ÌâÄ¿ÄÚÈÝ

£¨1£© 20ÊÀ¼Í30Äê´ú£¬EyringºÍPzerÔÚÅöײÀíÂ۵Ļù´¡ÉÏÌá³ö»¯Ñ§·´Ó¦µÄ¹ý¶É̬ÀíÂÛ£º»¯Ñ§·´Ó¦²¢²»ÊÇͨ¹ý¼òµ¥µÄÅöײ¾ÍÄÜÍê³ÉµÄ£¬¶øÊÇÔÚ·´Ó¦Îïµ½Éú³ÉÎïµÄ¹ý³ÌÖо­¹ýÒ»¸ö¸ßÄÜÁ¿µÄ¹ý¶É̬¡£ÈçͼÊÇNO2ºÍCO·´Ó¦Éú³ÉCO2ºÍNO¹ý³ÌÖеÄÄÜÁ¿±ä»¯Ê¾Òâͼ£¬ËµÃ÷Õâ¸ö·´Ó¦ÊÇ          £¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©·´Ó¦£¬NO2ºÍCOµÄ ×ÜÄÜÁ¿       £¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©CO2ºÍNOµÄ×ÜÄÜÁ¿¡£


£¨2£©ÔÚijÌå»ýΪ2LµÄÃܱÕÈÝÆ÷ÖгäÈë1.5mol NO2ºÍ2mol CO£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£ºNO2+COCO2+NO£¬2 minʱ£¬²âµÃÈÝÆ÷ÖÐNOµÄÎïÖʵÄÁ¿Îª0.5 mol £¬Ôò£º¢Ù´Ë¶Îʱ¼äÄÚ£¬ÓÃCO2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ               ¢Ú2 minʱ£¬ÈÝÆ÷ÄÚÆøÌåµÄ×ÜÎïÖʵÄÁ¿Îª_________.        

£¨1£© ·ÅÈÈ   ´óÓÚ
£¨2£©0.125mol/Lmin    3.5mol  

½âÎöÊÔÌâ·ÖÎö£º£¨1£©·´Ó¦ÎïµÄ×ÜÄÜÁ¿¸ßÓÚÉú³ÉÎïµÄ×ÜÄÜÁ¿£¬Òò´Ë¸Ã·´Ó¦Ê±·ÅÈÈ·´Ó¦¡£ËùÒÔ·´Ó¦Îï¶þÑõ»¯µªÓëÒ»Ñõ»¯Ì¼µÄ×ÜÄÜÁ¿´óÓÚ²úÎï¶þÑõ»¯Ì¼ÓëÒ»Ñõ»¯µªµÄ×ÜÄÜÁ¿¡££¨2£©¾ÝÌâÒâ¿ÉËã³öÒ»Ñõ»¯µªµÄËÙÂÊΪ0.5³ýÒÔ2ÔÙ³ýÒÔ2µÈÓÚ0.125 mol/Lmin£¬»¯Ñ§ËÙÂÊÖ®±ÈµÈÓÚ»¯Ñ§¼ÆÁ¿ÊýÖ®±È£¬¶øÒ»Ñõ»¯Ì¼ÓëÒ»Ñõ»¯µªµÄ»¯Ñ§¼ÆÁ¿Êý±ÈΪ1£»1ËùÒÔ£¬¶þÑõ»¯Ì¼µÄËÙÂÊΪ0.125mol/Lmin ¡£NO2+COCO2+NO£¬2 minÉú³ÉÁË0.5Ħ¶ûµÄÒ»Ñõ»¯µª£¬ÔòÒ²»áÉú³É0.5Ħ¶ûµÄ¶þÑõ»¯Ì¼¡£Í¬Ê±Ò²»áÏûºÄ0.5Ħ¶ûµÄÒ»Ñõ»¯Ì¼ºÍ¶þÑõ»¯µª¡£ËùÒÔÊ£ÓàµÄÒ»Ñõ»¯Ì¼Îª2¼õÈ¥0.5µÈÓÚ1.5£¬Ê£ÓàµÄ¶þÑõ»¯µªÎª1.5¼õÈ¥0.5µÈÓÚ1Ħ¶û¡£ËùÒÔÈÝÆ÷ÄÚÆøÌåµÄ×ÜÎïÖʵÄÁ¿Îª1.5+1+0.5+0.5µÈÓÚ3.5Ħ¶û
¿¼µã£º¿¼²é»¯Ñ§·´Ó¦ÓëÄÜÁ¿ÒÔ¼°»¯Ñ§·´Ó¦ËÙÂʵÄÏà¹Ø֪ʶ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

£¨12·Ö£©£¨1£©Ô̲ØÔÚº£µ×µÄ¡°¿Éȼ±ù¡±ÊǸßѹÏÂÐγɵÄÍâ¹ÛÏñ±ùµÄ¼×ÍéË®ºÏÎï¹ÌÌ壨·Ö×ÓʽΪCH4¡¤9H2O£©£¬Ôò356g¡°¿Éȼ±ù¡±ÊͷųöµÄ¼×ÍéȼÉÕ£¬Éú³ÉҺ̬ˮʱÄܷųö1780.6 kJµÄÈÈÁ¿£¬Ôò¼×ÍéȼÉÕµÄÈÈ»¯Ñ§·½³Ìʽ¿É±íʾΪ£º_______________________________¡£
£¨2£© ÔÚ100¡æʱ£¬½«0£®100molµÄN2O4ÆøÌå³äÈë1 LºãÈݳé¿ÕµÄÃܱÕÈÝÆ÷ÖУ¬¸ôÒ»¶¨Ê±¼ä¶Ô¸ÃÈÝÆ÷ÄÚÎïÖʵÄŨ¶È½øÐзÖÎöµÃµ½Ï±íÊý¾Ý£º

