£¨7·Ö£©(1)ÒÑÖª£ºH¡ªH¼üÄÜΪ436kJ/mol£¬N¡ÔN¼üÄÜΪ945kJ/mol£¬N¡ªH¼üÄÜΪ391kJ/mol¡£¸ù¾Ý¼üÄܼÆË㹤ҵºÏ³É°±µÄ·´Ó¦ÊÇ___________ ·´Ó¦(Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±)£¬ÏûºÄ1mol N2ºÏ³É°±·´Ó¦µÄ=______________¡£                                                                                                                                                                                                                                                                                                        
(2) ÏÖ¸ù¾ÝÏÂÁÐ3¸öÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£º
¢ÙFe2O3(s)+3CO(g) £½2Fe(s)+3CO2(g)        ¡÷H£½¨D24.8kJ£¯mol
¢Ú3Fe2O3(s)+ CO(g) £½2Fe3O4(s)+ CO2(g)      ¡÷H£½¨D47.2kJ£¯mol
¢ÛFe3O4(s)+CO(g) £½3FeO(s)+CO2(g)         ¡÷H£½+640.5kJ£¯mol
ÔòCOÆøÌ廹ԭFeO¹ÌÌåµÃµ½Fe¹ÌÌåºÍCO2ÆøÌåµÄÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪ        ¡£
£¨1£© ·ÅÈÈ(2·Ö)£»£­93KJ/mol   (2·Ö)
£¨2£©CO (g)£«FeO(s)=" Fe(s)+" CO2 (g)   ¡÷H=£­218KJ/mol   (3·Ö)

ÊÔÌâ·ÖÎö£º£¨1£©Òò·´Ó¦ÈȵÈÓÚ·´Ó¦ÎïµÄ×ܼüÄܼõÈ¥Éú³ÉÎïµÄ×ܼüÄÜ945kJ/mol+3¡Á436KJ/mol-6¡Á391kJ/mol=£­93KJ/mol £¬Í¨¹ý¼ÆËã·´Ó¦µÄ¡÷H´óÓÚ0Ϊ·ÅÈÈ·´Ó¦¡££¨2£©ËùÒÔÈÈ»¯Ñ§·½³ÌʽΪ-¢Ù-¢Ú-2¡Á¢Û£¬¼´CO (g)£«FeO(s)=" Fe(s)+" CO2 (g)£¬¡÷H×öÏàÓ¦µÄ¼ÆË㣬¡÷H=£­218KJ/mol¡£
µãÆÀ£º±¾ÀàÌâÐÍ×¢Òâ¸Ç˹¶¨ÂɵÄÓ¦Óá£ÊôÓÚ¼òµ¥Ìâ¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÈÈ»¯Ñ§·½³ÌʽÕýÈ·µÄÊÇ£¨µÄ¾ø¶ÔÖµ¾ùÕýÈ·£©
A£®C2H5OH£¨l£©+3O2£¨g£©==2CO2£¨g£©+3H2O£¨g£©£»¡÷H=¡ª1367.0 kJ/mol£¨È¼ÉÕÈÈ£©
B£®NaOH£¨aq£©+HCl£¨aq£©==NaCl£¨aq£©+H2O£¨l£©£»¡÷H=+57.3kJ/mol£¨ÖкÍÈÈ£©
C£®S£¨s£©+O2£¨g£©===SO2£¨g£©£»¡÷H=¡ª269.8kJ/mol£¨·´Ó¦ÈÈ£©
D£®2NO2==O2+2NO£»¡÷H=+116.2kJ/mol£¨·´Ó¦ÈÈ£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l) ¦¤H1£½£­870.3 kJ/mol
¢ÚC(s)£«O2(g)===CO2(g)¡¡¦¤H2£½£­393.5 kJ/mol
¢ÛH2(g)£«O2(g)===H2O(l)¡¡¦¤H3£½£­285.8 kJ/mol
Ôò·´Ó¦¢Ü2C(s)£«2H2(g)£«O2(g)===CH3COOH(l)µÄìʱäΪ
A£®488.3 kJ/molB£®£­224.15 kJ/mol
C£®£­488.3 kJ/molD£®244.15 kJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

