
(3) H2(g)+O2(g)==H2O(l)                  ¡÷H3= £­285 kJ¡¤mol-1
Ôò·´Ó¦2C(s)+2H2(g)+O2(g) ==CH3COOH(l) µÄ·´Ó¦ÈÈ¡÷HΪ£¨  £©
A£®¡ª243.5 kJ¡¤mol-1B£®¡ª487 kJ¡¤mol-1C£®¡ª1113.5 kJ¡¤mol-1D£®¡ª2227 kJ¡¤mol-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£ºFe2O3(s)+3CO(g)=2Fe(s)+3CO2(g) H="-24.8" kJ¡¤mol-1
3Fe2O3(s)+CO(g)=2Fe3O4(s)+CO2(g) H="-47.2" kJ¡¤mol-1
Fe3O4(s)+CO(g)="3FeO" (s)+CO2(g) H="+640.5" kJ¡¤mol-1
Ôò14g COÆøÌåÓë×ãÁ¿FeO³ä·Ö·´Ó¦µÃµ½Feµ¥ÖʺÍCO2ÆøÌåʱµÄ·´Ó¦ÈÈΪ£¨  £©
A£®-218 kJ¡¤mol-1B£®-109kJ¡¤mol-1C£®+218 kJ¡¤mol-1D£®+109 kJ¡¤mol-1

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

(12·Ö) ¼×´¼ÊÇÒ»ÖÖ¿ÉÔÙÉúÄÜÔ´£¬¾ßÓй㷺µÄ¿ª·¢ºÍÓ¦ÓÃÇ°¾°¡£
£¨1£©¹¤ÒµÉÏÒ»°ã²ÉÓÃÏÂÁÐÁ½ÖÖ·´Ó¦ºÏ³É¼×´¼£º
·´Ó¦I£º  CO(g) £« 2H2(g)  CH3OH(g)  ¦¤H1
·´Ó¦II£º CO2(g) £« 3H2(g) CH3OH(g)  +  H2O(g)  ¦¤H2
¢ÙÉÏÊö·´Ó¦·ûºÏ¡°Ô­×Ó¾­¼Ã¡±Ô­ÔòµÄÊÇ     £¨Ìî¡°I¡±»ò¡°¢ò¡±£©¡£
¢ÚÒÑÖª·´Ó¦¢ñµÄÄÜÁ¿±ä»¯ÈçͼËùʾ£ºÓɱíÖÐÊý¾ÝÅжϠ ¦¤H1      0 £¨Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£©¡£

¢ÛijζÈÏ£¬½«2 mol COºÍ6 mol H2³äÈë2LµÄÃܱÕÈÝÆ÷ÖУ¬³ä·Ö·´Ó¦£¬´ïµ½Æ½ºâºó£¬²âµÃc(CO)£½ 0.2 mol£¯L£¬    ÔòCOµÄת»¯ÂÊΪ          
£¨2£©ÒÑÖªÔÚ³£Î³£Ñ¹Ï£º
¢Ù 2CH3OH(l) £« 3O2(g) £½ 2CO2(g) £« 4H2O(g)   ¦¤H £½£­1275.6 kJ£¯mol
¢Ú 2CO (g)+ O2(g) £½ 2CO2(g)  ¦¤H £½£­566.0 kJ£¯mol
¢Û H2O(g) £½ H2O(l)  ¦¤H £½£­44.0 kJ£¯mol
Çë¼ÆËã1 mol¼×´¼²»ÍêȫȼÉÕÉú³É1 molÒ»Ñõ»¯Ì¼ºÍҺ̬ˮ·Å³öµÄÈÈÁ¿Îª________
£¨3£©Ä³ÊµÑéС×éÒÀ¾Ý¼×´¼È¼Éյķ´Ó¦Ô­Àí£¬Éè¼ÆÈçÓÒͼËùʾµÄµç³Ø×°Öá£

¢Ù¸Ãµç³ØÕý¼«µÄµç¼«·´Ó¦Îª____________
¢Ú¹¤×÷Ò»¶Îʱ¼äºó£¬²âµÃÈÜÒºµÄpH¼õС£¬¸Ãµç³Ø×Ü·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________.

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

×î½üÒâ´óÀûÂÞÂí´óѧµÄFuNvio CacaceµÈÈË»ñµÃÁ˼«¾ßÀíÂÛÑо¿ÒâÒåµÄN4·Ö×Ó¡£N4·Ö×ӽṹÈçͼËùʾ£¬ÒÑÖª¶ÏÁÑ1molN-NÎüÊÕ193kJÈÈÁ¿£¬Éú³É1molN¡ÔN·Å³ö941kJÈÈÁ¿¡£¸ù¾ÝÒÔÉÏÐÅÏ¢ºÍÊý¾Ý£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ  (    )
A£®N4ÊôÓÚÒ»ÖÖÐÂÐ͵Ļ¯ºÏÎïB£®N4ÊÇN2µÄͬϵÎï
C£®N4ת±äΪN2ÊÇÎïÀí±ä»¯D£®1molN4ÆøÌåת±äΪN2·Å³ö724kJÄÜÁ¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÔNA´ú±í°¢·ü¼ÓµÂÂÞ³£Êý£¬Ôò¹ØÓÚ·´Ó¦£º
C2H2(g)+ 5/2 O2(g) =2CO2(g)+H2O(1)£»¡÷H=£­1300kJ£¯mol
ÏÂÁÐ˵·¨ÖУ¬ÕýÈ·µÄÊÇ
A£®ÓÐ10NA¸öµç×ÓתÒÆʱ£¬¸Ã·´Ó¦ÎüÊÕ1300kJµÄÄÜÁ¿
B£®ÓÐNA¸öË®·Ö×ÓÉú³ÉÇÒΪҺ̬ʱ£¬ÎüÊÕ1300kJµÄÄÜÁ¿
C£®ÓÐNA¸ö̼Ñõ¹²Óõç×Ó¶ÔÐγÉʱ£¬·Å³ö1300kJµÄÄÜÁ¿
D£®ÓÐ8NA¸ö̼Ñõ¹²Óõç×Ó¶ÔÐγÉʱ£¬·Å³ö1300kJµÄÄÜÁ¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÔÚ1¡Á105 Pa£¬298 KÌõ¼þÏ£¬2 molÇâÆøȼÉÕÉú³ÉË®ÕôÆø·Å³ö484 kJÈÈÁ¿£¬ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ      
A£®H2O ( g ) £½ H2 ( g ) + 1/2O2 ( g )¡÷H =" +242" kJ/mol
B£®2H2 ( g ) + O2 ( g ) = 2H2O ( l )¡÷H = £­484 kJ/mol
C£®H2 ( g ) + 1/2O2 ( g ) = H2O ( g )¡÷H =" +242" kJ/mo
D£®2H2 ( g ) + O2 ( g ) = 2H2O ( g )¡÷H =" +484" kJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