ºìÁ×£¨P£©ºÍCl2·¢Éú·´Ó¦Éú³ÉPCl3ºÍPCl5£¬·´Ó¦¹ý³ÌºÍÄÜÁ¿¹ØϵÈçÏÂͼËùʾ£¬Í¼Öеġ÷H±íʾÉú³É1 mol²úÎïµÄÊý¾Ý¡£ÒÑÖªPCl5·Ö½âΪPCl3ºÍCl2ÊÇ¿ÉÄæ·´Ó¦¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨   £©
A£®ÆäËûÌõ¼þ²»±ä£¬Éý¸ßζÈÓÐÀûÓÚPCl5Éú³É
B£®·´Ó¦2P£¨s£©+5Cl2£¨g£©===2PCl5£¨g£©¶ÔÓ¦µÄ·´Ó¦ÈÈ¡÷H="-798" kJ/mol
C£®PºÍCl2·´Ó¦Éú³ÉPCl3µÄÈÈ»¯Ñ§·½³ÌʽΪ2P£¨s£©+3Cl2£¨g£©=2PCl3£¨g£©¡÷H="-306" kJ/mol
D£®ÆäËûÌõ¼þ²»±ä£¬¶ÔÓÚPCl5·Ö½âÉú³É£®PCl3ºÍCl2µÄ·´Ó¦£¬Ôö´óѹǿ£¬PCl5µÄת»¯ÂʼõС£¬Æ½ºâ³£ÊýK¼õС
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

Ò»Ñõ»¯Ì¼ÊÇÒ»ÖÖÓÃ;Ï൱¹ã·ºµÄ»¯¹¤»ù´¡Ô­ÁÏ¡£
¢ÅÀûÓÃÏÂÁз´Ó¦¿ÉÒÔ½«´ÖÄøת»¯Îª´¿¶È´ï99.9%µÄ¸ß´¿Äø¡£
Ni(s)£«4CO(g) Ni(CO)4(g)£¬¸Ã·´Ó¦µÄ¡÷H______0 (Ñ¡Ìî¡°£¾¡±»ò¡°£½¡±»ò¡°£¼¡±)¡£
¢Æ½ðÊôÑõ»¯Îï±»Ò»Ñõ»¯Ì¼»¹Ô­Éú³É½ðÊôµ¥ÖʺͶþÑõ»¯Ì¼¡£ÏÂ×óͼÊÇËÄÖÖ½ðÊôÑõ»¯Îï(Cr2O3¡¢SnO2¡¢PbO2¡¢Cu2O)±»Ò»Ñõ»¯Ì¼»¹Ô­Ê±lgÓëζÈ(t)µÄ¹ØϵÇúÏßͼ¡£ÔòÒ»Ñõ»¯Ì¼»¹Ô­ÈýÑõ»¯¸õ·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽ¿É±íʾΪ£ºK£½______________¡£800¡æʱ£¬ÆäÖÐ×îÒ×±»»¹Ô­µÄ½ðÊôÑõ»¯ÎïÊÇ___________£¬¸Ã·´Ó¦µÄƽºâ³£ÊýÊýÖµ(K)µÈÓÚ__________¡£
¢ÇÏÂÓÒͼÊÇһ̼ËáÑÎȼÁϵç³Ø(MCFC)£¬ÒÔˮúÆø(CO¡¢H2)ΪȼÁÏ£¬Ò»¶¨±ÈÀýLi2CO3ºÍNa2CO3µÍÈÛ»ìºÏÎïΪµç½âÖÊ¡£ÒÀ´Îд³öA¡¢BÁ½µç¼«·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½
__________________________________¡¢___________________________________¡£
 
