¹ý¶ÈÅÅ·ÅCO2»áÔì³É¡°ÎÂÊÒЧӦ¡±£¬ÎªÁ˼õÉÙúȼÉÕ¶Ô»·¾³Ôì³ÉµÄÎÛȾ£¬ÃºµÄÆø»¯ÊǸßЧ¡¢Çå½àÀûÓÃú̿µÄÖØҪ;¾¶¡£Ãº×ÛºÏÀûÓõÄÒ»ÖÖ;¾¶ÈçͼËùʾ¡£

£¨1£©ÒÑÖª¢ÙC(s) £« H2O(g) = CO(g)£«H2(g)      ¦¤H1£½£«131.3 kJ¡¤mol£­1
¢ÚC(s) £« 2H2O(g) = CO2(g) £« 2H2(g) ¦¤H2£½£«90 kJ¡¤mol£­1
ÔòÒ»Ñõ»¯Ì¼ÓëË®ÕôÆø·´Ó¦Éú³É¶þÑõ»¯Ì¼ºÍÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽÊÇ   ________________________£¬
£¨2£©ÓÃÏÂͼԭµç³Ø×°ÖÿÉÒÔÍê³É¹ý³Ì¢ÝµÄת»¯£¬¸Ã×°ÖÃbµç¼«µÄµç¼«·´Ó¦Ê½ÊÇ_______________________¡£

£¨3£©ÔÚѹǿΪ0.1 MPaÌõ¼þÏ£¬ÈÝ»ýΪV LµÄÃܱÕÈÝÆ÷ÖÐa mol COÓë2a mol H2ÔÚ´ß»¯¼Á×÷ÓÃÏ·´Ó¦Éú³É¼×´¼£º
CO(g)£«2H2(g) CH3OH(g)£¬COµÄƽºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçÏÂͼËùʾ£¬Ôò£º

¢Ùp1________p2(Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±)¡£
¢ÚÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬ÏòÈÝÆ÷ÖÐÔÙÔö¼Óa mol COÓë2a mol H2£¬´ïµ½ÐÂƽºâʱ£¬COµÄƽºâת»¯ÂÊ________(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)¡£
¢ÛÔÚp1Ï£¬100 ¡æʱ£¬CO(g)£«2H2(g) CH3OH(g)·´Ó¦µÄƽºâ³£ÊýΪ________(Óú¬a¡¢VµÄ´úÊýʽ±íʾ)¡£
£¨4£©Èçͼ±íʾCO2ÓëH2·´Ó¦Éú³ÉCH3OHºÍH2OµÄ¹ý³ÌÖÐÄÜÁ¿(µ¥Î»ÎªkJ¡¤mol£­1)µÄ±ä»¯£º

¹ØÓڸ÷´Ó¦µÄÏÂÁÐ˵·¨ÖУ¬ÕýÈ·µÄÊÇ________(Ìî±àºÅ)¡£
A£®¦¤H£¾0£¬¦¤S£¾0              B£®¦¤H£¾0£¬¦¤S£¼0
C£®¦¤H£¼0£¬¦¤S£¼0              D£®¦¤H£¼0£¬¦¤S£¾0
£¨5£©ÎªÌ½¾¿·´Ó¦Ô­Àí£¬ÏÖ½øÐÐÈçÏÂʵÑ飬ÔÚÌå»ýΪ1 LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1 mol CO2ºÍ3 mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºCO2(g)£«3H2(g) CH3OH(g)£«H2O(g)£¬²âµÃCO2(g)ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯µÄÇúÏßÈçͼËùʾ£º

¢Ù´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬CO2µÄƽ¾ù·´Ó¦ËÙÂÊv(CO2)£½________¡£
¢ÚÏÂÁдëÊ©ÖÐÄÜʹ»¯Ñ§Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯µÄÊÇ________(Ìî±àºÅ)¡£
A£®Éý¸ßζȠ            B£®½«CH3OH(g)¼°Ê±Òº»¯ÒƳö
C£®Ñ¡Ôñ¸ßЧ´ß»¯¼Á       D£®ÔÙ³äÈë1 mol CO2ºÍ3 mol H2

