ÌâÄ¿ÄÚÈÝ

½ðÊôÎÙÓÃ;¹ã·º£¬Ö÷ÒªÓÃÓÚÖÆÔìÓ²ÖÊ»òÄ͸ßεĺϽð£¬ÒÔ¼°µÆÅݵĵÆË¿¡£¸ßÎÂÏ£¬ÔÚÃܱÕÈÝÆ÷ÖÐÓÃH2»¹Ô­WO3¿ÉµÃµ½½ðÊôÎÙ£¬Æä×Ü·´Ó¦Îª£º
WO3 (s) + 3H2 (g)W (s) + 3H2O (g)
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÅÉÏÊö·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý±í´ïʽΪ___________________________¡£
¢ÆijζÈÏ·´Ó¦´ïƽºâʱ£¬H2ÓëË®ÕôÆøµÄÌå»ý±ÈΪ2:3£¬ÔòH2µÄƽºâת»¯ÂÊΪ_____________________£»ËæζȵÄÉý¸ß£¬H2ÓëË®ÕôÆøµÄÌå»ý±È¼õС£¬Ôò¸Ã·´Ó¦Îª·´Ó¦_____________________£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©¡£
¢ÇÉÏÊö×Ü·´Ó¦¹ý³Ì´óÖ·ÖΪÈý¸ö½×¶Î£¬¸÷½×¶ÎÖ÷Òª³É·ÖÓëζȵĹØϵÈçϱíËùʾ£º
ζÈ
25¡æ  ~  550¡æ  ~  600¡æ  ~  700¡æ
Ö÷Òª³É·Ý
WO3      W2O5      WO2        W
 
µÚÒ»½×¶Î·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ___________________________£»580¡æʱ£¬¹ÌÌåÎïÖʵÄÖ÷Òª³É·ÖΪ________£»¼ÙÉèWO3Íêȫת»¯ÎªW£¬ÔòÈý¸ö½×¶ÎÏûºÄH2ÎïÖʵÄÁ¿Ö®±ÈΪ____________________________________¡£
¢È ÒÑÖª£ºÎ¶ȹý¸ßʱ£¬WO2 (s)ת±äΪWO2 (g)£»
WO2 (s) + 2H2 (g)  W (s) + 2H2O (g)£»¦¤H £½ +66.0 kJ¡¤mol£­1
WO2 (g) + 2H2(g)  W (s) + 2H2O (g)£»¦¤H £½ £­137.9 kJ¡¤mol£­1
ÔòWO2 (s)  WO2 (g) µÄ¦¤H £½ ______________________¡£
¢ÉÎÙË¿µÆ¹ÜÖеÄWÔÚʹÓùý³ÌÖлºÂý»Ó·¢£¬Ê¹µÆË¿±äϸ£¬¼ÓÈëI2¿ÉÑÓ³¤µÆ¹ÜµÄʹÓÃÊÙÃü£¬Æ乤×÷Ô­ÀíΪ£ºW (s) +2I2 (g)WI4 (g)¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÓÐ____________¡£
a£®µÆ¹ÜÄÚµÄI2¿ÉÑ­»·Ê¹ÓÃ
b£®WI4ÔÚµÆË¿ÉϷֽ⣬²úÉúµÄWÓÖ³Á»ýÔÚµÆË¿ÉÏ
c£®WI4ÔڵƹܱÚÉϷֽ⣬ʹµÆ¹ÜµÄÊÙÃüÑÓ³¤   
d£®Î¶ÈÉý¸ßʱ£¬WI4µÄ·Ö½âËÙÂʼӿ죬WºÍI2µÄ»¯ºÏËÙÂʼõÂý
£¨1£©    £¨2£©60%  ÎüÈÈ
£¨3£©2WO3+H2=W2O5+H2O£»    W2O5ºÍWO2¡¢   1:1:4£»
£¨4£©+203.9KJ.mol-1         £¨5£© a¡¢b¡£

