ÌâÄ¿ÄÚÈÝ

£¨1£©¼ÒÓÃÒº»¯Ê¯ÓÍÆøµÄÖ÷Òª³É·ÖÖ®Ò»ÊǶ¡Íé(C4H10)£¬µ±10 kg¶¡ÍéÍêȫȼÉÕ²¢Éú³É¶þÑõ»¯Ì¼ÆøÌåºÍҺ̬ˮʱ£¬·Å³öµÄÈÈÁ¿Îª5¡Á105 kJ¡£ÊÔд³ö¶¡ÍéȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ£º                             ¡£ÒÑÖª1molҺ̬ˮÆû»¯Ê±ÐèÒªÎüÊÕ44 kJÈÈÁ¿£¬Ôò·´Ó¦C4H10(g)+6.5O2(g)=4CO2(g)+5H2O(g)µÄ¦¤H=         ¡£
£¨2£©ÓÐͬѧÓö¡ÍéÓë¿ÕÆøΪԭÁÏÖÆ×÷һȼÉÕµç³Ø£¬Í¨È붡ÍéµÄÒ»¼«Îª      ¼«¡£ÈôÒÔÏ¡ÁòËáΪµç½âÖÊÈÜҺʱ£¬ÆäÕý¼«·´Ó¦Ê½Îª                                        ¡£
£¨3£©ÒÑÖª:Fe(s) +1/2O2(g)=FeO(s)  ¦¤H=£­272.0kJ¡¤mol-1
2Al(s)+3/2O2(g)=Al2O3(s) ¦¤H=£­1675.7kJ¡¤mol-1 
AlºÍFeO·¢ÉúÂÁÈÈ·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ                                      ¡£
£¨4£©ÒÑÖª²ð¿ª1 mol H£­H¼ü£¬1 molN£­H¼ü£¬1 molN¡ÔN¼ü·Ö±ðÐèÒªµÄÄÜÁ¿ÊÇakJ¡¢bkJ¡¢ckJ£¬ÔòN2ÓëH2·´Ó¦Éú³ÉNH3µÄÈÈ»¯Ñ§·½³ÌʽΪ                                ¡£


ÊÔÌâ·ÖÎö£º£¨1£©ÒÀ¾ÝÖÊÁ¿»»ËãÎïÖʵÄÁ¿£¬½áºÏ»¯Ñ§·½³Ìʽ¶ÔÓ¦µÄÎïÖʵÄÁ¿¼ÆËã·´Ó¦·Å³öµÄÈÈÁ¿£¬ÒÀ¾ÝÈÈ»¯Ñ§·½³ÌʽÊéд·½·¨£¬±ê×¢ÎïÖʾۼ¯×´Ì¬ºÍ·´Ó¦ìʱäд³ö¡£ÓÉÌâÒâÖªH2O(l)=H2O(g) ¦¤H=44 kJ¡¤mol-1,ÀûÓøÇ˹¶¨ÂɼÆËã¿ÉµÃ£»
£¨2£©¶¡Íé--¿ÕÆøȼÁϵç³Ø¹¤×÷ʱ£¬¶¡Íéʧµç×Ó·¢ÉúÑõ»¯·´Ó¦×÷¸º¼«£»ÒÔÏ¡ÁòËáΪµç½âÖÊÈÜҺʱ£¬½áºÏµç¼«·´Ó¦Ê½µÄÊéд¹æÂɿɵã»
£¨3£©ÒÀ¾ÝÌâ¸ÉÈÈ»¯Ñ§·½³Ìʽ½áºÏ¸Ç˹¶¨ÂÉд³ö¸ÃÈÈ»¯Ñ§·´Ó¦·½³Ìʽ¡£½«·½³Ìʽ¢Ú-¢Ù¡Á3µÃ2Al£¨s£©+3FeO£¨s£©¨TAl2O3£¨s£©+3Fe£¨s£©¡÷H=-859.7 kJ?mol-1£»
(4)ÒÀ¾Ý¡÷H=·´Ó¦ÎïµÄ¼üÄÜÖ®ºÍ¡ªÉú³ÉÎïµÄ¼üÄÜÖ®ºÍ¿ÉÇóµÄºÏ³É°±·´Ó¦µÄìʱäΪ(3a+b-6c) kJ¡¤mol-1£¬ÒÀ¾ÝÈÈ»¯Ñ§·½³ÌʽÊéд·½·¨£¬±ê×¢ÎïÖʾۼ¯×´Ì¬ºÍ·´Ó¦ìʱäд³ö¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÏÂͼÊÇú»¯¹¤²úÒµÁ´µÄÒ»²¿·Ö£¬ÊÔÔËÓÃËùѧ֪ʶ£¬½â¾öÏÂÁÐÎÊÌ⣺

I£®ÒÑÖª¸Ã²úÒµÁ´ÖÐij·´Ó¦µÄƽºâ±í³£Êý´ïʽΪ£ºK£½£¬ËüËù¶ÔÓ¦·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                                            ¡£