£¨1£© CO(g)£«H2O(g)=CO2(g)£«H2(g)¡¡¦¤H£½£­41.3 kJ¡¤mol£­1¡¡£¨2£© O2 +4e£­ +2H2O = 4OH¡ª¡¡              £¨3£©¢Ù£¼¢Ú    V 2 / a2¡¡¡¡¢ÛÔö´ó    £¨4£©C¡¡  £¨5£©¢Ù0.075 mol/( L¡¤min). ¢ÚBD

½âÎöÊÔÌâ·ÖÎö£º£¨1£©¢Ú-¢Ù¿ÉµÃ£ºCO(g)£«H2O(g)=CO2(g)£«H2(g)¡¡¦¤H£½£­41.3 kJ/mol.. £¨2£©ÔÚȼÁϵç³ØÖУ¬Í¨ÈëȼÁϵĵ缫×÷¸º¼«£¬Í¨ÈëÑõÆøµÄµç¼«×÷Õý¼«¡£aµç¼«ÊǸº¼«£¬bµç¼«ÊÇÕý¼«¡£bµç¼«µÄµç¼«·´Ó¦Ê½ÊÇO2 +4e£­ +2H2O = 4OH¡ª¡¡£¨3£© ¢ÙÓÉͼ¿ÉÒÔ¿´³ö£ºÔÚζÈÏàͬʱ£¬×ª»¯ÂÊP2>P1¡£¸ù¾ÝƽºâÒƶ¯Ô­Àí£ºÔÚÆäËüÌõ¼þ²»±äµÄÇé¿öÏ¡£Ôö´óѹǿ£¬»¯Ñ§Æ½ºâÏòÆøÌåÌå»ý¼õСµÄ·½ÏòÒƶ¯¡£¼´ÏòÕý·´Ó¦·½ÏòÒƶ¯¡£Õâʱ·´Ó¦ÎïµÄת»¯ÂÊÌá¸ß¡£ËùÒÔP1<P2. ¢ÚÔÚÆäËûÌõ¼þ²»±äµÄÇé¿öÏ£¬ÏòÈÝÆ÷ÖÐÔÙÔö¼Óa mol COÓë2a mol H2¡£¼´Ôö´óÁËÌåϵµÄѹǿ£¬Õâʱ»¯Ñ§Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯¡£¹Ê´ïµ½ÐÂƽºâʱ£¬COµÄƽºâת»¯ÂÊÔö´ó¡£¢ÛÔÚp1Ï£¬100 ¡æʱ£¬CO(g)£«2H2(g) CH3OH(g)·´Ó¦µÄƽºâ³£ÊýΪK="C" (CH3OH)/ { C(CO)¡¤C2(H2)} ="(" a/2V)£º{(a/2V) ¡¤(a/V)}2=" V" 2 / a2.£¨4£©CO2(g)+3H2£¨g£©CH3OH(g)+CO(g).ÓÉͼ¿É¿´³ö¸Ã·´Ó¦µÄÕý·´Ó¦ÊÇ·ÅÈÈ·´Ó¦¡£¸Ã·´Ó¦µÄÕý·´Ó¦ÊǸöÆøÌåÌå»ý¼õСµÄ·ÅÈÈ·´Ó¦¡£ËùÒÔ¡÷H<0£»¡÷S<0¡£Ñ¡ÏîΪ£ºC¡££¨5£©¢Ù´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬CO2µÄƽ¾ù·´Ó¦ËÙÂÊv(CO2)=£¨1-0.25£©mol/L¡Â10min="0.075" mol/( L¡¤min). ¢ÚAÉý¸ßζȻ¯Ñ§Æ½ºâÏòÎüÈÈ·½ÏòÒƶ¯¡£ÓÉÓڸ÷´Ó¦µÄÕý·´Ó¦ÊÇ·ÅÈÈ·´Ó¦ËùÒÔÉý¸ßζȻ¯Ñ§Æ½ºâÏòÄæ·´Ó¦·½ÏòÒƶ¯¡£B£®¼õСÉú³ÉÎïµÄŨ¶È£¬»¯Ñ§Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯¡£¹Ê½«CH3OH(g)¼°Ê±Òº»¯ÒƳö¿ÉʹƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬C£®´ß»¯¼Á ¶Ô»¯Ñ§Æ½ºâÎÞÓ°Ïì¡£D£®´ïµ½Æ½ºâʱÈôÔÙ³äÈë1 mol CO2ºÍ3 mol H2£¬¼´Ôö´óÁËѹǿ£¬»¯Ñ§Æ½ºâÏòÆøÌåÌå»ý¼õСµÄ·½Ïò¼°Õý·´Ó¦·½ÏòÒƶ¯¡£ËùÒÔÄÜʹ»¯Ñ§Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯µÄ´ëÊ©ÊÇBD¡£
¿¼µã£º¿¼²éÈÈ»¯Ñ§·½³ÌʽµÄÊéд¡¢»¯Ñ§Æ½ºâ³£ÊýµÄ¼ÆËã¼°Íâ½çÌõ¼þ¶Ô»¯Ñ§Æ½ºâµÄÓ°ÏìµÈ֪ʶ¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

