ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿Ò»¶¨Î¶ÈÏ£¬ÏàͬÌå»ýµÄËĸö¸ÕÐÔÈÝÆ÷Öзֱð´æÔÚÒÔÏÂËĸöƽºâ£º

¢ÙN2(g)+3H2(g)2NH3(g) K1

¢ÚH2(g)+I2(g)2HI(g) K2

¢Û2NO2(g)N2O4(g) K3

¢ÜC(s)+H2O(g)CO(g)+H2(g) K4

ÇëÍê³ÉÏÂÁи÷Ì⣺

£¨1£©Ð´³ö·´Ó¦¢ÜµÄƽºâ³£ÊýµÄ±í´ïʽK4=____________£»

£¨2£©ÏÖÓÐÏàͬζÈϵÄÒÔÏÂƽºâ£º

¢Ý 2N2(g)+6H2(g)4NH3(g) K5

¢Þ2HI(g)H2(g)+I2(g) K6

ÔòK5=__ £»K6=_ £»£¨ÓÃK1¡¢K2¡¢K3¡¢K4±íʾ£©

£¨3£©Èôƽºâ¢ÛÖÐNO2µÄÌå»ý·ÖÊýΪa£¬Ä³Ê±¿ÌÔÙ¼ÓÈëÒ»¶¨Á¿µÄN2O4£¬´Ëʱ¦Ô£¨Õý£© ¦Ô£¨Ä棩£»Ôٴδﵽƽºâºó£¬NO2µÄÌå»ý·ÖÊý a¡££¨Ìî¡°©ƒ¡±¡¢¡°©‚¡±»ò¡°©„¡±£©

¡¾´ð°¸¡¿£¨1£©K4=£»

£¨2£©K5=K12£»K6=1/K2£»

£¨3£©¦Ô£¨Õý£©©‚¦Ô£¨Ä棩£»©‚¡£

¡¾½âÎö¡¿

ÊÔÌâ·ÖÎö£º£¨1£©»¯Ñ§Æ½ºâ³£ÊýµÈÓÚÉú³ÉÎïƽºâŨ¶ÈϵÊý´ÎÃݵĻýÓë·´Ó¦ÎïƽºâŨ¶ÈϵÊý´ÎÃݵĻýµÄ±ÈÖµ¡£Ôò·´Ó¦¢ÜµÄƽºâ³£ÊýµÄ±í´ïʽK4=£»

£¨2£©ÒÑÖª¢ÙN2(g)+3H2(g)2NH3(g) £¬¢Ù¡Á5µÃ2N2(g)+6H2(g)4NH3(g) ÔòK5=K12£»¢ÚºÍ¢Þ¹ý³Ì»¥Ä棬ƽºâ³£Êý»¥Îªµ¹Êý£¬Ôò¢Þ2HI(g) H2(g)+I2(g)µÄƽºâ³£ÊýK6=1/K2£»

£¨3£©Èôƽºâ¢ÛÖÐNO2µÄÌå»ý·ÖÊýΪa£¬Ä³Ê±¿ÌÔÙ¼ÓÈëÒ»¶¨Á¿µÄN2O4£¬Æ½ºâÄæÏòÒƶ¯£¬´Ëʱ¦Ô£¨Õý£©<¦Ô£¨Ä棩£»ÔÙ¼ÓÈëÒ»¶¨Á¿µÄN2O4£¬Ï൱ÓÚÔö´óѹǿ£¬Æ½ºâÕýÏòÒƶ¯£¬Ôٴδﵽƽºâºó£¬NO2µÄÌå»ý·ÖÊý<a¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø