µªÊǵØÇòÉϺ¬Á¿·á¸»µÄÒ»ÖÖÔªËØ£¬µª¼°Æ仯ºÏÎïÔÚ¹¤Å©ÒµÉú²ú¡¢Éú»îÖÐÓÐ×ÅÖØÒª×÷Óá£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÓÒͼÊÇN2ºÍH2·´Ó¦Éú³É2 mol NH3¹ý³ÌÖÐÄÜÁ¿±ä
»¯Ê¾Òâͼ£¬Ð´³öÉú³ÉNH3µÄÈÈ»¯Ñ§·½³Ìʽ£º
_____________________________________________
___________________________¡£
£¨2£©ÓÉÆø̬»ù̬ԭ×ÓÐγÉ1 mol»¯Ñ§¼üÊͷŵÄ×îµÍÄÜÁ¿½Ð¼üÄÜ¡£´Ó»¯Ñ§¼üµÄ½Ç¶È·ÖÎö£¬»¯
ѧ·´Ó¦µÄ¹ý³Ì¾ÍÊÇ·´Ó¦ÎïµÄ»¯Ñ§¼üµÄÆÆ»µºÍÉú³ÉÎïµÄ»¯Ñ§¼üµÄÐγɹý³Ì¡£ÔÚ»¯Ñ§·´Ó¦¹ý
³ÌÖУ¬²ð¿ª»¯Ñ§¼üÐèÒªÏûºÄÄÜÁ¿£¬Ðγɻ¯Ñ§¼üÓÖ»áÊÍ·ÅÄÜÁ¿¡£
ÒÑÖª·´Ó¦N2(g)£«3H2(g)??2NH3(g)¡¡¦¤H£½a kJ¡¤mol£­1¡£
ÊÔ¸ù¾Ý±íÖÐËùÁмüÄÜÊý¾Ý¹ÀËãaµÄÊýÖµ£º________¡£
»¯Ñ§¼ü
H¡ªH
N¡ªH
N¡ÔN
¼üÄÜkJ¡¤mol£­1
436
391
945
 
£¨1£©N2(g)£«3H2(g)??2NH3(g)   ¦¤H£½£­92.2 kJ¡¤mol£­1¡¡£¨2£©£­93

ÊÔÌâ·ÖÎö£º£¨1£©¸ù¾ÝÄÜÁ¿¹Øϵͼ£¬¸Ã·´Ó¦µÄ¦¤H£½E1£­E2£½(335£­427.2)kJ¡¤mol£­1£½£­92.2 kJ¡¤mol£­1¡££¨2£©¦¤H£½ÎüÊÕµÄÄÜÁ¿£­ÊͷŵÄÄÜÁ¿£½(945£«3¡Á436£­6¡Á391)kJ¡¤mol£­1£½£­93 kJ¡¤mol£­1¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÒÔÏÂÊÇһЩÎïÖʵÄÈ۷еãÊý¾Ý£¨³£Ñ¹£©£º
 
¼Ø
ÄÆ
Na2CO3
½ð¸Õʯ
ʯī
È۵㣨¡æ£©
63£®65
97£®8
851
3550
3850
·Ðµã£¨¡æ£©
774
882£®9
1850£¨·Ö½â²úÉúCO2£©
----
4250
 
½ðÊôÄƺÍCO2ÔÚ³£Ñ¹¡¢890¡æ·¢ÉúÈçÏ·´Ó¦£º4 Na£¨g£©+ 3CO2£¨g£© 2 Na2CO3£¨l£©+  C(s,½ð¸Õʯ) ¡÷H=£­1080£®9kJ/mol
£¨1£©Èô·´Ó¦ÔÚ10LÃܱÕÈÝÆ÷¡¢³£Ñ¹Ï½øÐУ¬Î¶ÈÓÉ890¡æÉý¸ßµ½1860¡æ£¬Èô·´Ó¦Ê±¼äΪ10min, ½ðÊôÄƵÄÎïÖʵÄÁ¿¼õÉÙÁË0£®2mol£¬Ôò10minÄÚCO2µÄƽ¾ù·´Ó¦ËÙÂÊΪ                 ¡£
£¨2£©¸ßѹÏÂÓÐÀûÓÚ½ð¸ÕʯµÄÖƱ¸£¬ÀíÓÉÊÇ                                            ¡£
£¨3£©ÓÉCO2£¨g£©+ 4Na£¨g£©=2Na2O£¨s£©+ C£¨s£¬½ð¸Õʯ£© ¡÷H=£­357£®5kJ/mol£»ÔòNa2O¹ÌÌåÓëC£¨½ð¸Õʯ£©·´Ó¦µÃµ½Na£¨g£©ºÍҺ̬Na2CO3£¨l£©µÄÈÈ»¯Ñ§·½³Ìʽ                     ¡£
£¨4£©ÏÂͼ¿ª¹ØK½ÓMʱ£¬Ê¯Ä«µç¼«·´Ó¦Ê½Îª                           ¡£

