¿ª·¢Ê¹ÓÃÇå½àÄÜÔ´£¬·¢Õ¹¡°µÍ̼¾­¼Ã¡±Õý³ÉΪ¿Æѧ¼ÒÑо¿µÄÖ÷Òª¿ÎÌâ¡£ÇâÆø¡¢¼×´¼ÊÇÓÅÖʵÄÇå½àȼÁÏ£¬¿ÉÖÆ×÷ȼÁϵç³Ø¡£
£¨1£©¼×ÍéË®ÕôÆøת»¯·¨ÖÆH2µÄÖ÷Ҫת»¯·´Ó¦ÈçÏ£º
CH4(g) + H2O(g)CO(g) + 3H2(g) ¡÷H=+206£®2 kJ¡¤mol£­1
CH4(g) + 2H2O(g)CO2(g) + 4H2(g) ¡÷H=+165£®0 kJ¡¤mol£­1
ÉÏÊö·´Ó¦ËùµÃÔ­ÁÏÆøÖеÄCOÄÜʹºÏ³É°±µÄ´ß»¯¼ÁÖж¾£¬±ØÐë³ýÈ¥¡£¹¤ÒµÉϳ£²ÉÓô߻¯¼Á´æÔÚÏÂCOÓëË®ÕôÆø·´Ó¦Éú³ÉÒ׳ýÈ¥µÄCO2£¬Í¬Ê±¿ÉÖƵõÈÌå»ýµÄÇâÆøµÄ·½·¨¡£´Ë·´Ó¦³ÆΪһÑõ»¯Ì¼±ä»»·´Ó¦£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ       ¡£
£¨2£©Éú²ú¼×´¼µÄÔ­ÁÏCOºÍH2À´Ô´ÓÚ£ºCH4(g) + H2O(g) CO(g) + 3H2(g)  ¦¤H>0
¢ÙÒ»¶¨Ìõ¼þÏÂCH4µÄƽºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼa¡£ÔòA¡¢B¡¢CÈýµã´¦¶ÔӦƽºâ³£Êý£¨KA¡¢KB¡¢KC£©µÄ´óС¹ØϵΪ___________¡£(Ìî¡°<¡±¡¢¡°>¡±¡¢¡°="¡±" )£»

¢Ú100¡æʱ£¬½«1 mol CH4ºÍ2 mol H2OͨÈëÈÝ»ýΪ1 LµÄ¶¨ÈÝÃÜ·âÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£¬ÄÜ˵Ã÷¸Ã·´Ó¦ÒѾ­´ïµ½Æ½ºâ״̬µÄÊÇ__________
a£®ÈÝÆ÷ÄÚÆøÌåÃܶȺ㶨  
b£®µ¥Î»Ê±¼äÄÚÏûºÄ0£®1 mol CH4ͬʱÉú³É0£®3 mol H2
c£®ÈÝÆ÷µÄѹǿºã¶¨      
d£®3vÕý(CH4) = vÄæ(H2)
£¨3£©25¡æʱ£¬ÔÚ20mL0£®1mol/LÇâ·úËáÖмÓÈëVmL0£®1mol/LNaOHÈÜÒº£¬²âµÃ»ìºÏÈÜÒºµÄpH±ä»¯ÇúÏßÈçͼËùʾ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ_____¡£

A£®pH£½3µÄHFÈÜÒººÍpH£½11µÄNaFÈÜÒºÖУ¬ ÓÉË®µçÀë³öµÄc(H+)ÏàµÈ
B£®¢ÙµãʱpH£½6£¬´ËʱÈÜÒºÖУ¬c(F£­)£­c(Na+)£½9£®9¡Á10-7mol/L
C£®¢Úµãʱ£¬ÈÜÒºÖеÄc(F£­)£½c(Na+)
D£®¢ÛµãʱV£½20mL£¬´ËʱÈÜÒºÖÐc(Na+)£½0£®1mol/L
£¨4£©³¤ÆÚÒÔÀ´£¬Ò»Ö±ÈÏΪ·úµÄº¬ÑõËá²»´æÔÚ¡£1971ÄêÃÀ¹ú¿Æѧ¼ÒÓ÷úÆøͨ¹ýϸ±ùĩʱ»ñµÃHFO£¬Æä½á¹¹Ê½ÎªH¡ªO¡ªF¡£HFOÓëË®·´Ó¦µÃµ½HFºÍ»¯ºÏÎïA£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ                     ¡£
£¨1£© CO(g) + H2O(g)CO2(g) + H2(g) ¡÷H=£­41£®2 kJ¡¤mol£­1      £¨2·Ö£©
£¨2£©¢Ù KC = KB >KA     £¨2·Ö£©
¢Ú cd    £¨2·Ö£©
£¨3£©BC   £¨2·Ö£©
£¨4£©H2O£«HFO=HF£«H2O2 £¨2·Ö£©