¢ÈÒÑÖª£º¢ÙCO(g)£«2H2(g)CH3OH(g)  ¡÷H£½£­90.7 kJ¡¤mol£­1
¢Ú2CH3OH(g)CH3OCH3(g)£«H2O(g) ¡÷H£½£­23.5 kJ¡¤mol£­1
¢ÛCO(g)£«H2O(g)CO2(g)£«H2(g)  ¡÷H£½£­41.2 kJ¡¤mol£­1
Ôò3CO(g)£«3H2(g)CH3OCH3(g)£«CO2(g)µÄ¡÷H£½_______________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨4·Ö£©
ÒÑÖªÏÂÁÐÁ½¸öÈÈ»¯Ñ§·½³Ìʽ£º
H2£¨g£©£«O2£¨g£©====H2O£¨l£©¡÷H£½£­285kJ¡¤mol£­1
C3H8£¨g£©£«5O2£¨g£©====3CO2£¨g£©£«4H2O£¨l£©¡÷H£½£­2220.0kJ¡¤mol£­1
£¨1£©ÊµÑé²âµÃH2ºÍC3 H8µÄ»ìºÏÆøÌå¹²5 mol£¬ÍêȫȼÉÕÉú³ÉҺ̬ˮʱ·ÅÈÈ6262.5kJ£¬Ôò»ìºÏÆøÌåÖÐH2ºÍC3 H8µÄÌå»ý±ÈÊÇ_______£»
£¨2£©ÒÑÖª£ºH2O£¨l£©====H2O£¨g£© ¡÷H£½£«44.0kJ¡¤mol£­1£¬Ð´³ö±ûÍéȼÉÕÉú³ÉCO2ºÍÆø̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ_________________________________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨4·Ö£©£¨1£©ÒÑÖªC2H5OH(g)+3O2(g)==2CO2(g)+3H2O(g) ¡÷H1="a" kJ¡¤mol£­1£»
H2O(g)==H2O(1) ¡÷H2="b" kJ¡¤mol£­1£»
C2H5OH(g)==C2H5OH(1) ¡÷H3="c" kJ¡¤mol£­1£¬
д³ö C2H5OH(1)ÍêȫȼÉÕÉú³ÉCO2(g)ºÍH2O(1)µÄÈÈ»¯Ñ§·½³Ìʽ                  
£¨2£©ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢Ù H2(g)£«1/2O2(g) £½H2O(g) ¦¤H£½£­241.8 kJ/mol
¢Ú C(s)£«1/2O2(g) £½CO(g) ¦¤H£½£­110.5 kJ/moL 
ÔòˮúÆøÄÜÔ´µÄÖÆÈ¡Ô­Àí£ºC(s)£«H2O(g)£½H2(g)£«CO (g) ¦¤H£½      kJ/moL

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ


ÏÂÁи÷×é±ä»¯ÖУ¬¦¤H»òQÇ°ÕßСÓÚºóÕßµÄÒ»×éÊÇ
¢ÙCH4(g)£«2O2(g)£½CO2(g)£«2H2O(l)¡¡¦¤H1£»  CH4(g)£«2O2(g)£½CO2(g)£«2H2O(g)¡¡¦¤H2
¢ÚÔÚÏ¡ÈÜÒºÖУºº¬0.5 mol H2SO4ÁòËáÓ뺬1 mol NaOHµÄÈÜÒº»ìºÏ£ºH£«(aq)£«OH£­(aq)£½H2O¦¤H1
NaOH(aq)£«CH3COOH(aq) £½CH3COONa(aq)£«H2O(l)  ¦¤H2
¢Ût ¡æʱ£¬ÔÚÒ»¶¨Ìõ¼þÏ£¬½«1 mol SO2ºÍ1 mol O2·Ö±ðÖÃÓÚºãÈݺͺãѹµÄÁ½¸öÃܱÕÈÝÆ÷ÖУ¬´ïµ½Æ½ºâ״̬ʱ·Å³öµÄÈÈÁ¿·Ö±ðΪQ1¡¢Q2
¢ÜCaCO3 (s)£½CaO(s)£«CO2(g)¡¡¦¤H1    CaO(s)£«H2O(l)£½Ca(OH)2(s)¡¡¦¤H2
A£®¢Ù¢Ú¢ÛB£®¢Ú¢ÛC£®¢Ù¢ÚD£®¢Ù¢Û¢Ü