°±ÓÐ׏㷺µÄÓÃ;£¬¿ÉÓÃÓÚ»¯·Ê¡¢ÏõËá¡¢ºÏ³ÉÏËάµÈ¹¤ÒµÉú²ú¡£
£¨1£©¸ù¾Ý×îС°È˹¤¹Ìµª¡±µÄÑо¿±¨µÀ£¬ÔÚ³£Î¡¢³£Ñ¹¡¢¹âÕÕÌõ¼þÏ£¬N2ÔÚ´ß»¯¼Á£¨²ôÓÐÉÙÁ¿Fe2O3µÄTiO2£©±íÃæÓëË®·¢Éú·´Ó¦¿ÉÉú³É°±Æø£º

¸Ã·´Ó¦Ôڹ̶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖнøÐУ¬ÓйØ˵·¨ÕýÈ·µÄÊÇ_____________£¨ÌîÐòºÅ×Öĸ£©¡£

A£®·´Ó¦´¦ÓÚƽºâ״̬ʱ£¬
B£®·´Ó¦´ïµ½Æ½ºâºó£¬
C£®ÌåϵµÄ×Üѹǿ²»±ä£¬ËµÃ÷·´Ó¦ÒÑ´ïƽºâ
D£®»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä£¬ËµÃ÷·´Ó¦ÒÑ´ïƽºâ
£¨2£©Òº°±×÷ΪһÖÖDZÔÚµÄÇå½àÆû³µÈ¼ÁÏÒÑÔ½À´Ô½±»Ñо¿ÈËÔ±ÖØÊÓ¡£ËüÔÚ°²È«ÐÔ¡¢¼Û¸ñµÈ·½Ãæ½Ï»¯Ê¯È¼ÁϺÍÇâȼÁÏÓÐ׎ϴóµÄÓÅÊÆ¡£°±ÔÚȼÉÕÊÔÑé»úÖÐÏà¹ØµÄ·´Ó¦ÓУº
   ¢Ù
 ¢Ú
   ¢Û
Çëд³öÉÏÊöÈý¸ö·´Ó¦ÖС¢¡¢ÈýÕßÖ®¼ä¹ØϵµÄ±í´ïʽ£¬£½_________¡£
£¨3£©¹¤ÒµÖÆÏõËáµÄÖ÷Òª·´Ó¦ÊÇ£º £½
¢ÙÉý¸ßζȣ¬·´Ó¦µÄKÖµ¼õС£¬ÔòQ______£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°£½¡±£©0¡£
¢ÚÈô·´Ó¦ÆðʼµÄÎïÖʵÄÁ¿Ïàͬ£¬ÏÂÁйØϵͼ´íÎóµÄÊÇ________£¨ÌîÐòºÅ£©¡£

¢ÛÔÚÈÝ»ý¹Ì¶¨µÄÃܱÕÈÝÆ÷Öз¢ÉúÉÏÊö·´Ó¦£¬ÈÝÆ÷ÄÚ²¿·ÖÎïÖʵÄŨ¶ÈÈçÏÂ±í£º
ʱ¼ä/Ũ¶È








Æðʼ
4.0
5.5
0
0
µÚ2min
3.2
a
0.8
1.2
µÚ4min
2.0
3.0
2.0
3.0
µÚ6min
2.0
3.0
2.0
3.0
 