ÊÔÌâ·ÖÎö£º£¨1£© ÏȶÔÒÑÖªÈÈ»¯Ñ§·½³Ìʽ±àºÅΪ¢Ù¢Ú£¬¹Û²ì·¢ÏÖ¢Ú-¢Ù¿ÉµÃ£¬CO(g) + H2O(g)CO2(g) + H2(g) ¡÷H=£­41£®2 kJ¡¤mol£­1£»
£¨2£©¢Ù ƽºâ³£ÊýÖ¸¸÷Éú³ÉÎïŨ¶ÈµÄ»¯Ñ§¼ÆÁ¿Êý´ÎÃݵij˻ý³ýÒÔ¸÷·´Ó¦ÎïŨ¶ÈµÄ»¯Ñ§¼ÆÁ¿Êý´ÎÃݵij˻ýËùµÃµÄ±ÈÖµ£¬»¯Ñ§Æ½ºâ³£ÊýÖ»ÓëζÈÓйأ¬ËùÒÔB¡¢CÁ½µãÊÇÏàͬµÄζȣ¬ËùÒÔKC = KB£¬È»ºóA¡¢BÁ½µãÏà±È£¬´ÓAµ½B£¬¼×ÍéµÄת»¯Âʱä´ó£¬ËµÃ÷ÏòÕýÏòÒƶ¯£¬»¯Ñ§Æ½ºâ³£Êý±ä´ó£¬ËùÒÔKB>KA £¬ËùÒÔ´ð°¸ÊÇKC = KB >KA£»
¢Ú ¸ÃÈÝÆ÷µÄÌå»ý±£³Ö²»±ä£¬¸ù¾ÝÖÊÁ¿Êغ㶨ÂÉÖª£¬·´Ó¦Ç°ºó»ìºÏÆøÌåµÄÖÊÁ¿²»±ä£¬ËùÒÔÈÝÆ÷ÄÚÆøÌåµÄÃܶȲ»±ä£¬ËùÒÔ²»ÄܱíÃ÷´ïµ½»¯Ñ§Æ½ºâ״̬£¬¹Êa´íÎó£»µ¥Î»Ê±¼äÄÚÏûºÄ0£®1 mol CH4ͬʱÉú³É0£®3 mol H2£¬¶¼´ú±ívÕý£¬ËùÒÔ²»ÄܱíÃ÷´ïµ½»¯Ñ§Æ½ºâ״̬£¬¹Êb´íÎó£»Ëæ·´Ó¦½øÐлìºÏÆøÌå×ÜÎïÖʵÄÁ¿Ôö´ó£¬ÈÝÆ÷ÈÝ»ý²»±ä£¬Ñ¹Ç¿Ôö´ó£¬µ±ÈÝÆ÷µÄѹǿºã¶¨Ê±£¬ËµÃ÷µ½´ïƽºâ£¬¹ÊcÕýÈ·£»3vÕý(CH4) = vÄæ(H2)£¬ÎïÖʵÄÕýÄæËÙÂÊÖ®±ÈµÈÓÚ»¯Ñ§¼ÆÁ¿ÊýÖ®±È£¬·´Ó¦µ½´ïƽºâ£¬¹ÊdÕýÈ·£¬ËùÒÔÑ¡cd£»
£¨3£©pH£½3µÄHFÈÜÒº£¬ÓÉË®µçÀë³öµÄc(H+)£½£½10-11mol¡¤L£­1£¬pH£½11µÄNaFÈÜÒºÖУ¬ÓÉË®µçÀë³öµÄc(H+)£½10-14/10-11£½10-3mol¡¤L£­1£¬A´í£»¢Ùµãʱ£¬¸ù¾Ýµç×ÓÊغ㣬c(OH£­)£«c(F£­)£½c(Na+)£«c(H£«)£¬c(F£­)£­c(Na+)£½c(H£«)£­c(OH£­)£½10-6£­£½9£®9¡Á10-7mol/L£¬B¶Ô£»¸ù¾Ýµç×ÓÊغ㣬c(OH£­)£«c(F£­)£½c(Na+)£«c(H£«)£¬¢ÚµãʱpH=7£¬ÈÜÒºÖеÄc(H£«)£­c(OH£­)£¬ËùÒÔc(F£­)£½c(Na+)£¬C¶Ô£»¢ÛµãʱV£½20mL£¬´ËʱÈÜÒºÖÐc(Na+)£½£½0£®05mol/L£¬ËùÒÔÑ¡BC£»
£¨4£©µç¸ºÐÔF£¾O£¬¹ÊH-O-FÖÐFÔªËرíÏÖ-1¼Û£¬ÓëË®·´Ó¦Éú³ÉHF£¬·¢ÉúË®½â·´Ó¦£¬¿ÉÍÆÖªAΪH-O-O-H£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºH2O+HFO=HF+H2O2¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