£¨5£©ÇëÔËÓÃÔ­µç³ØÔ­ÀíÉè¼ÆʵÑ飬ÑéÖ¤Cu2£«¡¢Ag+Ñõ»¯ÐÔµÄÇ¿Èõ¡£
ÔÚ·½¿òÄÚ»­³öʵÑé×°ÖÃͼ£¬ÒªÇóÓÃÉÕ±­ºÍÑÎÇÅ(ÔÚͬһÉÕ±­ÖУ¬
µç¼«ÓëÈÜÒºº¬ÏàͬµÄ½ðÊôÔªËØ)£¬²¢±ê³öÍâµç·µç×ÓÁ÷Ïò¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

¾Ý±¨µÀ£¬Ò»¶¨Ìõ¼þÏÂÓɶþÑõ»¯Ì¼ºÍÇâÆøºÏ³ÉÒÒ´¼ÒѳÉΪÏÖʵ¡£
ÒÑÖª£º¢ÙCH3CH2OH(l) +3 O2 (g) = 2CO2(g) +3H2O(l) ¡÷H=£­1366.8 kJ/ mol
¢Ú2H2 (g) + O2 (g) = 2H2O(l) ¡÷H=" -571.6" kJ/ mol
£¨1£©Ð´³öÓÉCO2ºÍH2 ·´Ó¦ºÏ³ÉCH3CH2OH (l)ºÍH2O(l)µÄÈÈ»¯Ñ§·½³Ìʽ                                                              ¡£
£¨2£©¼îÐÔÒÒ´¼È¼Áϵç³ØÒ×´¢´æ£¬Ò×Íƹ㣬¶Ô»·¾³ÎÛȾС£¬¾ßÓзdz£¹ãÀ«µÄ·¢Õ¹Ç°¾°¡£¸ÃȼÁϵç³ØÖУ¬Ê¹Óò¬×÷µç¼«£¬KOHÈÜÒº×öµç½âÖÊÈÜÒº¡£Çëд³ö¸ÃȼÁϵç³Ø¸º¼«Éϵĵ缫·´Ó¦Ê½Îª                                                                 ¡£
£¨3£©ÓÃÒÒ´¼È¼Áϵç³Øµç½â400 mL ±¥ºÍʳÑÎË®×°Öÿɼòµ¥±íʾÈçÏÂͼ£º

¸Ã×°ÖÃÖз¢Éúµç½â·´Ó¦µÄ·½³ÌʽΪ                                                       £»ÔÚÌú°ô¸½½ü¹Û²ìµ½µÄÏÖÏóÊÇ                                                       £»µ±Òõ¼«²úÉú448 mLÆøÌ壨Ìå»ýÔÚ±ê×¼×´¿öϲâµÃ£©Ê±£¬Í£Ö¹µç½â£¬½«µç½âºóµÄÈÜÒº»ìºÏ¾ùÔÈ£¬ÈÜÒºµÄpHΪ          ¡££¨²»¿¼ÂÇÆøÌåµÄÈܽ⼰ÈÜÒºÌå»ýµÄ±ä»¯£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

2012Äêʼ£¬Îíö²ÌìÆøÎÞÊý´ÎËÁÅ°¼ÒÏ纪µ¦¡£ÆäÖУ¬È¼ÃººÍÆû³µÎ²ÆøÊÇÔì³É¿ÕÆøÎÛȾµÄÔ­ÒòÖ®Ò»¡£
£¨1£©Æû³µÎ²Æø¾»»¯µÄÖ÷ÒªÔ­ÀíΪ£º2NO(g) + 2CO(g)2CO2(g)+ N2(g)¡£¡÷H£¼0
¢Ù¸Ã·´Ó¦Æ½ºâ³£Êý±í´ïʽ                        
¢ÚÈô¸Ã·´Ó¦ÔÚ¾øÈÈ¡¢ºãÈݵÄÃܱÕÌåϵÖнøÐУ¬ÏÂÁÐʾÒâͼÕýÈ·ÇÒÄÜ˵Ã÷·´Ó¦ÔÚ½øÐе½t1ʱ¿Ì´ïµ½Æ½ºâ״̬µÄÊÇ               £¨Ìî´úºÅ£©¡£

£¨2£©Ö±½ÓÅÅ·ÅúȼÉÕ²úÉúµÄÑÌÆø»áÒýÆðÑÏÖصĻ·¾³ÎÊÌâ¡£
úȼÉÕ²úÉúµÄÑÌÆøº¬µªµÄÑõ»¯ÎÓÃCH4´ß»¯»¹Ô­NOX¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£
ÒÑÖª£º¢ñ CH4(g)+2NO2(g)£½N2(g)£«CO2(g)+2H2O(g)¡¡¡÷H£½£­867 kJ/mol
¢ò 2NO2(g)N2O4(g)  ¡÷H£½£­56.9 kJ/mol
¢ó H2O(g) £½ H2O(l)  ¦¤H £½ £­44.0 kJ£¯mol
д³öCH4´ß»¯»¹Ô­N2O4(g)Éú³ÉN2ºÍH2O(l)µÄÈÈ»¯Ñ§·½³Ìʽ£º                   ¡£
£¨3£©¼×ÍéȼÁϵç³Ø¿ÉÒÔÌáÉýÄÜÁ¿ÀûÓÃÂÊ¡£ÏÂͼÊÇÀûÓü×ÍéȼÁϵç³Øµç½â100mL1mol/LʳÑÎË®,µç½âÒ»¶Îʱ¼äºó£¬ÊÕ¼¯µ½±ê×¼×´¿öϵÄÇâÆø2.24L£¨Éèµç½âºóÈÜÒºÌå»ý²»±ä£©.
¢Ù¼×ÍéȼÁϵç³ØµÄ¸º¼«·´Ó¦Ê½£º                                          ¡£
¢Úµç½âºóÈÜÒºµÄpH=        (ºöÂÔÂÈÆøÓëÇâÑõ»¯ÄÆÈÜÒº·´Ó¦)
¢ÛÑô¼«²úÉúÆøÌåµÄÌå»ýÔÚ±ê×¼×´¿öÏÂÊÇ       L

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£¬Ôò¢ÛÖеÄQ3ֵΪ
Zn(s)+O2£¨g£©= ZnO (s) ¡÷H=" -" Q1 kJ? mol-1   ¢Ù
Hg(l) +O2£¨g£©= HgO (s) ¡÷H= -Q2 kJ? mol-1   ¢Ú
Zn(s) +HgO (s) = Hg(l)+ ZnO (s)   ¡÷H= -Q3 kJ? mol-1  ¢Û
A£®Q2 -Q1B£®Q1 +Q2 C£®Q- Q2D£®-Q1- Q2

