Ç¿ËáºÍÇ¿¼îÔÚÏ¡ÈÜÒºÖеÄÖкÍÈȿɱíʾΪ£º
H£«(aq)+OH£­(aq)£½H2O(l)£» ¡÷H£½£­57.3 kJ¡¤mol-1£¬ÓÖÖªÔÚÈÜÒºÖз´Ó¦ÓУº
CH3COOH(aq)+NaOH(aq)£½CH3COONa(aq)+H2O(l)£»¡÷H£½£­Q1kJ¡¤mol-1,
1/2H2SO4(Ũ)+NaOH(aq)£½1/2Na2SO4(aq)+H2O(l) £»¡÷H£½£­Q2 kJ¡¤mol-1
HNO3(aq)+KOH(aq)£½ KNO3(aq)+H2O(l)£»¡÷H£½£­Q3 kJ¡¤mol-1
ÔòQ1¡¢Q2¡¢Q3µÄ¹ØϵÕýÈ·µÄÊÇ
A£®Q2 > Q3 > Q1B£®Q2 > Q1 > Q3C£®Q1 = Q2 = Q3D£®Q2 = Q3 > Q1
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÈÈ»¯Ñ§·½³Ìʽ£º2SO2(g)£«O2(g)  2SO3(g)£»¦¤H£½£­Q kJ¡¤mol£­1(Q£¾0)¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
A£®ÏàͬÌõ¼þÏ£¬2 mol SO2(g)ºÍ1 mol O2(g)Ëù¾ßÓеÄÄÜÁ¿Ð¡ÓÚ2 mol SO3(g)Ëù¾ßÓеÄÄÜÁ¿
B£®½«2 mol SO2(g)ºÍ1 mol O2(g)ÖÃÓÚÒ»ÃܱÕÈÝÆ÷Öгä·Ö·´Ó¦ºó£¬·Å³öÈÈÁ¿ÎªQ kJ
C£®Ôö´óѹǿ»òÉý¸ßζȣ¬¸Ãƽºâ¶¼ÏòÄæ·´Ó¦·½ÏòÒƶ¯
D£®È罫һ¶¨Á¿SO2(g)ºÍO2(g)ÖÃÓÚijÃܱÕÈÝÆ÷Öгä·Ö·´Ó¦·ÅÈÈQ kJ£¬Ôò´Ë¹ý³ÌÖÐÓÐ2 mol SO2(g)±»Ñõ»¯

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

£¨1£©ÒÑÖª£ºÔÚ1¡Á105PaÌõ¼þÏ£¬ÇâÆøµÄ±ê׼ȼÉÕÈÈÊÇ285.8 kJ¡¤mol£­1,ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ £¨       £©
A£®H2O£¨g£©£½H2£¨g£©£«1/2O2£¨g£©¦¤H£½£«285.8kJ¡¤mol£­1
B£®2H2£¨g£©£«O2£¨g£©£½2H2O£¨l£©¦¤H£½¡ª517.6kJ¡¤mol£­1
C£®H2£¨g£©£«1/2 O2£¨g£©£½H2O£¨g£©¦¤H£½¡ª285.8kJ¡¤mol£­1-
D£®2H2£¨g£©£«O2£¨g£©£½2H2O£¨l£©¦¤H£½£«517.6J¡¤mol£­1
£¨2£©¿Æѧ¼Ò¸Ç˹ÔøÌá³ö£º¡°²»¹Ü»¯Ñ§¹ý³ÌÊÇÒ»²½Íê³É»ò·Ö¼¸²½Íê³É£¬Õâ¸ö×ܹý³ÌµÄÈÈЧӦÊÇÏàͬµÄ¡£¡±ÀûÓøÇ˹¶¨ÂɿɲâijЩÌرð·´Ó¦µÄÈÈЧӦ¡£
¢Ù P4£¨s£¬°×Á×£©+ 5O2£¨g£©=P4O10(s)£»          ¦¤H£½¡ª2983.2kJ¡¤mol£­1  
¢Ú  P£¨s£¬ºìÁ×£©+  5/4 O2 (g) ="1/4" P4O10(s)   ¦¤H£½¡ª738.5kJ¡¤mol£­1-
Ôò°×Á×ת»¯ÎªºìÁ×µÄÈÈ»¯Ñ§·½³Ìʽ_________________________________¡£ÏàͬµÄ×´¿öÏ£¬ÄÜÁ¿½ÏµÍµÄÊÇ_________£»°×Á×µÄÎȶ¨ÐԱȺìÁ×___________£¨Ìî¡°¸ß¡±»ò¡°µÍ¡±£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

