ÌâÄ¿ÄÚÈÝ

£¨5·Ö£©½«1 molI2(g)ºÍ2 mol H2(g)ÖÃÓÚ2LÃܱÕÈÝÆ÷ÖУ¬ÔÚÒ»¶¨Î¶ÈÏ·¢Éú·´Ó¦£º

I2(g)£«H2(g)2HI(g)   DH<0£¬²¢´ïµ½Æ½ºâ¡£HIµÄÌå»ý·ÖÊýj(HI) ËæÊ±¼äµÄ±ä»¯ÈçÇúÏß(II)Ëùʾ¡£

£¨1£©´ïµ½Æ½ºâºó£¬I2(g)µÄÎïÖʵÄÁ¿Å¨¶ÈΪ                £»

£¨2£©Èô¸Ä±ä·´Ó¦Ìõ¼þ£¬ÔÚijÌõ¼þÏÂj(HI)µÄ±ä»¯ÈçÇúÏß(¢ñ)Ëùʾ£¬Ôò¸ÃÌõ¼þ¿ÉÄÜÊÇ                  £¨ÌîÈëÏÂÁÐÌõ¼þµÄÐòºÅ£©¡£

¢ÙºãÈÝÌõ¼þÏ£¬Éý¸ßζÈ

¢ÚºãÈÝÌõ¼þÏ£¬½µµÍζÈ

¢ÛºãÎÂÌõ¼þÏ£¬ËõС·´Ó¦ÈÝÆ÷Ìå»ý

¢ÜºãÎÂÌõ¼þÏ£¬À©´ó·´Ó¦ÈÝÆ÷Ìå»ý

¢ÝºãΡ¢ºãÈÝÌõ¼þÏ£¬¼ÓÈëÊʵ±µÄ´ß»¯¼Á

£¨1£©0.05mol?L-1   £¨2£©¢Û ¢É 
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
¸ù¾ÝµâÓëÇâÆø·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£¬ÏÂÁÐÅжÏÕýÈ·µÄÊÇ
¢ÙI2£¨g£©+H2£¨g£©?2HI£¨g£©¡÷H=-9.48kJ/mol
¢ÚI2£¨s£©+H2£¨g£©?2HI£¨g£©¡÷H=+26.48kJ/mol
£¨1£©Ð´³ö¹Ì̬µâÉú³ÉÆøÌ¬µâµÄÈÈ»¯Ñ§·½³Ìʽ£º
I2£¨g£©=I2£¨s£©¡÷H=-35.96kJ/mol
I2£¨g£©=I2£¨s£©¡÷H=-35.96kJ/mol
£®
£¨2£©Èô·´Ó¦¢ÚÎüÊÕ52.96kJÈÈÁ¿Ê±£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿ÊÇ
4
4
mol£®
£¨3£©¶ÔÓÚÔÚºãΡ¢ºãÈÝÃܱÕÈÝÆ÷ÖнøÐеķ´Ó¦¢Ù£¬ÄÜ˵Ã÷Æä´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ
CD
CD
£®
A£®ÈÝÆ÷ÖÐÆøÌåѹǿ²»±ä
B£®ÈÝÆ÷ÖÐÆøÌåÃܶȲ»±ä
C£®ÈÝÆ÷ÖÐÆøÌåÑÕÉ«µÄÉîdz²»±ä
D£®ÓÐn¸öH-H¼ü¶ÏÁѵÄͬʱÓÐ2n¸öH-I¼ü¶ÏÁÑ
£¨4£©¿Éͨ¹ý·´Ó¦2NO+O2=2NO2ºÍNO2+2H++2I-=NO+I2+H2OÀ´ÖÆÈ¡µâ£¬NOÔÚÖÆµâ¹ý³ÌÖеÄ×÷ÓÃÊÇ
´ß»¯¼Á
´ß»¯¼Á
£®
£¨5£©ÁòËá¹¤ÒµÎ²ÆøÖжþÑõ»¯ÁòµÄº¬Á¿³¬¹ý0.05%£¨Ìå»ý·ÖÊý£©Ê±Ðè¾­´¦Àíºó²ÅÄÜÅÅ·Å£®Ä³Ð£»¯Ñ§ÐËȤС×éÓû²â¶¨Ä³ÁòËṤ³§ÅÅ·ÅÎ²ÆøÖжþÑõ»¯ÁòµÄº¬Á¿£¬²ÉÓÃÒÔÏ·½°¸£ºÈçͼ2Ëùʾ£¬Í¼ÖÐÆøÌåÁ÷Á¿¼ÆBÓÃÓÚ׼ȷ²âÁ¿Í¨¹ýµÄÎ²ÆøÌå»ý£®½«Î²ÆøÍ¨ÈëÒ»¶¨Ìå»ýÒÑ֪Ũ¶ÈµÄµâË®ÖвⶨSO2µÄº¬Á¿£®µ±Ï´ÆøÆ¿CÖÐÈÜÒºµÄÀ¶É«Ïûʧʱ£¬Á¢¼´¹Ø±Õ»îÈûA£®
¢ÙÓõâË®²â¶¨SO2µÄº¬Á¿µÄ»¯Ñ§·½³ÌʽÊÇ
SO2+I2+2H2O=H2SO4+2HI
SO2+I2+2H2O=H2SO4+2HI
£®
¢ÚÏ´ÆøÆ¿CÖе¼¹ÜÄ©¶ËÁ¬½ÓÒ»¸ö¶à¿×ÇòÅÝD£¬¿ÉÒÔÌá¸ßʵÑéµÄ׼ȷ¶È£¬ÆäÀíÓÉÊÇ
Ôö´óSO2ÓëµâË®µÄ½Ó´¥Ãæ»ý£¬Ê¹SO2ºÍµâË®³ä·Ö·´Ó¦
Ôö´óSO2ÓëµâË®µÄ½Ó´¥Ãæ»ý£¬Ê¹SO2ºÍµâË®³ä·Ö·´Ó¦
£®
¢ÛÏ´ÆøÆ¿CÖÐÈÜÒºµÄÀ¶É«Ïûʧºó£¬Ã»Óм°Ê±¹Ø±Õ»îÈûA£¬²âµÃµÄSO2º¬Á¿
Æ«µÍ
Æ«µÍ
£¨Ìî¡°Æ«¸ß¡±¡¢¡°Æ«µÍ¡±»ò¡°ÎÞÓ°Ï족£©£®

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

¾«Ó¢¼Ò½ÌÍø