·´Ó¦ÔÚµÚ2 minµ½µÚ4 minʱ£¬O2µÄƽ¾ù·´Ó¦ËÙÂÊΪ________¡£
·´Ó¦ÔÚµÚ2 minʱ¸Ä±äÁËÌõ¼þ£¬¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇ______________________________¡£
¸ÃÌõ¼þÏ£¬·´Ó¦µÄƽºâ³£ÊýK£½________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

Æû³µÎ²ÆøÀﺬÓеÄNOÆøÌåÊÇÓÉÓÚÄÚȼ»úȼÉյĸßÎÂÒýÆðµªÆøºÍÑõÆø·´Ó¦ËùÖ£º
N2(g)£«O2(g) 2NO(g) ¦¤H£¬ÒÑÖª¸Ã·´Ó¦ÔÚ T ¡æʱ£¬Æ½ºâ³£ÊýK£½9.0¡£
Çë»Ø´ð£º
£¨1£©ÒÑÖª£ºN2(g)+2O2(g) 2NO2(g) ¦¤H1     2NO2(g) O2+2NO(g) ¦¤H2  ¦¤H=             £¨Óú¬¦¤H1¡¢¦¤H2µÄ±í´ïʽ±íʾ£©£»
£¨2£©Ä³Î¶ÈÏ£¬Ïò2 LµÄÃܱÕÈÝÆ÷ÖгäÈëN2ºÍO2¸÷1 mol£¬5·ÖÖÓºóO2µÄÎïÖʵÄÁ¿Îª0.5 mol£¬ÔòNOµÄ·´Ó¦ËÙÂÊ                £»
£¨3£©¼Ù¶¨¸Ã·´Ó¦ÊÇÔÚºãÈÝÌõ¼þϽøÐУ¬ÏÂÁÐÄÜÅжϸ÷´Ó¦ÒѴﵽƽºâµÄÊÇ________£»

A£®ÏûºÄ1 mol N2ͬʱÉú³É1 mol O2
B£®»ìºÏÆøÌåÃܶȲ»±ä
C£®»ìºÏÆøÌåƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»±ä
D£®2vÕý(N2)£½vÄæ(NO)
£¨4£©ÏÂͼÊÇ·´Ó¦N2(g)£«O2(g) 2NO(g)µÄ¡°K-T¡±¡¢¡°c(NO)-t¡±Í¼£¬ÓÉͼA¿ÉÒÔÍÆÖª¸Ã·´Ó¦Îª        ·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©¡£ÓÉͼB¿ÉÖª£¬Óëa¶ÔÓ¦µÄÌõ¼þÏà±È£¬b¸Ä±äµÄÌõ¼þ¿ÉÒÔÊÇ        £»

£¨5£©T ¡æʱ£¬Ä³Ê±¿Ì²âµÃÈÝÆ÷ÄÚN2¡¢O2¡¢NOµÄŨ¶È·Ö±ðΪ0.20 mol¡¤L£­1¡¢0.20mol¡¤L£­1ºÍ0.50mol¡¤L£­1£¬´Ëʱ·´Ó¦N2(g)£«O2(g) 2NO(g)________________(Ìî¡°´¦ÓÚ»¯Ñ§Æ½ºâ״̬¡±¡¢¡°ÏòÕý·´Ó¦·½Ïò½øÐС±»ò¡°ÏòÄæ·´Ó¦·½Ïò½øÐС±)£¬Æ½ºâʱ£¬N2ÔÚ»ìºÏÆøÌåµÄÌå»ý°Ù·ÖÊýΪ¶àÉÙ£¿£¨ÔÚ´ðÌ⿨ÉÏд³ö¾ßÌå¼ÆËã¹ý³Ì£¬½á¹û±£Áô2λÓÐЧÊý×Ö£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¢ñÔÚ´ß»¯¼Á×÷ÓÃÏ£¬CO2ºÍH2¿ÉÒÔÖÆÈ¡¼×´¼¡£Óù¤Òµ·ÏÆøÖеĿÉÖÆÈ¡¼×´¼£¬Æ䷴ӦΪ£ºCO2+3H2CH3OH+H2O ³£Î³£Ñ¹ÏÂÒÑÖªÏÂÁз´Ó¦µÄÄÜÁ¿±ä»¯Èçͼʾ£º