ÊÔÌâ·ÖÎö£º
£¨1£©¸ù¾Ý·½³Ìʽ¿ÉµÃ¡£
£¨2£©H2ÓëË®ÕôÆøµÄÌå»ý±È¼õС£¬ËµÃ÷·´Ó¦ÕýÏòÒƶ¯£¬ÔòÕý·´Ó¦ÊÇÎüÈÈ·´Ó¦¡£
£¨3£©Óɱí¸ñÐÅÏ¢¿ÉÖª£¬µÚÒ»½×¶Î·´Ó¦Ê±£º2WO3+H2=W2O5+H2O£¬Ö÷Òª³É·ÖΪ£ºW2O5ºÍWO2¡£
£¨4£©¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¦¤H £½+203.9 KJ.mol-1
£¨5£©a¡¢µâµ¥ÖÊÒ×Éý»ª£¬¿ÉÒÔÖظ´Ê¹Óá£
b¡¢WI4ÔÚµÆË¿ÉÏ·Ö½âºó¿ÉÒÔÖØгÁµí¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
I£®ÒÑÖª£º·´Ó¦H2(g) + Cl2(g) = 2HCl(g)   ¦¤H=" ¡ª184" kJ/mol
4HCl(g)+O2(g)  2Cl2(g)+2H2O(g)     ¦¤H=" ¡ª115.6" kJ/mol 
      
Çë»Ø´ð£º
£¨1£©H2ÓëO2·´Ó¦Éú³ÉÆø̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ                                     
£¨2£©¶Ï¿ª1 mol H¡ªO ¼üËùÐèÄÜÁ¿Ô¼Îª                kJ
II£®ÊÔÔËÓÃËùѧ֪ʶ£¬½â¾öÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑ֪ij·´Ó¦µÄƽºâ±í´ïʽΪ£º£¬ËüËù¶ÔÓ¦µÄ»¯Ñ§·½³ÌʽΪ£º         
£¨2£©ÒÑÖªÔÚ400¡æʱ£¬N2 (g)+ 3H2(g) 2NH3(g) ¡÷H<0 µÄK=0.5£¬Ôò400¡æʱ£¬ÔÚ0.5LµÄ·´Ó¦ÈÝÆ÷ÖнøÐкϳɰ±·´Ó¦£¬Ò»¶Îʱ¼äºó£¬²âµÃN2¡¢H2¡¢NH3µÄÎïÖʵÄÁ¿·Ö±ðΪ2mol¡¢1mol¡¢2mol£¬Ôò´Ëʱ·´Ó¦v(N2)Õý          v(N2)Ä棨Ì£¾¡¢£¼¡¢£½¡¢²»ÄÜÈ·¶¨£©£¨1·Ö£©
ÓûʹµÃ¸Ã·´Ó¦µÄ»¯Ñ§·´Ó¦ËÙÂʼӿ죬ͬʱʹƽºâʱNH3µÄÌå»ý°Ù·ÖÊýÔö¼Ó£¬¿É²ÉÈ¡µÄÕýÈ·´ëÊ©ÊÇ       £¨ÌîÐòºÅ£©£¨1·Ö£©
A£®ËõСÌå»ýÔö´óѹǿ    B£®Éý¸ßζȠ  C£®¼Ó´ß»¯¼Á   D£®Ê¹°±ÆøÒº»¯ÒÆ×ß
£¨3£©ÔÚÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖУ¬½øÐÐÈçÏ»¯Ñ§·´Ó¦£ºA(g) + 3B(g)  2C(g) + D£¨s£© ¦¤H£¬Æ仯ѧƽºâ³£ÊýKÓëζÈtµÄ¹ØϵÈçÏÂ±í£º
t/K
300
400
500
¡­
K/(mol¡¤L¡ª1)2
4¡Á106
8¡Á107
K1
¡­
 