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ú×÷ΪȼÁÏÓÐÁ½ÖÖ;¾¶£º
¢ñ.C(s)£«O2(g)===CO2(g)¡¡¦¤H1<0
¢ò.C(s)£«H2O(g)===CO(g)£«H2(g)¡¡¦¤H2>0
2CO(g)£«O2(g)===2CO2(g)¡¡¦¤H3<0
2H2(g)£«O2(g)===2H2O(g)¡¡¦¤H4<0
Çë»Ø´ð£º(1);¾¶¢ñ·Å³öµÄÈÈÁ¿________;¾¶¢ò·Å³öµÄÈÈÁ¿(Ìî¡°>¡±¡¢¡°<¡±»ò¡°£½¡±)¡£
(2)¦¤H1¡¢¦¤H2¡¢¦¤H3¡¢¦¤H4Ö®¼ä¹ØϵµÄÊýѧ±í´ïʽÊÇ________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨9·Ö£©¼×´¼ÊÇÒ»ÖÖ¿ÉÔÙÉúÄÜÔ´£¬¾ßÓй㷺µÄ¿ª·¢ºÍÓ¦ÓÃÇ°¾°¡£
¢Å¹¤ÒµÉÏÒ»°ã²ÉÓÃÏÂÁз´Ó¦ºÏ³É¼×´¼£ºCO(g)£«2H2(g)CH3OH(g)£»¦¤H
¢ÙÏÂÁÐÊý¾ÝÊÇÔÚ²»Í¬Î¶ÈϵĻ¯Ñ§Æ½ºâ³£Êý£¨K£©¡£
250¡æ:  K1=__________
300¡æ:  K2=0.270
350¡æ:  K3=0.012
ÓɱíÖÐÊý¾ÝÅжϦ¤H    0£¨Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±£½¡££©
¢Ú250¡æ£¬½«2molCOºÍ6molH2³äÈë2LµÄÃܱÕÈÝÆ÷ÖУ¬³ä·Ö·´Ó¦£¬´ïµ½Æ½ºâºó£¬²âµÃ:
COÊ£Óà0.4mol£¬ÇóK1¡£
¢ÆÒÑÖªÔÚ³£Î³£Ñ¹Ï£º¢ÙCH3OH£¨l£©+O2£¨g£©=CO£¨g£©+2H2O£¨l£©£»¡÷H=-442.8KJ/mol
¢Ú2CO(g)+O2(g)£½2CO2(g)£»¦¤H2 £½£­566.0kJ£¯mol
д³ö¼×´¼È¼ÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ____________________________________________£¬
¢ÇijʵÑéС×éÒÀ¾Ý¼×´¼È¼Éյķ´Ó¦Ô­Àí£¬Éè¼ÆÈçͼ(A)ËùʾµÄȼÁϵç³Ø×°Öá£Ôò£º
¢Ù¸ÃȼÁϵç³Ø¸º¼«µÄµç¼«·´Ó¦Îª£º___________________________£¬
¢ÚÓøü״¼È¼Áϵç³Ø¶ÔB³Ø½øÐеç½â£¬¼ºÖªc¡¢dÊÇÖÊÁ¿ÏàͬµÄÍ­°ô£¬µç½â2minºó£¬È¡³öc¡¢d£¬Ï´¾»¡¢ºæ¸É¡¢³ÆÁ¿£¬ÖÊÁ¿²îΪ0.64g£¬ÔÚͨµç¹ý³ÌÖУ¬µç·ÖÐͨ¹ýµÄµç×ÓΪ_________mol¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨3·Ö£©½«Ãº×ª»¯ÎªË®ÃºÆø£¨COºÍH2µÄ»ìºÏÆøÌ壩ÊÇͨ¹ý»¯Ñ§·½·¨½«Ãº×ª»¯Îª½à¾»È¼Áϵķ½·¨Ö®Ò»¡£Ãº×ª»¯ÎªË®ÃºÆøµÄÖ÷Òª»¯Ñ§·´Ó¦Îª£ºC(s) + H2O(g)£½CO(g) + H2(g)£»¡÷H1¡£  ÒÑÖª£º
¢Ù2H2(g) + O2(g) £½ 2H2O(g)£»¡÷H2£½£­483.6kJ¡¤mol£­1   
¢Ú2C(s) + O2(g) £½ 2 CO(g)£»¡÷H3£½£­221.0kJ¡¤mol£­1
½áºÏÉÏÊöÈÈ»¯Ñ§·½³Ìʽ£¬¼ÆËãµÃ³ö¡÷H1£½                        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨2·Ö£©°×Á×£¨P4£©ÊÇÕýËÄÃæÌå¹¹Ð͵ķÖ×Ó£¬µ±ÓëÑõÆø×÷ÓÃÐγÉP4O10ʱ£¬Ã¿Á½¸öÁ×Ô­×ÓÖ®¼ä²åÈëÒ»¸öÑõÔ­×Ó£¬´ËÍ⣬ÿ¸öÁ×Ô­×ÓÓÖÒÔË«¼üÔÙ½áºÏÒ»¸öÑõÔ­×Ó¡£»¯Ñ§·´Ó¦¿ÉÊÓΪ¾É¼ü¶ÏÁѺÍмüÐγɵĹý³Ì¡£ÏÖÌṩÒÔÏ»¯Ñ§¼üµÄ¼üÄÜ£¨kJ/mol£©¡£P-P¼ü£º198 kJ /mol £»P-O¼ü£º360 kJ /mol £»O=O¼ü£º498 kJ /mol £»P=O¼ü£º585 kJ /mol¡£ÊÔ¸ù¾ÝÕâЩÊý¾Ý£¬¼ÆËã³öÒÔÏ·´Ó¦µÄ·´Ó¦ÈÈ£ºP4£¨S£¬°×Á×£©+ 5O2£¨g£©= P4O10£¨s£©£»
¦¤H=                                                                     ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