д³öÓɶþÑõ»¯Ì¼ºÍÇâÆøÖƱ¸¼×´¼µÄÈÈ»¯Ñ§·½³Ìʽ£º                              ¡£
¢òÅðÇ⻯ÄÆ£¨NaBH4£©ÊÇÓлúºÏ³ÉÖеÄÖØÒª»¹Ô­¼Á¡£×îÐÂÑо¿·¢ÏÖ£¬ÒÔNaBH4ºÍH2O2ΪԭÁÏ£¬NaOHÈÜÒº×÷µç½âÖÊÈÜÒº£¬¿ÉÒÔÉè¼Æ³ÉÈ«ÒºÁ÷µç³Ø£¬Æ乤×÷Ô­ÀíÈçͼËùʾ£¬¼ÙÉèµç³Ø¹¤×÷Ç°×óÓÒÁ½²ÛÈÜÒºµÄÌå»ý¸÷Ϊ1L£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©µç¼«bΪ           £¨Ìî¡°Õý¼«¡±»ò¡°¸º¼«¡±£©£¬µç¼«aÉÏ·¢Éú·´Ó¦µÄµç¼«·´Ó¦Ê½Îª                                     ¡£
£¨2£©µç³Ø¹¤×÷ʱ£¬Na+Ïò     ¼«£¨Ìî¡°a¡±»ò¡°b¡±£©Òƶ¯£¬µ±×ó²Û²úÉú0£®0125molBO2¡ªÀë×Óʱ£¬ÓÒ²ÛÈÜÒºpH=           
£¨3£©Óøõç³Øµç½âÒ»¶¨Å¨¶ÈµÄCuSO4ÈÜÒºÖÁÎÞÉ«ºó¼ÌÐøµç½âÒ»¶Îʱ¼ä¡£¶Ï¿ªµç·£¬ÏòÈÜÒºÖмÓÈë0£®1molCu(OH)2£¬ÈÜÒº»Ö¸´µ½µç½â֮ǰ״̬£¬Ôòµç½â¹ý³ÌÖÐתÒƵç×ÓÊýĿΪ_________

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¶þ¼×ÃÑ£¨CH3OCH3£©ÊÇÒ»ÖÖÖØÒªµÄ¾«Ï¸»¯¹¤²úÆ·£¬±»ÈÏΪÊǶþʮһÊÀ¼Í×îÓÐDZÁ¦µÄȼÁÏ[ ÒÑÖª£ºCH3OCH3(g)+3O2(g)£½2CO2(g)+3H2O(1) ¡÷H£½£­1455kJ/mol ]¡£Í¬Ê±ËüÒ²¿ÉÒÔ×÷ΪÖÆÀä¼Á¶øÌæ´ú·úÂÈ´úÌþ¡£¹¤ÒµÉÏÖƱ¸¶þ¼×ÃѵÄÖ÷Òª·½·¨¾­ÀúÁËÈý¸ö½×¶Î£º
¢Ù¼×´¼ÒºÌåÔÚŨÁòËá×÷ÓÃÏ»ò¼×´¼ÆøÌåÔÚ´ß»¯×÷ÓÃÏÂÖ±½ÓÍÑË®Öƶþ¼×ÃÑ£»2CH3OHCH3OCH3£«H2O
¢ÚºÏ³ÉÆøCOÓëH2Ö±½ÓºÏ³É¶þ¼×ÃÑ£º3H2(g)£«3CO(g)CH3OCH3(g)£«CO2(g)  ¡÷H£½£­247kJ/mol
¢ÛÌìÈ»ÆøÓëË®ÕôÆø·´Ó¦ÖƱ¸¶þ¼×ÃÑ¡£ÒÔCH4ºÍH2OΪԭÁÏÖƱ¸¶þ¼×ÃѺͼ״¼¹¤ÒµÁ÷³ÌÈçÏ£º