ÇëÍê³ÉÏÂÁÐÎÊÌ⣺
¢ÙÅжϸ÷´Ó¦µÄ¦¤H        0£¨Ìî¡°>¡±»ò¡°<¡±£© £¨1·Ö£©
¢ÚÔÚÒ»¶¨Ìõ¼þÏ£¬ÄÜÅжϸ÷´Ó¦Ò»¶¨´ï»¯Ñ§Æ½ºâ״̬µÄÊÇ         £¨ÌîÐòºÅ£©
A£®3v(B)£¨Õý£©=2v(C)£¨Ä棩      B£®AºÍBµÄת»¯ÂÊÏàµÈ
C£®ÈÝÆ÷ÄÚѹǿ±£³Ö²»±ä    D£®»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
£¨4£©ÒÔÌìÈ»Æø(¼ÙÉèÔÓÖʲ»²ÎÓë·´Ó¦)ΪԭÁϵÄȼÁϵç³ØʾÒâͼÈçͼËùʾ¡£

¢Ù·Åµçʱ£¬¸º¼«µÄµç¼«·´Ó¦Ê½Îª                             
¢Ú¼ÙÉè×°ÖÃÖÐÊ¢×°100.0 mL 3.0 mol¡¤L¡ª1 KOHÈÜÒº£¬·Åµçʱ²ÎÓë·´Ó¦µÄÑõÆøÔÚ±ê×¼×´¿öÏÂÌå»ýΪ8 960 mL¡£·ÅµçÍê±Ïºó£¬µç½âÖÊÈÜÒºÖи÷Àë×ÓŨ¶ÈµÄ´óС¹ØϵΪ                                  
ÓÉÓÚ´ß»¯¼Á¿ÉÒÔΪ»¯Ñ§¹¤ÒµÉú²ú´øÀ´¾Þ´óµÄ¾­¼ÃЧÒ棬´ß»¯¼ÁÑо¿ºÍÑ°ÕÒÒ»Ö±ÊÇÊܵ½ÖØÊӵĸ߿Ƽ¼ÁìÓò¡£
£¨1£©V2O5ÊǽӴ¥·¨ÖÆÁòËáµÄ´ß»¯¼Á¡£ÏÂͼΪÁòËáÉú²ú¹ý³ÌÖÐ2SO2 (g) + O2(g)2SO3(g) ¦¤H£½£­196.6 kJ¡¤mol£­1·´Ó¦¹ý³ÌµÄÄÜÁ¿±ä»¯Ê¾Òâͼ¡£

¢ÙV2O5µÄʹÓûáʹͼÖÐBµã         £¨Ìî¡°Éý¸ß¡±¡¢¡°½µµÍ¡±£©¡£
¢ÚÒ»¶¨Ìõ¼þÏ£¬SO2Óë¿ÕÆø·´Ó¦tminºó£¬SO2ºÍSO3ÎïÖʵÄÁ¿Å¨¶È·Ö±ðΪa mol/LºÍb mol/L£¬ÔòSO2ÆðʼÎïÖʵÄÁ¿Å¨¶ÈΪ     mol/L£»Éú³ÉSO3µÄ»¯Ñ§·´Ó¦ËÙÂÊΪ        mol/(L¡¤min)¡£
£¨2£©ÏÂͼÊÇÒ»ÖÖÒÔÍ­¡¢Ï¡ÁòËáΪԭÁÏÉú²úÀ¶·¯µÄÉú²úÁ÷³ÌʾÒâͼ¡£