Ëæ×Å´óÆøÎÛȾµÄÈÕÇ÷ÑÏÖØ£¬¹ú¼ÒÄâÓÚ¡°Ê®¶þÎ塱Æڼ䣬½«¶þÑõ»¯Áò(SO2)ÅÅ·ÅÁ¿¼õÉÙ8%£¬µªÑõ»¯Îï(NOx)ÅÅ·ÅÁ¿¼õÉÙ10%¡£Ä¿Ç°£¬Ïû³ý´óÆøÎÛȾÓжàÖÖ·½·¨¡£
¢ñ£®´¦ÀíNOxµÄÒ»ÖÖ·½·¨ÊÇÀûÓü×Íé´ß»¯»¹Ô­NOx¡£
CH4(g)+4NO2(g)£½4NO(g)+CO2(g)+2H2O(g) ¡÷H1£½£­574kJ¡¤mol­£­1
CH4(g)+4NO(g)£½2N2(g)+CO2(g)+2H2O(g)   ¡÷H2
CH4(g)+2NO2 (g)£½N2(g) + CO2(g)+2H2O(g)     ¡÷H3£½£­867kJ¡¤mol£­1
Ôò¡÷H2£½                 ¡£
¢ò£®»¯Ê¯È¼ÁϵÄȼÉÕ¡¢º¬Áò½ðÊô¿óʯµÄÒ±Á¶ºÍÁòËáµÄÉú²ú¹ý³ÌÖвúÉúµÄSO2ÊÇ´óÆøÖÐSO2µÄÖ÷ÒªÀ´Ô´¡££¨1£©½«Ãº×ª»¯ÎªË®ÃºÆøÊǽ«Ãº×ª»¯Îª½à¾»È¼Áϵķ½·¨Ö®Ò»£¬·´Ó¦Îª   C(s) + H2O(g)£½ CO(g) + H2(g)£¬
¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪK£½                    ¡£ 800¡æʱ£¬½«1molCO¡¢3mol H2O¡¢1mol H2³äÈëÈÝ»ýΪ1LµÄÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£ºCO(g) + H2O(g)  CO2(g) + H2(g)£¬·´Ó¦¹ý³ÌÖи÷ÎïÖʵÄŨ¶ÈÈçÓÒͼt1Ç°Ëùʾ±ä»¯¡£Èô±£³ÖζȲ»±ä£¬t2ʱÔÙÏòÈÝÆ÷ÖгäÈëCO¡¢H2¸÷1mol£¬Æ½ºâ½«     Òƶ¯£¨Ìî¡°Ïò×󡱡¢ ¡°ÏòÓÒ¡±»ò¡°²»¡±£©¡£t2ʱ£¬Èô¸Ä±ä·´Ó¦Ìõ¼þ£¬µ¼ÖÂH2Ũ¶È·¢ÉúÈçÓÒͼt2ºóËùʾµÄ±ä»¯£¬Ôò¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇ       £¨Ìî·ûºÅ£©¡£
a¼ÓÈë´ß»¯¼Á      b½µµÍζȠ    cËõСÈÝÆ÷Ìå»ý      d¼õÉÙCO2µÄÁ¿

