ÌâÄ¿ÄÚÈÝ

¼×´¼ÊÇÖØÒªµÄ»¯Ñ§¹¤Òµ»ù´¡Ô­ÁϺÍÇå½àÒºÌåȼÁÏ¡£¹¤ÒµÉÏ¿ÉÀûÓÃCO»òCO2À´Éú²úȼÁϼ״¼¡£ÒÑÖª¼×´¼ÖƱ¸µÄÓйػ¯Ñ§·´Ó¦ÒÔ¼°ÔÚ²»Í¬Î¶ÈϵĻ¯Ñ§·´Ó¦Æ½ºâ³£ÊýÈçϱíËùʾ£º

»¯Ñ§·´Ó¦

ƽºâ³£Êý

ζȡæ

500

800

¢Ù2H2(g)+CO(g) CH3OH(g)

K1

2.5

0.15

¢ÚH2(g)+CO2(g) H2O (g)+CO(g)

K2

1.0

2.50

¢Û3H2(g)+CO2(g) CH3OH(g)+H2O (g)

K3

£¨1£©·´Ó¦¢ÚÊÇ________________£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©·´Ó¦¡£

£¨2£©Ä³Î¶ÈÏ·´Ó¦¢ÙÖÐH2µÄƽºâת»¯ÂÊ£¨a£©ÓëÌåϵ×Üѹǿ(P)µÄ¹Øϵ£¬Èç×óÏÂͼËùʾ¡£Ôòƽºâ״̬ÓÉA±äµ½Bʱ£¬Æ½ºâ³£ÊýK(A)_____________K(B)£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£¾Ý·´Ó¦¢ÙÓë¢Ú¿ÉÍƵ¼³öK1¡¢K2ÓëK3Ö®¼äµÄ¹Øϵ£¬ÔòK3=_______£¨ÓÃK1¡¢K2±íʾ£©¡£

£¨3£©ÔÚ3 LÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦¢Ú£¬ÒÑÖªc(CO)Ó뷴Ӧʱ¼ät±ä»¯ÇúÏߢñÈçÓÒÉÏͼËùʾ£¬ÈôÔÚt0ʱ¿Ì·Ö±ð¸Ä±äÒ»¸öÌõ¼þ£¬ÇúÏߢñ±äΪÇúÏߢòºÍÇúÏߢó¡£

µ±ÇúÏߢñ±äΪÇúÏߢòʱ£¬¸Ä±äµÄÌõ¼þÊÇ_____________________¡£

µ±ÇúÏߢñ±äΪÇúÏߢóʱ£¬¸Ä±äµÄÌõ¼þÊÇ_____________________¡£

£¨4£©Ò»ÖÖ¼×´¼È¼Áϵç³Ø£¬Ê¹Óõĵç½âÖÊÈÜÒºÊÇ2mol¡¤L£­1µÄKOHÈÜÒº¡£

Çëд³ö¼ÓÈë(ͨÈë)bÎïÖÊÒ»¼«µÄµç¼«·´Ó¦Ê½_________________£»

ÿÏûºÄ6.4g¼×´¼×ªÒƵĵç×ÓÊýΪ_______________¡£

£¨5£©Ò»¶¨Ìõ¼þϼ״¼ÓëÒ»Ñõ»¯Ì¼·´Ó¦¿ÉÒԺϳÉÒÒËᡣͨ³£×´¿öÏ£¬½«a mol/LµÄ´×ËáÓëb mol/LBa(OH)2ÈÜÒºµÈÌå»ý»ìºÏºó£¬ÈÜÒºÖУº2c(Ba2£«)= c(CH3COO-)£¬Óú¬aºÍbµÄ´úÊýʽ±íʾ¸Ã»ìºÏÈÜÒºÖд×ËáµÄµçÀë³£ÊýKaΪ________________¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

µªÑõ»¯ÎNOx£©ÊÇ´óÆøÖ÷ÒªÎÛȾÎïÖ®Ò»£¬Çë¸ù¾ÝÒÔÏ·½·¨µÄÑ¡ÓûشðÏàÓ¦ÎÊÌ⣺

I.ÀûÓÃCH4´ß»¯»¹Ô­µªÑõ»¯ÎïÏû³ýµªÑõ»¯ÎïµÄÎÛȾ¡£ÒÑÖª£º

¢ÙCH4(g)+4NO2(g)4NO(g)+CO2(g)+2H2O(l) ¡÷H=-662kJ/mol

¢ÚCH4(g)+2NO2(g)N2(g)+CO2(g)+2H2O(l) ¡÷H=-955kJ/mol

ÆäÖТÙʽµÄƽºâ³£ÊýΪK1£¬¢ÚʽµÄƽºâ³£ÊýΪK2£¬Ôò£º(1) CH4 (g)+4NO (g£©=2N2(g) + CO2 (g )+H2O £¨1£©¡÷H=_________ £»¸Ä·´Ó¦µÄƽºâ³£ÊýK=_________£¨Óú¬Kl¡¢K2µÄ´úÊýʽ±íʾ£©¡£

IIÀûÓÃȼÁϵç³ØµÄÔ­ÀíÀ´´¦ÀíµªÑõ»¯ÎïÊÇÒ»ÖÖз½Ïò¡£×°ÖÃÈçͼËùʾ£¬ÔÚ´¦Àí¹ý³ÌÖÐʯīµç¼«IÉÏ·´Ó¦Éú³ÉÒ»ÖÖÑõ»¯ÎïY¡£

(2£©Ð´³öÑõ»¯ÎïYµÄ»¯Ñ§Ê½_____________£»

(3)ʯīIIµç¼«Îª______£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©¼«£¬¸Ãµç¼«·´Ó¦Îª______________£»

¢ó.ÀûÓûîÐÔÌ¿»¹Ô­·¨´¦ÀíµªÑõ»¯Îï¡£Óйط´Ó¦ÎªC(s)+2NO(g)N2(g)+ CO2(g) ¡÷H=akJ/mol¡£Ä³Ñо¿Ð¡×éÏòijºãÈÝÃܱÕÈÝÆ÷ÖмÓÈëÒ»¶¨Á¿µÄ»îÐÔÌ¿ºÍNO£¬ºãΣ¨T1¡æ£©Ìõ¼þÏ·´Ó¦£¬ÒÑÖªÔÚ²»Í¬Ê±¼ä²âµÃ¸÷ÎïÖʵÄŨ¶ÈÈçÏ£º

£¨4£©¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽK=_______£¬ÔÚ0¡«10sÄÚ£¬N2µÄƽ¾ù·´Ó¦ËÙÂÊΪ______mol/(L¡¤s)£¬NOµÄת»¯ÂÊΪ________£»

£¨5£©30sºó£¬¸Ä±äijһÌõ¼þ£¬·´Ó¦ÖØдﵽƽºâ£¬Ôò¸Ä±äµÄÌõ¼þ¿ÉÄÜÊÇ________£¬Èô30sºóÉý¸ßζÈÖÁT2¡æ£¬Æ½ºâʱ£¬ÈÝÆ÷ÖÐNOµÄŨ¶ÈÓÉ0.060mol/L±äΪ0.072mol/L£¬Ôò¸Ã·´Ó¦µÄa___0 £¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£

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

¾«Ó¢¼Ò½ÌÍø