¢ÙÉú³ÉCuSO4µÄ×Ü·´Ó¦Îª2Cu+O2+2H2SO4£½2 CuSO4+2H2O£¬ÉÙÁ¿Æð´ß»¯×÷Óã¬Ê¹·´Ó¦°´ÒÔÏÂÁ½²½Íê³É£º
µÚÒ»²½£ºCu£«2Fe3£«£½2Fe2£«£«Cu2£«
µÚ¶þ²½£º                                   ¡££¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©
¢Úµ÷½ÚÈÜÒºpHΪ3¡«4µÄÄ¿µÄÊÇ                                £¬µ÷½Úʱ¼ÓÈëµÄÊÔ¼Á¿ÉÒÔΪ        ¡££¨Ñ¡ÌîÐòºÅ£©
a£®NaOHÈÜÒº     b£®CuO·ÛÄ©     c£®Cu2(OH)2CO3      d£®°±Ë®
£¨3£©ÄÉÃ×TiO2ÊÇÓÅÁ¼µÄ¹âÃô´ß»¯¼Á¡£¹¤ÒµÉÏÓÃîÑÌú¿óÖƵôÖTiO2£»ÔÙת»¯ÎªTiCl4(l)£»ÓÉTiCl4(l)ÖÆÈ¡ÄÉÃ×TiO2µÄ·½·¨Ö®Ò»Êǽ«TiCl4ÆøÌåµ¼ÈëÇâÑõ»ðÑæÖУ¨700¡«1000¡æ£©½øÐÐË®½â¡£
ÒÑÖª£ºTiO2(s)£«2Cl2(g)£½TiCl4(l)£«O2(g)   ¦¤H£½£«140 kJ¡¤mol£­1
2C(s)£«O2(g)£½2CO(g)  ¦¤H£½£­221 kJ¡¤mol£­1
¢Ùд³öTiO2ºÍ½¹Ì¿¡¢ÂÈÆø·´Ó¦Éú³ÉTiCl4ºÍCOµÄÈÈ»¯Ñ§·½³Ìʽ£º                            ¡£
¢Úд³öÉÏÊöTiCl4(l)ÖÆÈ¡ÄÉÃ×TiO2µÄ»¯Ñ§·½³Ìʽ£º                                         ¡£
¢ñ£®¼×´¼ÊÇÒ»ÖÖÐÂÐ͵ÄÄÜÔ´¡£
£¨1£©ºÏ³ÉÆø£¨×é³ÉΪH2ºÍCO£©ÊÇÉú²ú¼×´¼µÄÖØÒªÔ­ÁÏ£¬Çëд³öÓɽ¹Ì¿ºÍË®ÔÚ¸ßÎÂÏÂÖÆÈ¡ºÏ³ÉÆøµÄ»¯Ñ§·½³Ìʽ           ¡£
£¨2£©ÒÑÖªH2£¨g£©¡¢CO£¨g£©ºÍCH3OH£¨l£©µÄȼÉÕÈÈ¡÷H·Ö±ðΪ-285.8kJ¡¤mol-1¡¢-283.0kJ¡¤mol-1ºÍ-726.5kJ¡¤mol-1£¬Ôò¼×´¼²»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼ºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽΪ      £»
£¨3£©ÔÚÈÝ»ýΪl LµÄÃܱÕÈÝÆ÷ÖУ¬ÓÉCOºÍH2ºÏ³É¼×´¼¡£ÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬¿¼²éζȶԷ´Ó¦µÄÓ°Ï죬ʵÑé½á¹ûÈçÏÂͼËùʾ£¨×¢£ºT1¡¢T2¾ù´óÓÚ300¡æ£©£»