£¨1£©Ð´³öCO(g)¡¢H2(g)¡¢O2(g)·´Ó¦Éú³ÉCO2(g)ºÍH2O(1)µÄÈÈ»¯Ñ§·½³Ìʽ£¨½á¹û±£ÁôһλСÊý£©                                                ¡£
£¨2£©¢Ù·½·¨ÖÐÓü״¼ÒºÌåÓëŨÁòËá×÷ÓÃÖ±½ÓÍÑË®Öƶþ¼×ÃÑ£¬¾¡¹Ü²úÂʸߣ¬µ«ÊÇÖð²½±»ÌÔÌ­µÄÖ÷ÒªÔ­ÒòÊÇ                                        ¡£
£¨3£©ÔÚ·´Ó¦ÊÒ2ÖУ¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦3H2(g)£«3CO(g)CH3OCH3(g)£«CO2(g)ÔÚÃܱÕÈÝÆ÷Öдﵽƽºâºó£¬ÒªÌá¸ßCOµÄת»¯ÂÊ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊÇ     ¡£
A£®µÍθßѹ    B£®¼Ó´ß»¯¼Á    C£®Ôö¼ÓCOŨ¶È    D£®·ÖÀë³ö¶þ¼×ÃÑ
£¨4£©ÔÚ·´Ó¦ÊÒ3ÖУ¬ÔÚÒ»¶¨Î¶ȺÍѹǿÌõ¼þÏ·¢ÉúÁË·´Ó¦£º3H2(g)£«CO2(g) CH3OH(g)£«H2O (g) ¡÷H£¼0·´Ó¦´ïµ½Æ½ºâʱ£¬¸Ä±äζȣ¨T£©ºÍѹǿ£¨P£©£¬·´Ó¦»ìºÏÎïCH3OH¡°ÎïÖʵÄÁ¿·ÖÊý¡±±ä»¯Çé¿öÈçͼËùʾ£¬¹ØÓÚζȣ¨T£©ºÍѹǿ£¨P£©µÄ¹ØϵÅжÏÕýÈ·µÄÊÇ   £¨ÌîÐòºÅ£©¡£

A£®P3£¾P2   T3£¾T2       B£®P2£¾P4   T4£¾T2
C£®P1£¾P3   T1£¾T3       D£®P1£¾P4   T2£¾T3
£¨5£©·´Ó¦ÊÒ1Öз¢Éú·´Ó¦£ºCH4(g)£«H2O(g)CO(g)£«3H2(g) ¡÷H£¾0д³öƽºâ³£ÊýµÄ±í´ïʽ£º                          ¡£Èç¹ûζȽµµÍ£¬¸Ã·´Ó¦µÄƽºâ³£Êý                  ¡££¨Ìî¡°²»±ä¡±¡¢¡°±ä´ó¡±¡¢¡°±äС¡±£©
£¨6£©ÈçͼΪÂÌÉ«µçÔ´¡°¶þ¼×ÃÑȼÁϵç³Ø¡±µÄ¹¤×÷Ô­ÀíʾÒâͼ¡£Ôòaµç¼«µÄ·´Ó¦Ê½Îª£º________________¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