£¨2£©µâÑ­»·¹¤ÒÕ²»½öÄÜÎüÊÕSO2½µµÍ»·¾³ÎÛȾ£¬Í¬Ê±ÓÖÄÜÖƵÃÇâÆø£¬¾ßÌåÁ÷³ÌÈçÏ£º

¢ÙÓÃÀë×Ó·½³Ìʽ±íʾ·´Ó¦Æ÷Öз¢ÉúµÄ·´Ó¦                        ¡£
¢ÚÓû¯Ñ§Æ½ºâÒƶ¯µÄÔ­Àí·ÖÎö£¬ÔÚ HI·Ö½â·´Ó¦ÖÐʹÓÃĤ·´Ó¦Æ÷·ÖÀë³öH2µÄÄ¿µÄÊÇ               ¡£
¢ó£®¿ª·¢ÐÂÄÜÔ´Êǽâ¾ö´óÆøÎÛȾµÄÓÐЧ;¾¶Ö®Ò»¡£¼×´¼È¼Áϵç³Ø(¼ò³ÆDMFC)ÓÉÓڽṹ¼òµ¥¡¢ÄÜÁ¿×ª»¯Âʸߡ¢¶Ô»·¾³ÎÞÎÛȾ,¿É×÷Ϊ³£¹æÄÜÔ´µÄÌæ´úÆ·¶øÔ½À´Ô½Êܵ½¹Ø×¢¡£DMFC¹¤×÷Ô­ÀíÈçͼËùʾ£º

ͨÈëaÆøÌåµÄµç¼«ÊÇÔ­µç³ØµÄ     £¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£¬
Æäµç¼«·´Ó¦Ê½Îª                  ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ͨ³£ÈËÃǰѲð¿ª1 molij»¯Ñ§¼üËùÎüÊÕµÄÄÜÁ¿¿´³É¸Ã»¯Ñ§¼üµÄ¼üÄÜ¡£¼üÄܵĴóС¿ÉÒÔºâÁ¿»¯Ñ§¼üµÄÇ¿Èõ,Ò²¿ÉÒÔ¹ÀË㻯ѧ·´Ó¦µÄ·´Ó¦ÈÈ(¦¤H),»¯Ñ§·´Ó¦µÄ¦¤HµÈÓÚ·´Ó¦ÖжÏÁѾɻ¯Ñ§¼üµÄ¼üÄÜÖ®ºÍÓë·´Ó¦ÖÐÐγÉл¯Ñ§¼üµÄ¼üÄÜÖ®ºÍµÄ²î¡£
»¯Ñ§¼ü
Si¡ªO
Si¡ªCl
H¡ªH
H¡ªCl
Si¡ªSi
Si¡ªC
¼üÄÜ/kJ¡¤mol-1
460
360
436
431
176
347
 
Çë»Ø´ðÏÂÁÐÎÊÌâ:
(1)±È½ÏÏÂÁÐÁ½×éÎïÖʵÄÈÛµã¸ßµÍ(Ìî¡°>¡±»ò¡°<¡±)¡£
SiC¡¡¡¡¡¡¡¡Si;SiCl4¡¡¡¡¡¡¡¡SiO2¡£ 
(2)ÈçͼÁ¢·½ÌåÖÐÐĵġ°¡±±íʾ¹è¾§ÌåÖеÄÒ»¸öÔ­×Ó,ÇëÔÚÁ¢·½ÌåµÄ¶¥µãÓá°¡±±íʾ³öÓëÖ®½ôÁڵĹèÔ­×Ó¡£

(3)¹¤ÒµÉÏÓøߴ¿¹è¿Éͨ¹ýÏÂÁз´Ó¦ÖÆÈ¡:SiCl4(g)+2H2(g)Si(s)+4HCl(g),¸Ã·´Ó¦µÄ·´Ó¦ÈȦ¤H=¡¡¡¡¡¡kJ/mol¡£ 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÓÃCl2Éú²úijЩº¬ÂÈÓлúÎïʱ»á²úÉú¸±²úÎïHCl¡£ÀûÓ÷´Ó¦A,¿ÉʵÏÖÂȵÄÑ­»·ÀûÓá£
·´Ó¦A:4HCl+O22Cl2+2H2O
ÒÑÖª:¢ñ.·´Ó¦AÖÐ,4 mol HCl±»Ñõ»¯,·Å³ö115.6 kJµÄÈÈÁ¿¡£
¢ò.

ÅжÏÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨  £©
A£®·´Ó¦AµÄ¦¤H>-115.6 kJ/mol
B£®¶Ï¿ª1 mol H¡ªO¼üÓë¶Ï¿ª1 mol H¡ªCl¼üËùÐèÄÜÁ¿Ïà²îԼΪ32 kJ
C£®H2OÖÐH¡ªO¼ü±ÈHClÖÐH¡ªCl¼üÈõ
D£®ÓÉ¢òÖеÄÊý¾ÝÅжÏÂÈÔªËصķǽðÊôÐÔ±ÈÑõÔªËØÇ¿

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºµ¥Ñ¡Ìâ

ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙC£¨s£©+O2£¨g£©==CO2£¨g£©¡÷H= ¡ª393.5kJ/mol
¢ÚCO£¨g£©+ 1/2 O2£¨g£©="=" CO2£¨g£©¡÷H= ¡ª283.0kJ/mol
¢Û2Fe£¨s£©+3CO£¨g£©==Fe2O3£¨s£©+3C£¨s£©¡÷H= ¡ª489.0kJ/mol
Ôò4Fe£¨s£©+3O2£¨g£©==2Fe2O3£¨s£©µÄ·´Ó¦ÈȦ¤HΪ£¨   £©
A£®-1641.0kJ/molB£®+3081kJ/mol
C£®+663.5kJ/molD£®-2507.0kJ/mol

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

̼¼°Æ仯ºÏÎïÓй㷺µÄÓÃ;¡£
(1)½«Ë®ÕôÆøͨ¹ýºìÈȵÄ̼¼´¿É²úÉúˮúÆø¡£·´Ó¦Îª
C(s)£«H2O(g)CO(g)£«H2(g)¡¡¦¤H£½£«131.3 kJ¡¤mol£­1£¬
ÒÔÉÏ·´Ó¦´ïµ½Æ½ºâºó£¬ÔÚÌå»ý²»±äµÄÌõ¼þÏ£¬ÒÔÏ´ëÊ©ÓÐÀûÓÚÌá¸ßH2OµÄƽºâת»¯ÂʵÄÊÇ________¡£(ÌîÐòºÅ)
A£®Éý¸ßζÈB£®Ôö¼Ó̼µÄÓÃÁ¿C£®¼ÓÈë´ß»¯¼ÁD£®ÓÃCOÎüÊÕ¼Á³ýÈ¥CO
(2)ÒÑÖª£ºC(s)£«CO2(g)2CO(g)¡¡¦¤H£½£«172.5 kJ¡¤mol£­1£¬ÔòCO(g)£«H2O(g)CO2(g)£«H2(g)µÄìʱ䦤H£½________¡£
(3)COÓëH2ÔÚÒ»¶¨Ìõ¼þÏ¿ɷ´Ó¦Éú³É¼×´¼£ºCO(g)£«2H2(g)CH3OH(g)¡£¼×´¼ÊÇÒ»ÖÖȼÁÏ£¬¿ÉÀûÓü״¼Éè¼ÆÒ»¸öȼÁϵç³Ø£¬ÓÃÏ¡ÁòËá×÷µç½âÖÊÈÜÒº£¬¶à¿×ʯī×÷µç¼«£¬¸Ãµç³Ø¸º¼«·´Ó¦Ê½Îª__________________________________¡£
ÈôÓøõç³ØÌṩµÄµçÄܵç½â60 mL NaClÈÜÒº£¬ÉèÓÐ0.01 mol CH3OHÍêÈ«·Åµç£¬NaCl×ãÁ¿£¬ÇÒµç½â²úÉúµÄCl2È«²¿Òݳö£¬µç½âÇ°ºóºöÂÔÈÜÒºÌå»ýµÄ±ä»¯£¬Ôòµç½â½áÊøºóËùµÃÈÜÒºµÄpH£½________¡£
(4)½«Ò»¶¨Á¿µÄCO(g)ºÍH2O(g)·Ö±ðͨÈëµ½Ìå»ýΪ2.0 LµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÒÔÏ·´Ó¦£ºCO(g)£«H2O(g)CO2(g)£«H2(g)¡£µÃµ½ÈçÏÂÊý¾Ý£º
ζÈ/¡æ
ÆðʼÁ¿/mol
ƽºâÁ¿/mol
´ïµ½Æ½ºâËùÐèʱ¼ä/min
H2O
CO
H2
CO
 