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ       £¨ÌîÐòºÅ£©
A£®Î¶ÈΪT1ʱ£¬´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬Éú³É¼×´¼µÄƽ¾ùËÙÂÊΪv(CH3OH)=(mol¡¤L-1¡¤min-1)
B£®¸Ã·´Ó¦ÔÚT1ʱµÄƽºâ³£Êý±ÈT2ʱµÄ´ó
C£®¸Ã·´Ó¦ÎªÎüÈÈ·´Ó¦
D£®´¦ÓÚAµãµÄ·´Ó¦Ìåϵ´ÓT1±äµ½T2£¬´ïµ½Æ½ºâʱ¼õС
£¨4£©ÔÚT1ζÈʱ£¬½«1 mol COºÍ2mol H2³äÈëÒ»ÃܱպãÈÝÈÝÆ÷ÖУ¬³ä·Ö·´Ó¦´ïµ½Æ½ºâºó£¬ÈôCOת»¯ÂÊΪa£¬ÔòÈÝÆ÷ÄÚµÄѹǿÓëÆðʼѹǿ֮±ÈΪ          £»
£¨5£©ÔÚÖ±½ÓÒÔ¼×´¼ÎªÈ¼Áϵĵç³ØÖУ¬µç½âÖÊÈÜҺΪ¼îÐÔ£¬¸º¼«µÄ·´Ó¦Ê½Îª        £»¼ÙÉèÔ­µç½âÖÊΪNaOH£¬ÇÒµç³Ø¹¤×÷Ò»¶Îʱ¼äºóÈÜÖÊÖ»ÓÐNa2CO3£¬´ËʱÈÜÒºÖи÷Àë×ÓŨ¶È´óС¹ØϵΪ          
¢ò£®ÒÑÖªKsp(AgCl)=1.56¡Á10-10£¬Ksp(AgBr)=7.7¡Á10-13£¬Ksp(Ag2CrO4)=9¡Á10-11¡£Ä³ÈÜÒºÖк¬ÓÐC1-£¬ Br-ºÍCrO42-£¬Å¨¶È¾ùΪ0.010mo1¡¤L-1£¬Ïò¸ÃÈÜÒºÖÐÖðµÎ¼ÓÈë0.010mol¡¤L-1µÄAgNO3ÈÜҺʱ£¬ÈýÖÖÒõÀë×Ó²úÉú³ÁµíµÄÏȺó˳ÐòΪ        ¡£
ÄòËØ(H2NCONH2)ÊÇÒ»Öַdz£ÖØÒªµÄ¸ßµª»¯·Ê£¬ÔÚ¹¤Å©ÒµÉú²úÖÐÓÐ×ŷdz£ÖØÒªµÄµØλ¡£
£¨1£©¹¤ÒµÉϺϳÉÄòËصķ´Ó¦ÈçÏ£º
2NH3(l)+CO2(g)H2O(l)+H2NCONH2(l)  ¡÷H=-103£®7 kJ¡¤mol-1
ÏÂÁдëÊ©ÖÐÓÐÀûÓÚÌá¸ßÄòËصÄÉú³ÉËÙÂʵÄÊÇ           
A£®²ÉÓøßÎÂ
B£®²ÉÓøßѹ
C£®Ñ°ÕÒ¸ü¸ßЧµÄ´ß»¯¼Á
D£®¼õСÌåϵÄÚCO2Ũ¶È
£¨2£©ºÏ³ÉÄòËصķ´Ó¦ÔÚ½øÐÐʱ·ÖΪÈçÏÂÁ½²½£º
µÚÒ»²½£º2NH3(l)+CO2(g) H2NCOONH4(°±»ù¼×Ëáï§)(l) ¡÷H1
µÚ¶þ²½£ºH2NCOONH4(l) H2O(l)+H2NCONH2(l)  ¡÷H2£®
ijʵÑéС×éÄ£Ä⹤ҵÉϺϳÉÄòËصÄÌõ¼þ£¬ÔÚÒ»Ìå»ýΪ0£®5 LÃܱÕÈÝÆ÷ÖÐͶÈë4 mol°±ºÍl mol¶þÑõ»¯Ì¼£¬ÊµÑé²âµÃ·´Ó¦Öи÷×é·ÖËæʱ¼äµÄ±ä»¯ÈçÏÂͼIËùʾ£º

¢ÙÒÑÖª×Ü·´Ó¦µÄ¿ìÂýÓÉÂýµÄÒ»²½¾ö¶¨£¬ÔòºÏ³ÉÄòËØ×Ü·´Ó¦µÄ¿ìÂýÓɵڠ   ²½·´Ó¦¾ö¶¨£¬×Ü·´Ó¦½øÐе½   minʱµ½´ïƽºâ¡£
¢ÚµÚ¶þ²½·´Ó¦µÄƽºâ³£ÊýKËæζȵı仯ÈçÉÏͼIIËùʾ£¬Ôò¡÷H2    0(Ìî¡°>¡± ¡°<¡± »ò ¡°=¡±)
£¨3£©ÔÚζÈ70-95¡æʱ£¬¹¤ÒµÎ²ÆøÖеÄNO¡¢NO2¿ÉÒÔÓÃÄòËØÈÜÒºÎüÊÕ£¬½«Æäת»¯ÎªN2
¢ÙÄòËØÓëNO¡¢NO2ÈýÕßµÈÎïÖʵÄÁ¿·´Ó¦£¬»¯Ñ§·½³ÌʽΪ
                                                            