¡°½ÚÄܼõÅÅ¡±£¬¼õÉÙÈ«ÇòÎÂÊÒÆøÌåÅÅ·Å£¬ÒâÒåÊ®·ÖÖش󡣶þÑõ»¯Ì¼µÄ²¶×½Óë·â´æÊÇʵÏÖÎÂÊÒÆøÌå¼õÅŵÄÖØҪ;¾¶Ö®Ò»£¬¿Æѧ¼ÒÀûÓÃÈÜÒºÅçÁÜ¡°²¶×½¡±¿ÕÆøÖеġ£
£¨1£©Ê¹ÓùýÁ¿ÈÜÒºÎüÊÕ£¬·´Ó¦µÄÀë×Ó·½³ÌʽΪ________£»Èôº¬ÓÐ3molNaOHµÄÈÜÒº¡°²¶×½¡±ÁË22£®4LÆøÌ壨±ê×¼×´¿ö£©£¬ÔòËùµÃÈÜÒºÖÐÄÆÓë̼ԪËصÄÎïÁÏÊغã¹ØϵʽΪ__________£¨ÓÃÀë×ÓŨ¶ÈµÄ¹Øϵʽ±íʾ£©¡£
£¨2£©¢ÙÒÔºÍΪԭÁϿɺϳɻ¯·ÊÄòËØ[]¡£ÒÑÖª£º
   ¢Ù
 ¢Ú
  ¢Û
ÊÔд³öºÍºÏ³ÉÄòËغÍҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ__________¡£
¢Úͨ¹ý·´Ó¦¿Éת»¯Îª£¬ÔÚ´ß»¯¼Á×÷ÓÃÏÂCOºÍ·´Ó¦Éú³É¼×´¼£ºÄ³ÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖгäÓÐ10molCOÓë20mol£¬COµÄƽºâת»¯ÂÊ£¨a£©Óëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçÏÂͼËùʾ¡£

A£®ÈôAµã±íʾÔÚijʱ¿Ì´ïµ½µÄƽºâ״̬£¬´ËʱÈÝÆ÷µÄÈÝ»ýΪVL£¬Ôò¸ÃζÈϵÄƽºâ³£ÊýK=__________£»Æ½ºâ״̬BµãʱÈÝÆ÷µÄÈÝ»ý_______VL¡££¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©
B£®ÈôA¡¢CÁ½µã¶¼±íʾ´ïµ½µÄƽºâ״̬£¬Ôò×Ô·´Ó¦¿ªÊ¼µ½´ïƽºâ״̬ËùÐèµÄʱ¼ä_______£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©
C£®ÔÚ²»¸Ä±ä·´Ó¦ÎïÓÃÁ¿µÄÇé¿öÏ£¬ÎªÌá¸ßCOµÄת»¯ÂʿɲÉÈ¡µÄ´ëÊ©ÊÇ________£¨Ð´³öÒ»ÖÖ¼´¿É£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º 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´£º ÌâÐÍ£ºÌî¿ÕÌâ

£¨8·Ö£©°Ñú×÷ΪȼÁÏ¿Éͨ¹ýÏÂÁÐÁ½ÖÖ;¾¶£º
;¾¶I£ºC(s) +O2 (g)=CO2(g)     ¡÷H1<0                    ¢Ù
;¾¶II£ºÏÈÖƳÉˮúÆø£ºC(s) +H2O(g)=CO(g)+H2(g)  ¡÷H2>0   ¢Ú
ÔÙȼÉÕˮúÆø£º2CO(g)+O2 (g)=2CO2(g)  ¡÷H3<0               ¢Û
2H2(g)+O2 (g)=2H2O(g)  ¡÷H4<0                ¢Ü
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Í¾¾¶I·Å³öµÄÈÈÁ¿  ( Ìî¡°´óÓÚ¡±¡°µÈÓÚ¡±»ò¡°Ð¡ÓÚ¡±) ;¾¶II·Å³öµÄÈÈÁ¿¡£
£¨2£©¡÷H1¡¢¡÷H2¡¢¡÷H3¡¢¡÷H4µÄÊýѧ¹ØϵʽÊÇ              ¡£
£¨3£©12gÌ¿·ÛÔÚÑõÆøÖв»ÍêȫȼÉÕÉú³ÉÒ»Ñõ»¯Ì¼£¬·Å³ö110.35kJÈÈÁ¿¡£ÆäÈÈ»¯Ñ§·½³ÌʽΪ                                             ¡£
£¨4£©ÃºÌ¿×÷ΪȼÁϲÉÓÃ;¾¶IIµÄÓŵãÓР                                             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