¢Ù´Ó±íÖзÖÎö£º¸Ã·´Ó¦µÄƽºâ³£ÊýΪ___________£»
¢ÚÔÚÉÏÊöÌõ¼þÏ£¬60sÄÚN2O4µÄƽ¾ù·´Ó¦ËÙÂÊΪ_____________£»   
¢Û´ïƽºâºóÏÂÁÐÌõ¼þµÄ¸Ä±ä¿ÉʹNO2Ũ¶ÈÔö´óµÄÊÇ_________¡£

A£®Ôö´óÈÝÆ÷µÄÈÝ»ý B£®ÔÙ³äÈëÒ»¶¨Á¿µÄN2O4
C£®ÔÙ³äÈëÒ»¶¨Á¿µÄNO2 D£®ÔÙ³äÈëÒ»¶¨Á¿µÄHe
£¨3£©³£ÎÂÏ¢ÙÓõÈŨ¶ÈµÄÑÎËá·Ö±ðÖк͵ÈÌå»ýpH=12µÄ°±Ë®ºÍNaOHÈÜÒº£¬ÏûºÄÑÎËáµÄÌå»ý·Ö±ðΪV1¡¢V2£¬ÔòV1_____V2(Ìî¡°>¡±¡°<¡±»ò¡°=¡±ÏÂͬ)£»
¢ÚÓõÈŨ¶ÈµÄÑÎËá·Ö±ðÖк͵ÈÌå»ýŨ¶È¾ùΪ0.01mol/LµÄ°±Ë®ºÍNaOHÈÜÒº£¬ÏûºÄÑÎËáµÄÌå»ý·Ö±ðΪV3¡¢V4£¬ÔòV3_____V4£»
¢ÛÓõÈŨ¶ÈµÄÑÎËá·Ö±ðºÍµÈÌå»ýŨ¶È¾ùΪ0.01mol/LµÄ°±Ë®ºÍNaOHÈÜÒº·´Ó¦£¬×îºóÈÜÒº¾ùΪÖÐÐÔ£¬ÏûºÄÑÎËáµÄÌå»ý·Ö±ðΪV5¡¢V6£¬ÔòV5_____V6¡£

£¨16·Ö£©
ÓÃCaSO4´úÌæO2ÓëȼÁÏCO·´Ó¦£¬¼È¿ÉÒÔÌá¸ßȼÉÕЧÂÊ£¬ÓÖÄܵõ½¸ß´¿CO2£¬ÊÇÒ»ÖÖ¸ßЧ¡¢Çå½à¡¢¾­¼ÃµÄÐÂÐÍȼÉÕ¼¼Êõ£¬·´Ó¦¢ÙΪÖ÷·´Ó¦£¬·´Ó¦¢ÚºÍ¢ÛΪ¸±·´Ó¦¡£
¢Ù1/4CaSO4(s)+CO(g)1/4CaS(s)+CO2(g) ¡÷H1=£­47.3kJ/mol
¢ÚCaSO4(s)+CO(g)CaO(s)+ CO2(g)+ SO2(g) ¡÷H2=+210.5kJ/mol
¢ÛCO(g)1/2C(s)+1/2CO2(g) ¡÷H3=-86.2kJ/mol
£¨1£©·´Ó¦2 CaSO4(s)+7CO(g)CaS(s)+CaO(s)+C(s)+6CO2(g)+SO2(g)µÄ¡÷H=       £¨Óá÷H1¡÷H2¡÷H3±íʾ£©¡£
£¨2£©·´Ó¦¢Ù¡«¢ÛµÄƽºâ³£ÊýµÄ¶ÔÊýlgKË淴ӦζÈTµÄ±ä»¯ÇúÏß¼ûͼ18.½áºÏ¸÷·´Ó¦µÄ¡÷H£¬¹éÄÉlgK¡«TÇúÏ߱仯¹æÂÉ£º

a£©                                  
b)                                   
£¨3£©ÏòÊ¢ÓÐCaSO4µÄÕæ¿ÕºãÈÝÈÝÆ÷ÖгäÈëCO£¬·´Ó¦¢ÙÓÚ900 ºC´ïµ½Æ½ºâ£¬cƽºâ£¨CO£©=8.0¡Á10-5mol¡¤L-1£¬¼ÆËãCOµÄת»¯ÂÊ£¨ºöÂÔ¸±·´Ó¦£¬½á¹û±£Áô2λÓÐЧÊý×Ö£©¡£
£¨4£©Îª¼õÉÙ¸±²úÎ»ñµÃ¸ü´¿¾»µÄCO2£¬¿ÉÔÚ³õʼȼÁÏÖÐÊÊÁ¿¼ÓÈë       ¡£
£¨5£©ÒÔ·´Ó¦¢ÙÖÐÉú³ÉµÄCaSΪԭÁÏ£¬ÔÚÒ»¶¨Ìõ¼þϾ­Ô­×ÓÀûÓÃÂÊ100%µÄ¸ßη´Ó¦£¬¿ÉÔÙÉú³ÉCaSO4£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ       £»ÔÚÒ»¶¨Ìõ¼þÏÂCO2¿ÉÓë¶Ô¶þ¼×±½·´Ó¦£¬ÔÚÆä±½»·ÉÏÒýÈëÒ»¸öôÈ»ù£¬²úÎïµÄ½á¹¹¼òʽΪ       ¡£

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

¾«Ó¢¼Ò½ÌÍø