¸Ç˹¶¨ÂÉÔÚÉú²úºÍ¿ÆѧÑо¿ÖÐÓкÜÖØÒªµÄÒâÒå¡£ÓÐЩ·´Ó¦µÄ·´Ó¦ÈÈËäÈ»ÎÞ·¨Ö±½Ó²âµÃ£¬µ«¿Éͨ¹ý¼ä½ÓµÄ·½·¨²â¶¨¡£ÏÖ¸ù¾ÝÏÂÁÐ3¸öÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º
Fe2O3(s)+3CO(g)=2Fe(s)+3CO2(g)       ¡÷H= ¨D24.8kJ£¯mol
3Fe2O3(s)+ CO(g)==2Fe3O4(s)+ CO2(g)   ¡÷H= ¨D47.2kJ£¯mol
Fe3O4(s)+CO(g)==3FeO(s)+CO2(g)     ¡¡¡÷H= +640.5kJ£¯mol
д³öCOÆøÌ廹ԭFeO¹ÌÌåµÃµ½Fe¹ÌÌåºÍCO2ÆøÌåµÄÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º
¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡           ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁзÖÎöÍƶÏÓдíÎóµÄÊÇ
A£®¡°±ù£¬Ë®ÎªÖ®£¬¶øº®ÓÚË®¡±ËµÃ÷ÏàͬÖÊÁ¿µÄË®ºÍ±ù£¬±ùµÄÄÜÁ¿½ÏµÍ
B£®Ä³Î¶ÈÏ£¬Ïò±¥ºÍʳÑÎË®ÖÐͨÈëHClÆøÌåûÓÐÃ÷ÏÔµÄÏÖÏó·¢Éú
C£®ÏàͬζÈÏ£¬°ÑË®ÃæÉϵĿÕÆø»»³ÉÏàͬѹÁ¦ÏµĴ¿Ñõ£¬100gË®ÖÐÈÜÈëÑõÆøµÄÖÊÁ¿Ôö¼Ó
D£®½«ÍâÐβ»¹æÔòµÄµ¨·¯¾§ÌåͶÈë±¥ºÍÁòËáÍ­ÈÜÒºÖУ¬Í¨¹ý¹Û²ì¾§ÌåÍâÐεı仯¶ø¾§ÌåÖÊÁ¿±£³Ö²»±ä£¬ÄÜÖ¤Ã÷ÈܽâƽºâÊÇÒ»¸ö¶¯Ì¬Æ½ºâ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐ˵·¨»ò±íʾ·½·¨ÕýÈ·µÄÊÇ
A£®·´Ó¦ÎïµÄ×ÜÄÜÁ¿µÍÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿Ê±£¬¸Ã·´Ó¦Ò»¶¨²»ÄÜ·¢Éú
B£®Ç¿Ëá¸úÇ¿¼î·´Ó¦·Å³öµÄÈÈÁ¿¾ÍÊÇÖкÍÈÈ
C£®ÓÉʯī±È½ð¸ÕʯÎȶ¨¿ÉÖª£º
D£®ÔÚ¡¢Ê±£¬ÍêȫȼÉÕÉú³ÉÆø̬ˮ£¬·Å³öÈÈÁ¿£¬ÔòÇâÆøµÄȼÉÕÈÈΪ241.8

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÈÈ»¯Ñ§·½³Ìʽ£º H2O(g)£½H2(g) +1/2O2(g) ¡÷H = +241.8kJ/mol
H2(g)+ 1/2O2(g)£½H2O(1) ¡÷H =" -" 285.8kJ/mol
µ±1gҺ̬ˮ±äΪˮÕôÆøʱ£¬ÆäÈÈÁ¿±ä»¯ÊÇ    £¨  £©
A£®ÎüÈÈ44kJB£®ÎüÈÈ2.44KJ
C£®·ÅÈÈ44kJD£®·ÅÈÈ2.44KJ

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖª£º C(s)+1£¯2 O2(g_)=CO(g)£»¡÷H1=" -110.35" kJ£¯mol
CO(g)+1£¯2 O2(g)=CO2(g)£»¡÷H2=" -282.57" kJ£¯mol
Ôò£ºC(s)+O2(g)=CO2(g)£»¡÷H3=
A£®+172.22 kJ£¯molB£®-172.22 kJ£¯mol
C£®+392.92 kJ£¯molD£®-392.92 kJ£¯mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

molÆø̬ÄÆÀë×ÓºÍ1molÆø̬ÂÈÀë×Ó½áºÏÉú³É1molÂÈ»¯Äƾ§ÌåÊͷųöµÄÈÈÄÜΪÂÈ»¯Äƾ§ÌåµÄ¾§¸ñÄÜ¡£ÏÂÁÐÈÈ»¯Ñ§·½³ÌÖУ¬ÄÜÖ±½Ó±íʾ³öÂÈ»¯Äƾ§Ìå¸ñÄܵÄÊÇ£¨   £©
A£®Na+(g)+Cl£­(g)NaCl(s); ¡÷HB£®Na(s)+Cl2(g)NaCl(s); ¡÷H1
C£®Na(s)Na(g); ¡÷H2D£®Na(g)£­e£­Na+(g); ¡÷H

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