(14·Ö)°±ÊÇÖØÒªµÄ»¯¹¤²úÆ·Ö®Ò»£¬Ñо¿ºÏ³É°±·´Ó¦¾ßÓÐÖØÒªÒâÒå¡£
(1) ÒÑÖª¶ÏÁÑÏÂÁл¯Ñ§¼üÐèÒªÎüÊÕµÄÄÜÁ¿·Ö±ðΪ£º
 £¬
д³öÒÔN2ºÍH2ΪԭÁϺϳÉNH3µÄÈÈ»¯Ñ§·½³Ìʽ________________________¡£
(2) ijС×éÑо¿ÁËÆäËûÌõ¼þ²»±äʱ£¬¸Ä±äijһÌõ¼þ¶ÔÉÏÊö·´Ó¦µÄÓ°Ï죬ʵÑé½á¹ûÈçÏÂͼËùʾ£º
¢Ùt1ʱ¿Ì¸Ä±äµÄÌõ¼þΪ__________________¡£
¢Út2ʱ¿Ì£¬ºãѹ³äÈ뺤Æø,t3ʱ¿Ì´ïµ½Æ½ºâ¡£ÔÚͼÖл­³öt2ʱ¿ÌºóµÄËÙÂʱ仯ͼÏñ¡£

(3) ÏàͬζÈÏ£¬A¡¢B¡¢CÈý¸öÃܱÕÈÝÆ÷£¬A¡¢BºãÈÝ£¬C´øÓпÉ×ÔÓÉÒƶ¯µÄ»îÈûK£¬¸÷ÏòÆäÖгäÈëÈçͼËùʾ·´Ó¦Î³õʼʱ¿ØÖÆ»îÈûKʹÈýÕßÌå»ýÏàµÈ£¬Ò»¶Îʱ¼äºó¾ù´ïµ½Æ½ºâ¡£

¢Ù´ïµ½Æ½ºâʱ£¬A¡¢CÁ½¸öÈÝÆ÷ÖÐNH3µÄŨ¶È·Ö±ðΪcl¡¢c2£¬Ôòc1______c2(Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±)¡£
¢Ú´ïµ½Æ½ºâʱ£¬ÈôA¡¢BÁ½ÈÝÆ÷Öз´Ó¦ÎïµÄת»¯ÂÊ·Ö±ðΪ¦Á(A),¦Á(B)£¬Ôò¦Á(A)+¦Á(B)______1(Ìî¡° >¡±¡¢¡°<¡±»ò¡°=¡±£©¡£
¢Û´ïµ½Æ½ºâʱ£¬ÈôÈÝÆ÷CµÄÌå»ýÊÇÆðʼʱµÄ3/4£¬ÔòƽºâʱÈÝÆ÷CÖÐH2µÄÌå»ý·ÖÊýΪ_______¡£
(4) Ö±½Ó¹©°±Ê½¼îÐÔȼÁϵç³Ø£¨DAFC)£¬ÒÔKOHÈÜҺΪµç½âÖÊÈÜÒº£¬Æäµç³Ø·´Ó¦Îª 4NH3+3O2=2N2+6H2O£¬Ôò¸º¼«µÄµç¼«·´Ó¦Ê½Îª__________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÏÂÁÐÓйØÈÈ»¯Ñ§·½³Ìʽ¼°ÆäÐðÊöÕýÈ·µÄÊÇ
A£®ÇâÆøµÄȼÉÕÈÈΪ-285.5kJ£¯mo1£¬ÔòË®µç½âµÄÈÈ»¯Ñ§·½³ÌʽΪ£º
2H2O£¨1£© =2H2£¨g£©+O2£¨g£©£»¡÷H=+285.5KJ£¯mo1
B£®1mol¼×ÍéÍêȫȼÉÕÉú³ÉCO2ºÍH2O£¨1£©Ê±·Å³ö890kJÈÈÁ¿£¬ËüµÄÈÈ»¯Ñ§·½³ÌʽΪ
1/2CH4£¨g£©+O2£¨g£©= 1/2CO2£¨g£©+H2O£¨1£©£»¡÷H= Ò»445kJ£¯mol
C£®ÒÑÖª2C£¨s£©+O2£¨g£©=2CO£¨g£©£»¡÷H= Ò»221kJ¡¤mol-1£¬ÔòCµÄȼÉÕÈÈΪһ110.5kJ£¯mo1
D£®HFÓëNaOHÈÜÒº·´Ó¦£ºH+£¨aq£©+OH¡ª£¨aq£©=H2O£¨1£©£»¡÷H= Ò»57.3kJ£¯mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