ÔËÓû¯Ñ§·´Ó¦Ô­ÀíÑо¿µª¡¢Áò¡¢ÂÈ¡¢µâµÈµ¥Öʼ°Æ仯ºÏÎïµÄ·´Ó¦ÓÐÖØÒªÒâÒå

£¨1£©ÁòËáÉú²úÖУ¬SO2´ß»¯Ñõ»¯Éú³ÉSO3£»   2SO2£¨g£©+O2£¨g£©2SO3£¨g£©£¬»ìºÏÌåϵÖÐSO3  µÄ°Ù·Öº¬Á¿ºÍζȵĹØϵÈçÏÂͼËùʾ£¨ÇúÏßÉÏÈκÎÒ»µã¶¼±íʾƽºâ״̬£©£®¸ù¾Ýͼʾ»Ø´ðÏÂÁÐÎÊÌ⣬
¢Ù2SO2£¨g£©+O2£¨g£©2SO3£¨g£©µÄ¡÷H     0
£¨Ìî¡°>¡±»ò¡°<¡±£©£ºÈôÔÚºãΡ¢ºãѹÌõ¼þÏÂÏòÉÏÊöƽºâÌåϵÖÐͨÈ뺤Æø£¬Æ½ºâ             Òƶ¯£¨Ìî¡°Ïò×󡱡°ÏòÓÒ¡±»ò¡°²»¡±£©
¢ÚÈôζÈΪT1¡¢T2£¬·´Ó¦µÄƽºâ³£Êý·Ö±ðΪK1£¬K2£¬ÔòK1        K2£»·´Ó¦½øÐе½×´Ì¬Dʱ£¬            £¨Ìî¡°>¡±¡°<¡±»ò¡°=¡±£©
£¨2£©µªÊǵØÇòÉϺ¬Á¿·á¸»µÄÒ»ÖÖÔªËØ£¬µª¼°Æ仯ºÏÎïÔÚ¹¤Å© ÒµÉú²ú¡¢Éú»îÖÐÓÐ×ÅÖØÒª×÷Óã¬

¢ÙÈçͼÊÇÒ»¶¨µÄζȺÍѹǿÏÂN2ºÍH2·´Ó¦Éú³ÉlmolNH3¹ý³ÌÖÐÄÜÁ¿±ä»¯Ê¾Òâͼ£¬Çëд³ö¹¤ÒµºÏ³É°±µÄÈÈ»¯Ñ§·½³Ìʽ£º                                                     
£¨¡÷HµÄÊýÖµÓú¬×ÖĸQ1¡¢Q2µÄ´úÊýʽ±íʾ£©
¢Ú°±ÆøÈÜÓÚË®µÃµ½°±Ë®£¬ÔÚ25¡æÏ£¬½«a mol¡¤L-1µÄ°±Ë®Óëb mol¡¤L-1µÄÑÎËáµÈÌå»ý»ìºÏ£¬·´Ó¦ºóÈÜÒºÏÔÖÐÐÔ£¬Ôòc£¨NH4+£©     c£¨Cl-£©£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©£»Óú¬aºÍbµÄ´úÊýʽ±íʾ³ö¸Ã»ìºÏÈÜÒºÖа±Ë®µÄµçÀëƽºâ³£Êý             .
£¨3£©º£Ë®Öк¬ÓдóÁ¿µÄÔªËØ£¬³£Á¿ÔªËØÈçÂÈ£¬Î¢Á¿ÔªËØÈçµâ£¬ÆäÔÚº£Ë®ÖоùÒÔ»¯ºÏ̬´æÔÚ£¬ÔÚ25¡æÏ£¬Ïò0£®1L0.002mol¡¤L-lµÄNaClÈÜÒºÖÐÖðµÎ¼ÓÈëÊÊÁ¿µÄ0£®1L0.002mol¡¤L-lÏõËáÒøÈÜÒº£¬Óа×É«³ÁµíÉú³É£¬´Ó³ÁµíÈܽâƽºâµÄ½Ç¶È½âÊͲúÉú³ÁµíµÄÔ­ÒòÊÇ                    £¬Ïò·´Ó¦ºóµÄ×ÇÒºÖмÌÐø¼ÓÈë0£®1L0.002mol¡¤L-1µÄNaIÈÜÒº£¬¿´µ½µÄÏÖÏóÊÇ              £¬²úÉú¸ÃÏÖÏóµÄÔ­ÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©                                        ¡£
£¨ÒÑÖª£º25¡æʱKSP£¨AgCl£©=1.6¡Ál0-10     KSP£¨AgI£©=1.5¡Ál0-16£©

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