900
1.0
2.0
0.4
1.6
3.0
 
ͨ¹ý¼ÆËãÇó³ö¸Ã·´Ó¦µÄƽºâ³£Êý(½á¹û±£ÁôÁ½Î»ÓÐЧÊý×Ö)________¡£¸Ä±ä·´Ó¦µÄijһÌõ¼þ£¬·´Ó¦½øÐе½t minʱ£¬²âµÃ»ìºÏÆøÌåÖÐCO2µÄÎïÖʵÄÁ¿Îª0.6 mol¡£ÈôÓÃ200 mL 5 mol/LµÄNaOHÈÜÒº½«ÆäÍêÈ«ÎüÊÕ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ(ÓÃÒ»¸öÀë×Ó·½³Ìʽ±íʾ)__________________________
(5)¹¤ÒµÉú²úÊÇ°ÑˮúÆøÖеĻìºÏÆøÌå¾­¹ý´¦Àíºó»ñµÃµÄ½Ï´¿H2ÓÃÓںϳɰ±¡£ºÏ³É°±·´Ó¦Ô­ÀíΪN2(g)£«3H2(g)2NH3(g)¡¡¦¤H£½£­92.4 kJ¡¤mol£­1¡£ÊµÑéÊÒÄ£Ä⻯¹¤Éú²ú£¬·Ö±ðÔÚ²»Í¬ÊµÑéÌõ¼þÏ·´Ó¦£¬N2Ũ¶ÈËæʱ¼ä±ä»¯Èçͼ¼×Ëùʾ¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÓëʵÑé¢ñ±È½Ï£¬ÊµÑé¢ò¸Ä±äµÄÌõ¼þΪ________________________________¡£
¢ÚʵÑé¢ó±ÈʵÑé¢ñµÄζÈÒª¸ß£¬ÆäËûÌõ¼þÏàͬ£¬ÇëÔÚͼÒÒÖл­³öʵÑé¢ñºÍʵÑé¢óÖÐNH3Ũ¶ÈËæʱ¼ä±ä»¯µÄʾÒâͼ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂͼÊÇú»¯¹¤²úÒµÁ´µÄÒ»²¿·Ö,ÊÔÔËÓÃËùѧ֪ʶ,½â¾öÏÂÁÐÎÊÌâ:

¢ñ.ÒÑÖª¸Ã²úÒµÁ´ÖÐij·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ:K=,д³öËüËù¶ÔÓ¦·´Ó¦µÄ»¯Ñ§·½³Ìʽ:                                  
                                                     ¡¡¡£ 
¢ò.¶þ¼×ÃÑ(CH3OCH3)ÔÚδÀ´¿ÉÄÜÌæ´ú²ñÓͺÍÒº»¯Æø×÷Ϊ½à¾»ÒºÌåȼÁÏʹÓ᣹¤ÒµÉÏÒÔCOºÍH2ΪԭÁÏÉú²úCH3OCH3¡£¹¤ÒµÖƱ¸¶þ¼×ÃÑÔÚ´ß»¯·´Ó¦ÊÒÖÐ(ѹÁ¦2.0¡«10.0 MPa,ζÈ230¡«280 ¡æ)½øÐÐÏÂÁз´Ó¦:
¢ÙCO(g)+2H2(g)CH3OH(g)
¦¤H1="-90.7" kJ¡¤mol-1
¢Ú2CH3OH(g)CH3OCH3(g)+H2O(g)
¦¤H2="-23.5" kJ¡¤mol-1
¢ÛCO(g)+H2O(g)CO2(g)+H2(g)
¦¤H3="-41.2" kJ¡¤mol-1
(1)д³ö´ß»¯·´Ó¦ÊÒÖÐÈý¸ö·´Ó¦µÄ×Ü·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ:                          ¡£
(2)ÔÚijζÈÏÂ,2 LÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦¢Ù,ÆðʼʱCO¡¢H2µÄÎïÖʵÄÁ¿·Ö±ðΪ2 molºÍ6 mol,3 minºó´ïµ½Æ½ºâ,²âµÃCOµÄת»¯ÂÊΪ60%,Ôò3 minÄÚCOµÄƽ¾ù·´Ó¦ËÙÂÊΪ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£ÈôͬÑùÌõ¼þÏÂÆðʼʱCOÎïÖʵÄÁ¿Îª4 mol,´ïµ½Æ½ºâºóCH3OHΪ2.4 mol,ÔòÆðʼʱH2Ϊ¡¡¡¡¡¡¡¡mol¡£
(3)ÏÂÁÐÓйط´Ó¦¢ÛµÄ˵·¨ÕýÈ·µÄÊÇ¡¡¡¡¡¡¡¡¡£
A.ÔÚÌå»ý¿É±äµÄÃܱÕÈÝÆ÷ÖÐ,ÔÚ·´Ó¦¢Û´ïµ½Æ½ºâºó,Èô¼Óѹ,Ôòƽºâ²»Òƶ¯¡¢»ìºÏÆøÌåƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä¡¢»ìºÏÆøÌåÃܶȲ»±ä
B.Èô830 ¡æʱ·´Ó¦¢ÛµÄK=1.0,ÔòÔÚ´ß»¯·´Ó¦ÊÒÖз´Ó¦¢ÛµÄK£¾1.0
C.ijζÈÏÂ,ÈôÏòÒÑ´ïƽºâµÄ·´Ó¦¢ÛÖмÓÈëµÈÎïÖʵÄÁ¿µÄCOºÍH2O(g),ÔòƽºâÓÒÒÆ¡¢Æ½ºâ³£Êý±ä´ó
(4)ΪÁËÑ°ÕÒºÏÊʵķ´Ó¦Î¶È,Ñо¿Õß½øÐÐÁËһϵÁÐʵÑé,ÿ´ÎʵÑé±£³ÖÔ­ÁÏÆø×é³É¡¢Ñ¹Ç¿¡¢·´Ó¦Ê±¼äµÈÒòËز»±ä,ʵÑé½á¹ûÈçͼ,

ÔòCOת»¯ÂÊËæζȱ仯µÄ¹æÂÉÊÇ                                        ¡¡¡£
ÆäÔ­ÒòÊÇ                                                   ¡¡¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

ÔËÓû¯Ñ§·´Ó¦Ô­ÀíÑо¿NH3µÄÐÔÖʾßÓÐÖØÒªÒâÒå¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©°±Æø¡¢¿ÕÆø¿ÉÒÔ¹¹³ÉȼÁϵç³Ø£®Æäµç³Ø·´Ó¦Ô­ÀíΪ4NH3£«3O2=2N2£«6H2O¡£Ôòµç½âÖÊÈÜÒºÓ¦¸ÃÏÔ             £¨Ìî¡°ËáÐÔ¡±¡°ÖÐÐÔ¡±»ò¡°¼îÐÔ¡±£©£®Õý¼«µÄµç¼«·´Ó¦Ê½Îª           ¡£
£¨2£©25¡æʱ£®½«amol¡¤L¡ª1µÄ°±Ë®Óë0.1mol¡¤L¡ª1µÄÑÎËáµÈÌå»ý»ìºÏ¡£
¢Ùµ±ÈÜÒºÖÐÀë×ÓŨ¶È¹ØϵÂú×ãc(NH4+)>c(Cl-)£©Ê±£®Ôò·´Ó¦µÄÇé¿ö¿ÉÄÜΪ          ¡£
A£®ÑÎËá²»×㣮°±Ë®Ê£Óà   B£®°±Ë®ÓëÑÎËáÇ¡ºÃÍêÈ«·´Ó¦    C£®ÑÎËá¹ýÁ¿
¢Úµ±ÈÜÒºÖÐc(NH4+)=c(Cl-)£©Ê±£®Óú¬¡°a¡±µÄ´úÊýʽ±íʾNH3¡¤H2OµÄµçÀëƽºâ³£ÊýKb=______________.
£¨3£©ÔÚ0.5LºãÈÝÃܱÕÈÝÆ÷ÖУ¬Ò»¶¨Á¿µÄN2ÓëH2½øÐз´Ó¦£ºN2(g)£«3H2(g)2NH3(g)  ?H=bkJ/mol£¬Æ仯ѧƽºâ³£ÊýKÓëζȵĹØϵÈçÏ£º
ζÈ/¡æ
200
300
400
K
1.0
0.86
0.5
 