(18·Ö)ÔÚÈÝ»ý²»Í¬µÄ¶à¸öÃܱÕÈÝÆ÷ÄÚ£¬·Ö±ð³äÈëͬÁ¿µÄN2ºÍH2£¬ÔÚ²»Í¬Î¶ÈÏ£¬Í¬Ê±·¢Éú·´Ó¦N2+3H22NH3£¬²¢·Ö±ðÔÚtÃëʱ²â¶¨ÆäÖÐNH3µÄÌå»ý·ÖÊý£¬»æͼÈçÓÒ£º

(1)A£¬B£¬C£¬D£¬EÎåµãÖУ¬ÉÐδ´ïµ½»¯Ñ§Æ½ºâ״̬µÄµãÊÇ                  ¡£
(2) ÏòÒ»ºãÈÝÈÝÆ÷ÖмÓÈë1 mol N2 ºÍ3 mol H2£¬T3ʱ£¬²âµÃÌåϵѹǿΪԭÀ´µÄ7/8£¬²¢·Å³ö23.1 kJµÄÈÈÁ¿£¬Ôò¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ              ¡£
(3)µ±ÉÏÊö·´Ó¦´¦ÓÚƽºâ״̬ʱ£¬ÔÚÌå»ý²»±äµÄÌõ¼þÏ£¬ÏÂÁдëÊ©ÖÐÓÐÀûÓÚÌá¸ßNH3²úÂʵÄÓР         £¨Ìî×Öĸ£©
A£®Éý¸ßζÈB£®½µµÍζÈC£®Ôö´óѹǿD£®¼õСѹǿ
E£®¼ÓÈë´ß»¯¼Á   F£®ÎüÊÕNH3    G£®Í¨ÈëN2
(4)AC¶ÎµÄÇúÏߺÍCE¶ÎÇúÏ߱仯Ç÷ÊÆÏà·´£¬ÊÔ´Ó·´Ó¦ËÙÂʺÍƽºâ½Ç¶È˵Ã÷ÀíÓÉ¡£
                                                                 ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨8·Ö£©£¨1£©ÒÑÖª¶Ï¿ª1 mol  H2ÖеĻ¯Ñ§¼üÐèÎüÊÕ436 kJÈÈÁ¿£¬¶Ï¿ª1 mol Cl2ÖеĻ¯Ñ§¼üÐèÒªÎüÊÕ243 kJÄÜÁ¿£¬¶øÐγÉ1 molHCl·Ö×ÓÖеĻ¯Ñ§¼üÒªÊÍ·Å431 kJÄÜÁ¿£¬ÊÔÇó1 mol H2Óë1 mol Cl2·´Ó¦µÄÄÜÁ¿±ä»¯¡£¡÷H=¡¡¡¡¡¡¡¡¡¡¡£
£¨2£©ÒÑÖª³£ÎÂʱ£¬0.1mol/LijһԪËáHAÔÚË®ÖÐÓÐ0.1%·¢ÉúµçÀ룬Ôò¸ÃÈÜÒºµÄPH=___¢Ù___£¬´ËËáµÄµçÀëƽºâ³£ÊýK=___¢Ú___£¬ÓÉHAµçÀë³öÀ´µÄH+µÄŨ¶ÈԼΪˮµçÀë³öÀ´µÄH+µÄŨ¶ÈµÄ__¢Û__±¶¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨14·Ö£©ÁòËáÑÎÖ÷ÒªÀ´×Եزã¿óÎïÖÊ£¬¶àÒÔÁòËá¸Æ¡¢ÁòËáþµÄÐÎ̬´æÔÚ¡£
£¨1£©ÒÑÖª£º¢ÙNa2SO4(s)=Na2S(s)+2O2(g)       ¦¤H1="+1011.0" kJ¡¤mol-1
¢ÚC(s)+O2(g)=CO2(g)             ¦¤H2=£­393.5 kJ¡¤mol-1
¢Û2C(s)+O2(g)=2CO(g)            ¦¤H3=£­221.0 kJ¡¤mol-1
Ôò·´Ó¦¢ÜNa2SO4(s)+4C(s)=Na2S(s)+4CO(g) ¦¤H4=________kJ¡¤mol-1£»¹¤ÒµÉÏÖƱ¸Na2SʱÍùÍù»¹Òª¼ÓÈë¹ýÁ¿µÄÌ¿£¬Í¬Ê±»¹ÒªÍ¨Èë¿ÕÆø£¬Ä¿µÄÓÐÁ½¸ö£¬ÆäÒ»ÊÇʹÁòËáÄƵõ½³ä·Ö»¹Ô­£¨»òÌá¸ßNa2S²úÁ¿£©£¬Æä¶þÊÇ_____________________________________________¡£
£¨2£©ÖÇÄܲÄÁÏÊǵ±½ñ²ÄÁÏÑо¿µÄÖØÒª·½ÏòÖ®Ò»£¬ÄÉÃ×Fe3O4ÓÉÓÚ¾ßÓиߵıȱíÃæ¡¢¸ßµÄ±È±¥ºÍ´Å»¯Ç¿¶ÈºÍ˳´ÅΪÁãµÄ³¬Ë³´ÅÐÔ¶ø±»¹ã·ºµØÓÃ×÷´ÅÁ÷ÌåµÄ´ÅÐÔÁ£×Ó¡£Ë®ÈÈ·¨ÖƱ¸Fe3O4ÄÉÃ׿ÅÁ£µÄ·´Ó¦ÊÇ£º
3Fe2+ + 2S2O32- + O2 + xOH-=Fe3O4+S4O62-+2H2O
Çë»Ø´ðÏÂÁÐÎÊÌ⣺

¢ÙË®ÈÈ·¨ËùÖƵõÄË®»ù´ÅÁ÷Ì峬¹ý30Ì춼δ³öÏÖ·Ö²ãºÍ»ì×ÇÏÖÏó£¬ÒòΪ¸Ã·ÖɢϵÊÇ________¡£
¢ÚÉÏÊö·´Ó¦·½³Ìʽx=___________________¡£
¢Û¸Ã·´Ó¦ÖÐ1molFe2+±»Ñõ»¯Ê±£¬±»Fe2+»¹Ô­µÄO2µÄÎïÖʵÄÁ¿Îª_____¡£
£¨3£©¸ßÎÂʱ£¬ÓÃCO»¹Ô­MgSO4¿ÉÖƱ¸¸ß´¿MgO¡£
¢Ù750¡æʱ£¬²âµÃÆøÌåÖꬵÈÎïÖʵÄÁ¿SO2ºÍSO3£¬´Ëʱ·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ____________¡£
¢ÚÓÉMgO¿ÉÖƳɡ°Ã¾¡ª´ÎÂÈËáÑΡ±È¼Áϵç³Ø£¬Æä×°ÖÃʾÒâͼÈçÉÏͼ£¬¸Ãµç³Ø·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_____________________________________________________¡£

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=" -870.3" kJ¡¤mol£­1
£¨2£©C(s) + O2(g) ="=" CO2(g)  ¡÷H2=" -393.5" kJ¡¤mol£­1
£¨3£©H2(g) + 1/2 O2(g) ="=" H2O(l)  ¡÷H3= -285.8kJ¡¤mol£­1
Ôò·´Ó¦£º2C(s) + 2H2(g) + O2(g) ="=" CH3COOH(l)µÄ·´Ó¦ÈÈΪ£¨  £©
A£®-488.3 kJ¡¤mol£­1 B£®-244.15 kJ¡¤mol£­1 C£®+488.3 kJ¡¤mol£­1D£®+244.15 kJ¡¤mol£­1

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