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

£¨11·Ö£©Ñо¿È¼ÁϵÄȼÉպͶÔÎÛȾÆøÌå²úÎïµÄÎÞº¦»¯´¦Àí£¬¶ÔÓÚ·ÀÖ¹´óÆøÎÛȾÓÐÖØÒªÒâÒå¡£
£¨1£©½«Ãº×ª»¯ÎªÇå½àÆøÌåȼÁÏ£º
ÒÑÖª£ºH2(g)+1/2O2(g)=H2O(g) H= ?241£®8kJ/mol 
C(s)+1/2O2(g)=CO(g)  H= ?110£®5kJ/mol
д³ö½¹Ì¿ÓëË®ÕôÆø·´Ó¦ÖÆH2ºÍCOµÄÈÈ»¯Ñ§·½³Ìʽ                         ¡£
£¨2£©Ò»¶¨Ìõ¼þÏ£¬ÔÚÃܱÕÈÝÆ÷ÄÚ£¬SO2±»Ñõ»¯³ÉSO3µÄÈÈ»¯Ñ§·½³ÌʽΪ£º2SO2(g)+O2(g)  2SO3(g)£»
¡÷H=?a kJ/mo1£¬ÔÚÏàͬÌõ¼þÏÂÒªÏëµÃµ½2akJÈÈÁ¿£¬¼ÓÈë¸÷ÎïÖʵÄÎïÖʵÄÁ¿¿ÉÄÜÊÇ            
A£®4mo1 SO2ºÍ2mol O2¡¡¡¡¡¡¡¡¡¡ B£®4mol SO2¡¢2mo1 O2ºÍ2mol SO3
C£®4mol SO2ºÍ4mo1 O2¡¡¡¡¡¡     D£®6mo1 SO2ºÍ4mo1 O2
£¨3£©Æû³µÎ²ÆøÖÐNOxºÍCOµÄÉú³É¼°×ª»¯£º
¢ÙÒÑÖªÆø¸×ÖÐÉú³ÉNOµÄ·´Ó¦Îª£ºN2(g)+O2(g)  2NO(g) H£¾0
ÔÚÒ»¶¨Î¶ÈϵĶ¨ÈÝÃܱÕÈÝÆ÷ÖУ¬ÄÜ˵Ã÷´Ë·´Ó¦ÒÑ´ïƽºâµÄÊÇ          
A£®Ñ¹Ç¿²»±ä                   B£®»ìºÏÆøÌåƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä
C£®2vÕý(N2)£½vÄæ(NO)           D£® N2µÄÌå»ý·ÖÊý²»Ôٸıä
¢ÚÆû³µÈ¼ÓͲ»ÍêȫȼÉÕʱ²úÉúCO£¬ÓÐÈËÉèÏë°´ÏÂÁз´Ó¦³ýÈ¥CO£º2CO(g)=2C(s)+O2(g) H£¾0£¬
¼òÊö¸ÃÉèÏëÄÜ·ñʵÏÖµÄÒÀ¾Ý                                                             ¡£
£¨4£©È¼ÁÏCO¡¢H2ÔÚÒ»¶¨Ìõ¼þÏ¿ÉÒÔÏ໥ת»¯£ºCO(g)£«H2O(g)  CO2(g)£«H2(g)¡£ÔÚ420¡æʱ£¬Æ½ºâ³£ÊýK=9¡£Èô·´Ó¦¿ªÊ¼Ê±£¬CO¡¢H2OµÄŨ¶È¾ùΪ0£®1mol/L£¬ÔòCOÔÚ´Ë·´Ó¦Ìõ¼þϵÄת»¯ÂÊΪ              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

Íê³ÉÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£¨»¯Ñ§·½³Ìʽ¡¢µç¼«·´Ó¦Ê½¡¢±í´ïʽµÈ£©µÄÊéд£º
£¨1£©ÒÑÖª£º2Cu(s)£«1/2O2(g)=Cu2O(s)£»¡÷H=-169kJ¡¤mol-1£¬
C(s)£«1/2O2(g)=CO(g)£»¡÷H=-110.5kJ¡¤mol-1£¬
Cu(s)£«1/2O2(g)=CuO(s)£»¡÷H=-157kJ¡¤mol-1
ÓÃÌ¿·ÛÔÚ¸ßÎÂÌõ¼þÏ»¹Ô­CuOÉú³ÉCu2OµÄÈÈ»¯Ñ§·½³ÌʽÊÇ£º                
£¨2£©ÔÚÒ»¶¨Ìõ¼þÏ£¬¶þÑõ»¯ÁòºÍÑõÆø·¢ÉúÈçÏ·´Ó¦£º2SO2(g)+O2(g)2SO3(g)£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽ£º             
£¨3£©ÒÔ¼×Íé¡¢¿ÕÆøΪ·´Ó¦ÎKOHÈÜÒº×÷µç½âÖÊÈÜÒº¹¹³ÉȼÁϵç³Ø£¬Ôò¸º¼«·´Ó¦Ê½Îª£º           ¡£
£¨4£©ÌúÔÚ³±ÊªµÄ¿ÕÆøÖз¢ÉúÎüÑõ¸¯Ê´µÄµç³Ø·´Ó¦·½³ÌʽΪ                     ¡£
£¨5£©¡°Ã¾¡ª´ÎÂÈËáÑΡ±È¼Áϵç³Ø£¬Æä×°ÖÃʾÒâͼÈçͼ£¬¸Ãµç³Ø·´Ó¦µÄ×Ü·´Ó¦·½³ÌʽΪ_______________¡£