¢ÚÒÑÖª£ºN2(g)+O2(g)=2NO(g)£®¡÷H=180£®6 kJ¡¤mol-1
N2(g)+3H2(g)=2NH3(g) ¡÷H=-92£®4 kJ¡¤mol-1
2H2(g)+O2(g)=2H2O(g)  ¡÷H=-483£®6 kJ¡¤mol-1
Ôò4NO(g)+4NH3(g)+O2(g)=4N2(g)+6H2O(g)  ¡÷H=    kJ¡¤mol-1
£¨4£©ÄòËØȼÁϵç³Ø½á¹¹ÈçÉÏͼIIIËùʾ¡£Æ乤×÷ʱ¸º¼«µç¼«·´Ó¦Ê½
¿É±íʾΪ                                                   ¡£
Æû³µÄÚȼ»ú¹¤×÷ʱ²úÉúµÄµç»ð»¨ºÍ¸ßλáÒýÆð·´Ó¦£ºN2(g)£«O2(g)=2NO(g)£¬µ¼ÖÂÆû³µÎ²ÆøÖеÄNOºÍNO2¶Ô´óÆøÔì³ÉÎÛȾ¡£
£¨1£©ÔÚ²»Í¬Î¶ȣ¨T1£¬T2£©Ï£¬Ò»¶¨Á¿µÄNO·Ö½â²úÉúN2ºÍO2µÄ¹ý³ÌÖÐN2µÄÌå»ý·ÖÊýËæʱ¼ät±ä»¯ÈçÓÒͼËùʾ¡£¸ù¾ÝͼÏñÅжϷ´Ó¦N2(g)£«O2(g)£½2NO(g)Ϊ_________·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©£¬Ëæ×ÅζȵÄÉý¸ß£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK________£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£¬Æ½ºâÏò________Òƶ¯£¨Ìî¡°Ïò×󡱡°ÏòÓÒ¡±»ò¡°²»¡±£©¡£

£¨2£©Ä³Î¶Èʱ£¬ÏòÈÝ»ýΪ1LµÄÃܱÕÈÝÆ÷ÖгäÈë5mol N2Óë2£®5molO2,·¢ÉúN2(g)£«O2(g)=2NO(g)·´Ó¦£¬2minºó´ïµ½Æ½ºâ״̬£¬NOµÄÎïÖʵÄÁ¿Îª1mol£¬Ôò2minÄÚÑõÆøµÄƽ¾ù·´Ó¦ËÙÂÊΪ_________£¬¸ÃζÈÏ£¬·´Ó¦µÄƽºâ³£ÊýK=________¡£¸ÃζÈÏ£¬Èô¿ªÊ¼Ê±ÏòÉÏÊöÈÝÆ÷ÖмÓÈëµÄN2ÓëO2¾ùΪ1mol£¬ÔòN2µÄƽºâŨ¶ÈΪ_______mol/L¡£
£¨3£©Îª±ÜÃâÆû³µÎ²ÆøÖеÄÓк¦ÆøÌå¶Ô´óÆøµÄÎÛȾ£¬¸øÆû³µ°²×°Î²Æø¾»»¯×°Ö᣾»»¯×°ÖÃÀï×°Óк¬PdµÈ¹ý¶ÉÔªËصĴ߻¯¼Á£¬ÆøÌåÔÚ´ß»¯¼Á±íÃæÎü¸½Óë½âÎü×÷ÓõĻúÀíÈçÓÒͼËùʾ
¡£
д³öÉÏÊö±ä»¯ÖеÄ×Ü»¯Ñ§·´Ó¦·½³Ìʽ£º________________________________________¡£
£¨4£©Óô߻¯»¹Ô­µÄ·½·¨Ò²¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£ÀýÈ磺
CH4(g)£«4NO2(g)£½4NO(g)£«CO2(g)£«2H2O(g)  ¡÷H£½£­574kJ/mol
CH4(g)£«4NO(g)£½2N2(g)£«CO2(g)£«2H2O(g)  ¡÷H£½£­1160kJ/mol
д³öCH4»¹Ô­NO2ÖÁN2µÄÈÈ»¯Ñ§·½³Ìʽ_______________________________________¡£

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

¾«Ó¢¼Ò½ÌÍø