¢Ùд³ö¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýµÄ±í´ïʽ£º__________£¬b________£¨Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©0
¢Ú400¡æʱ£¬²âµÃijʱ¿Ì°±Æø¡¢µªÆø¡¢ÇâÆøµÄÎïÖʵÄÁ¿·Ö±ðΪ3mol¡¢2mol¡¢1molʱ£¬´Ëʱ¿Ì¸Ã·´Ó¦µÄvÕý£¨N2£©_________£¨Ìî¡°´óÓÚ¡±¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©vÄ棨N2£©.
£¨4£©ÒÑÖª£º¢Ù4NH3(g)£«3O2(g)=2N2(g)£«6H2O(g) ?H="-1266.8KJ/mol" £»¢ÚN2(g)£«O2(g)=2NO(g)  ?H=+180.5KJ/mol£¬Ð´³ö°±¸ßδ߻¯Ñõ»¯µÄÈÈ»¯Ñ§·½³Ìʽ£º                    ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º²»Ïê ÌâÐÍ£ºÌî¿ÕÌâ

¹¤ÒµÉϳ£ÀûÓÃCOºÍH2ºÏ³É¿ÉÔÙÉúÄÜÔ´¼×´¼¡£
£¨1£©ÒÑÖªCO(g)¡¢CH3OH(l)µÄȼÉÕÈÈ·Ö±ðΪ283.0 kJ¡¤mol-1ºÍ726.5 kJ¡¤mol-1£¬ÔòCH3OH(l)²»ÍêȫȼÉÕÉú³ÉCO(g)ºÍH2O(l)µÄÈÈ»¯Ñ§·½³ÌʽΪ                       ¡£
£¨2£©ºÏ³É¼×´¼µÄ·½³ÌʽΪCO(g)+2H2(g)CH3OH(g)    ¦¤H <0¡£
ÔÚ230 ¡æ?270 ¡æ×îΪÓÐÀû¡£ÎªÑо¿ºÏ³ÉÆø×îºÏÊʵÄÆðʼ×é³É±Èn(H2):n(CO)£¬·Ö±ðÔÚ230 ¡æ¡¢250 ¡æºÍ270 ¡æ½øÐÐʵÑ飬½á¹ûÈçÏÂ×óͼËùʾ¡£ÆäÖÐ270 ¡æµÄʵÑé½á¹ûËù¶ÔÓ¦µÄÇúÏßÊÇ_____£¨Ìî×Öĸ£©£»µ±ÇúÏßX¡¢Y¡¢Z¶ÔÓ¦µÄͶÁϱȴﵽÏàͬµÄCOƽºâת»¯ÂÊʱ£¬¶ÔÓ¦µÄ·´Ó¦Î¶ÈÓëͶ½Ï±ÈµÄ¹ØϵÊÇ                    ¡£

£¨3£©µ±Í¶ÁϱÈΪ1¡Ã1£¬Î¶ÈΪ230 ¡æ£¬Æ½ºâ»ìºÏÆøÌåÖУ¬CH3OHµÄÎïÖʵÄÁ¿·ÖÊýΪ        £¨±£Áô1λСÊý£©£»Æ½ºâʱCOµÄת»¯ÂÊ           ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