£¨6£©¹¤ÒµÉϵç½âÈÛÈÚÂÈ»¯ÄƵķ½³ÌʽΪ                                        ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

Ñо¿NO2¡¢SO2 ¡¢COµÈ´óÆøÎÛȾÆøÌåµÄ²âÁ¿¼°´¦Àí¾ßÓÐÖØÒªÒâÒå¡£
£¨1£©I2O5¿ÉʹH2S¡¢CO¡¢HC1µÈÑõ»¯£¬³£ÓÃÓÚ¶¨Á¿²â¶¨COµÄº¬Á¿¡£ÒÑÖª£º
2I2(s)+5O2(g)£½2I2O5(s)         ¡÷H£½£­75.56  kJ¡¤mol£­1
2CO(g)+O2(g)£½2CO2(g)       ¡÷H£½£­566.0  kJ¡¤mol£­1
д³öCO(g)ÓëI2O5(s)·´Ó¦Éú³ÉI2(s)ºÍCO2(g)µÄÈÈ»¯Ñ§·½³Ìʽ£º                                     ¡£
£¨2£©Ò»¶¨Ìõ¼þÏ£¬NO2ÓëSO2·´Ó¦Éú³ÉSO3ºÍNOÁ½ÖÖÆøÌ壺NO2(g)+SO2(g)SO3(g)+NO(g)½«Ìå»ý±ÈΪ1¡Ã2µÄNO2¡¢SO2ÆøÌåÖÃÓÚÃܱÕÈÝÆ÷Öз¢ÉúÉÏÊö·´Ó¦£¬ÏÂÁÐÄÜ˵Ã÷·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ                     ¡£
a£®Ìåϵѹǿ±£³Ö²»±ä
b£®»ìºÏÆøÌåÑÕÉ«±£³Ö²»±ä
c£®SO3ºÍNOµÄÌå»ý±È±£³Ö²»±ä
d£®Ã¿ÏûºÄ1molSO2µÄͬʱÉú³É1molNO
²âµÃÉÏÊö·´Ó¦Æ½ºâʱNO2ÓëSO2Ìå»ý±ÈΪ1¡Ã6£¬Ôòƽºâ³£ÊýK£½                 ¡£
£¨3£©´ÓÍÑÏõ¡¢ÍÑÁòºóµÄÑÌÆøÖлñÈ¡¶þÑõ»¯Ì¼£¬ÓöþÑõ»¯Ì¼ºÏ³É¼×´¼ÊÇ̼¼õÅŵÄз½Ïò¡£½«CO2ת»¯Îª¼×´¼µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCO2  (g)£«3H2(g)  CH3OH(g)£«H2O(g)   ¡÷H3
¢ÙÈ¡Îå·ÝµÈÌåÌå»ýCO2ºÍH2µÄµÄ»ìºÏÆøÌå £¨ÎïÖʵÄÁ¿Ö®±È¾ùΪ1£º3£©£¬·Ö±ð¼ÓÈëζȲ»Í¬¡¢ÈÝ»ýÏàͬµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÉÏÊö·´Ó¦£¬·´Ó¦Ïàͬʱ¼äºó£¬²âµÃ¼×´¼µÄÌå»ý·ÖÊý¦Õ(CH3OH) Ó뷴ӦζÈTµÄ¹ØϵÇúÏßÈçͼËùʾ£¬ÔòÉÏÊöCO2ת»¯Îª¼×´¼·´Ó¦µÄ¡÷H3                     0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£

¢ÚÔÚÈÝ»ýΪ1LµÄºãÎÂÃܱÕÈÝÆ÷ÖгäÈë1molCO2ºÍ3molH2£¬½øÐÐÉÏÊö·´Ó¦¡£²âµÃCO2ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçÏÂ×óͼËùʾ¡£ÈôÔÚÉÏÊöƽºâÌåϵÖÐÔÙ³ä0.5molCO2ºÍ1.5molË®ÕôÆø£¨±£³ÖζȲ»±ä£©£¬Ôò´Ëƽºâ½«                       Òƶ¯£¨Ìî¡°ÏòÕý·´Ó¦·½Ïò¡±¡¢¡°²»¡±»ò¡°Äæ·´Ó¦·½Ïò¡±£©¡£
         
¢ÛÖ±½Ó¼×´¼È¼Áϵç³Ø½á¹¹ÈçÉÏÓÒͼËùʾ¡£Æ乤×÷ʱ¸º¼«µç¼«·´Ó¦Ê½¿É±íʾΪ                      ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

¶þ¼×ÃÑÊÇ¡ªÖÖÖØÒªµÄÇå½àȼÁÏ£¬Ò²¿ÉÌæ´ú·úÀû°º×÷ÖÆÀä¼ÁµÈ£¬¶Ô³ôÑõ²ãÎÞÆÆ»µ×÷Ó᣹¤ÒµÉÏ¿ÉÀûÓÃúµÄÆø»¯²úÎˮúÆø£©ºÏ³É¶þ¼×ÃÑ¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÃºµÄÆø»¯µÄÖ÷Òª»¯Ñ§·´Ó¦·½³ÌʽΪ_______________________________________¡£
£¨2£©ÃºµÄÆø»¯¹ý³ÌÖвúÉúµÄÓк¦ÆøÌåÓÃÈÜÒºÎüÊÕ£¬Éú³ÉÁ½ÖÖËáʽÑΣ¬¸Ã·´Ó¦µÄ
»¯Ñ§·½³ÌʽΪ__________________________________________________________¡£
£¨3£©ÀûÓÃˮúÆøºÏ³É¶þ¼×ÃѵÄÈý²½·´Ó¦ÈçÏ£º
¢Ù2H2£¨g£©+CO£¨g£©CH3OH£¨g£©£»¡÷H=-90.8kJ¡¤mol£­1
¢Ú2CH3OH£¨g£©CH3OCH3£¨g£©+H2O£¨g£©£»¡÷H=-23.5kJ¡¤mol£­1
¢ÛCO£¨g£©+H2O£¨g£©CO2£¨g£©+H2£¨g£©£»¡÷H=-41.3kJ¡¤mol£­1
×Ü·´Ó¦£º3H2£¨g£©+3CO£¨g£©CH3OCH3£¨g£©+CO2£¨g£©µÄ¡÷H=             £»
Ò»¶¨Ìõ¼þϵÄÃܱÕÈÝÆ÷ÖУ¬¸Ã×Ü·´Ó¦´ïµ½Æ½ºâ£¬ÒªÌá¸ßCOµÄת»¯ÂÊ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊÇ
________________£¨Ìî×Öĸ´úºÅ£©¡£
a£®¸ßÎÂb£®¼ÓÈë´ß»¯¼Ác£®¼õÉÙCO2µÄŨ¶Èd£®Ôö¼ÓCOµÄŨ¶Èe£®·ÖÀë³ö¶þ¼×ÃÑ
£¨4£©ÒÑÖª·´Ó¦¢Ú2CH3OH£¨g£©CH3OCH3£¨g£©+H2O(g)ijζÈϵÄƽºâ³£ÊýΪ400¡£
´ËζÈÏ£¬ÔÚÃܱÕÈÝÆ÷ÖмÓÈëCH3OH£¬·´Ó¦µ½Ä³Ê±¿Ì²âµÃ¸÷×é·ÖµÄŨ¶ÈÈçÏ£º
ÎïÖÊ
CH3OH
CH3OCH3
H2O
Ũ¶È/£¨mol?L£©
0.44
0.6
0.6
¢Ù±ÈʱÕý¡¢Äæ·´Ó¦ËÙÂʵĴóС£º_______£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£
¢ÚÈô¼ÓÈëCH3OHºó£¬¾­l0min·´Ó¦´ïµ½Æ½ºâ£¬´Ëʱc(CH3OH)=__________£»¸Ãʱ
¼äÄÚ·´Ó¦ËÙÂÊv(CH3OH)=__________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