II£®¶þ¼×ÃÑ£¨CH3OCH3£©ÔÚδÀ´¿ÉÄÜÌæ´ú²ñÓͺÍÒº»¯Æø×÷Ϊ½à¾»ÒºÌåȼÁÏʹÓ㬹¤ÒµÉÏÒÔCOºÍH2ΪԭÁÏÉú²úCH3OCH3¡£¹¤ÒµÖƱ¸¶þ¼×ÃÑÔÚ´ß»¯·´Ó¦ÊÒÖУ¨Ñ¹Á¦2£®0¡«10£®0Mpa£¬Î¶È230¡«280¡æ£©½øÐÐÏÂÁз´Ó¦£º
¢ÙCO(g)£«2H2(g)CH3OH(g) ¡÷H1£½£­90£®7kJ¡¤mol£­1
¢Ú2CH3OH(g)CH3OCH3(g)£«H2O(g) ¡÷H2£½£­23£®5kJ¡¤mol£­1
¢ÛCO(g)£«H2O(g)CO2(g)£«H2(g) ¡÷H3£½£­41£®2kJ¡¤mol£­1
£¨1£©´ß»¯·´Ó¦ÊÒÖÐ×Ü·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ                                         ¡£
830¡æʱ·´Ó¦¢ÛµÄK£½1£®0£¬ÔòÔÚ´ß»¯·´Ó¦ÊÒÖз´Ó¦¢ÛµÄK    1£®0£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£
£¨2£©ÔÚijζÈÏ£¬Èô·´Ó¦¢ÙµÄÆðʼŨ¶È·Ö±ðΪ£ºc(CO)£½1 mol/L£¬c(H2)£½2£®4 mol/L£¬5 minºó´ïµ½Æ½ºâ£¬COµÄת»¯ÂÊΪ50£¥£¬Ôò5 minÄÚCOµÄƽ¾ù·´Ó¦ËÙÂÊΪ      ¡¡     £»Èô·´Ó¦ÎïµÄÆðʼŨ¶È·Ö±ðΪ£ºc(CO)£½4 mol/L£¬c(H2)£½a mol/L£»´ïµ½Æ½ºâºó£¬c(CH3OH)£½2 mol/L£¬a£½        mol/L¡£
£¨3£©·´Ó¦¢ÚÔÚt¡æʱµÄƽºâ³£ÊýΪ400£¬´ËζÈÏ£¬ÔÚ0£®5LµÄÃܱÕÈÝÆ÷ÖмÓÈëÒ»¶¨µÄ¼×´¼£¬·´Ó¦µ½Ä³Ê±¿Ì²âµÃ¸÷×é·ÖµÄÎïÖʵÄÁ¿Å¨¶ÈÈçÏ£º
ÎïÖÊ
CH3OH
CH3OCH3
H2O
c(mol/L)
0£®8
1£®24
1£®24
 
¢Ù´Ëʱ¿ÌvÕý   vÄ棨Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±
¢Úƽºâʱ¶þ¼×ÃѵÄÎïÖʵÄÁ¿Å¨¶ÈÊÇ                    ¡£
ÒÔ¶þ¼×ÃÑ¡¢¿ÕÆø¡¢KOH ÈÜҺΪԭÁÏ£¬ÒÔʯīΪµç¼«¿ÉÖ±½Ó¹¹³ÉȼÁϵç³Ø£¬Ôò¸Ãµç³ØµÄ¸º¼«·´Ó¦Ê½Îª                                              £»ÈôÒÔ1£®12L/min£¨±ê×¼×´¿ö£©µÄËÙÂÊÏòµç³ØÖÐͨÈë¶þ¼×ÃÑ£¬Óøõç³Øµç½â500mL2mol/L CuSO4ÈÜÒº£¬Í¨µç0£®50 minºó£¬¼ÆËãÀíÂÛÉÏ¿ÉÎö³ö½ðÊôÍ­µÄÖÊÁ¿Îª                         
¢ñ£º¹¤ÒµÉÏÓÃCO2ºÍH2ÔÚÒ»¶¨Ìõ¼þ·¢ÉúÈçÏ·´Ó¦ºÏ³É¼×´¼²¢·Å³ö´óÁ¿µÄÈÈ£ºCO2(g)+3H2(g)CH3OH(g)+H2O(g) ¦¤H1   »Ø´ðÏÂÁÐÎÊÌâ¡£
£¨1£©ÒÑÖª£º2H2(g)+O2(g)=2H2O(g)   ¦¤H2
Ôò·´Ó¦2CH3OH(g)+3O2(g)=2CO2(g)+4H2O(g)  ¦¤H=           £¨Óú¬¦¤H1¡¢¦¤H2±íʾ£©
£¨2£©Èô·´Ó¦Î¶ÈÉý¸ß£¬CO2µÄת»¯ÂÊ       (Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©¡£
£¨3£©Ð´³öÔÚËáÐÔ»·¾³ÖУ¬¼×´¼È¼Áϵç³ØÖеÄÕý¼«·´Ó¦·½³Ìʽ                             
¢ò£ºÉú²ú¼×´¼µÄÔ­ÁÏH2¿ÉÓÃÈçÏ·½·¨ÖƵãºCH4(g) + H2O(g)  CO(g) + 3H2(g)£¬Ò»¶¨Î¶ÈÏ£¬½«2 mol CH4ºÍ4 mol H2OͨÈëÈÝ»ýΪ10LµÄÃܱշ´Ó¦ÊÒÖУ¬·´Ó¦ÖÐCOµÄÎïÖʵÄÁ¿Å¨¶ÈµÄ±ä»¯Çé¿öÈçͼËùʾ£¬Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨4£©·´Ó¦½øÐе½4·ÖÖÓµ½´ïƽºâ¡£Çë¼ÆËã´Ó·´Ó¦¿ªÊ¼µ½¸Õ¸Õƽºâ£¬Æ½¾ù·´Ó¦ËÙÂÊv(H2)Ϊ               £»²¢Çó´Ë·´Ó¦ÔÚ´ËζÈϵÄƽºâ³£Êý£¨ÔÚ´ðÌ⿨¶ÔÓ¦µÄ·½¿òÄÚд³ö¼ÆËã¹ý³Ì£©¡£
£¨5£©ÔÚµÚ5·ÖÖÓʱ½«ÈÝÆ÷µÄÌå»ý˲¼äËõСһ°ëºó£¬ÈôÔÚµÚ8·ÖÖÓʱ´ïµ½ÐµÄƽºâ£¨´ËʱCOµÄŨ¶ÈԼΪ0.25 mol¡¤L¡ª1 £©£¬ÇëÔÚͼÖл­³öµÚ5·ÖÖÓºóH2Ũ¶ÈµÄ±ä»¯ÇúÏß¡£
¶þ¼×ÃÑÊÇÒ»ÖÖÖØÒªµÄÇå½àȼÁÏ£¬Ò²¿ÉÌæ´ú·úÀû°º×÷ÖÆÀä¼ÁµÈ£¬¶Ô³ôÑõ²ãÎÞÆÆ»µ×÷Ó᣹¤ÒµÉÏ¿ÉÀûÓÃúµÄÆø»¯²úÎˮúÆø£©ºÏ³É¶þ¼×ÃÑ( CH3OCH3)¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)ÀûÓÃˮúÆøºÏ³É¶þ¼×ÃѵÄÈý²½·´Ó¦ÈçÏ£º
¢Ù2H2(g)+CO(g)CH3OH(g)    ¡÷H= £­90£®8kJ/mol
¢Ú2CH3OH(g)CH3OCH3(g)+H2O(g) ¡÷H=£­23£®5kJ/mol
¢ÛCO(g)+H2O(g)CO2(g)+H2(g)  ¡÷H=£­41£®3kJ/mol
×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£º
________£¨Ìî×Öĸ´úºÅ£©¡£
a£®Ñ¹ËõÌå»ý     b£®¼ÓÈë´ß»¯¼Á  c£®¼õÉÙCO2µÄŨ¶È  d£®Ôö¼ÓCOµÄŨ¶È
e£®·ÖÀë³ö¶þ¼×ÃÑ£¨CH3OCH3£©
(2)ÒÑÖª·´Ó¦¢Ú2CH3OH(g) CH3OCH3(g)+H2O(g) ¡÷H=£­23£®5kJ/mol
ijζÈϵÄƽºâ³£ÊýΪ400¡£´ËζÈÏ£¬ÔÚÃܱÕÈÝÆ÷ÖмÓÈëCH3OH£¬·´Ó¦µ½Ä³Ê±¿Ì²âµÃ¸÷×é·ÖµÄŨ¶ÈÈçÏ£º
ÎïÖÊ
CH3OH
CH3OCH3
H2O
Ũ¶È£¨mol¡¤L-1£©
0£®40
0£®6
0£®6
¢Ù±È½Ï´ËʱÕý¡¢Äæ·´Ó¦ËÙÂʵĴóС±È½Ï£º_________£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£
¢Ú¸Ã·´Ó¦µÄƽºâ³£ÊýµÄ±í´ïʽΪK=_____,ζÈÉý¸ß£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK____£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø