ÌâÄ¿ÄÚÈÝ

ÔÚ80¡æÊ±£¬½«0.4molµÄËÄÑõ»¯¶þµªÆøÌå³äÈë2LÒѳé¿ÕµÄ¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷ÖУ¬¸ôÒ»¶Îʱ¼ä¶Ô¸ÃÈÝÆ÷ÄÚµÄÎïÖʽøÐзÖÎö£¬µÃµ½ÈçÏÂÊý¾Ý£º
C£¨mol/L£©         Ê±¼ä£¨s£©¡¡0¡¡20¡¡40¡¡60¡¡80¡¡100
C£¨N2O4£©0.20a0.10cde
C£¨NO2£©0.000.12b0.220.220.22
·´Ó¦½øÐÐÖÁ100sºó½«·´Ó¦»ìºÏÎïµÄζȽµµÍ£¬·¢ÏÖÆøÌåµÄÑÕÉ«±ädz£®
£¨1£©¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨2£©±íÖÐb
 
c£¨Ìî¡°£¼¡±¡¢¡°=¡±¡¢¡°£¾¡±£©£®
£¨3£©20sʱ£¬N2O4µÄŨ¶Èc=
 
mol/L£®
£¨4£©¼ÆËãÆ½ºâʱ»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿£¨½á¹û±£ÁôÕûÊý£©£®
¿¼µã£º»¯Ñ§Æ½ºâ½¨Á¢µÄ¹ý³Ì
רÌ⣺»¯Ñ§Æ½ºâרÌâ
·ÖÎö£º£¨1£©ÒÀ¾Ý·´Ó¦Ìõ¼þ·ÖÎöÅжϣ¬ËÄÑõ»¯¶þµª·´Ó¦Éú³É¶þÑõ»¯µª£»
£¨2£©½áºÏ»¯Ñ§Æ½ºâµÄÈý¶ÎʽÁÐʽ¼ÆËã·ÖÎö±È½Ï£»
£¨3£©»¯Ñ§Æ½ºâµÄÈý¶ÎʽÁÐʽ¼ÆËãN2O4µÄŨ¶È£»
£¨4£©ÒÀ¾Ý80¡ãCƽºâ״̬ϽáºÏÈý¶ÎʽÁÐʽ¼ÆËãÆ½ºâÎïÖʵÄÁ¿£¬½áºÏ»ìºÏÆøÌ寽¾ùĦ¶ûÖÊÁ¿=
m
n
£®
½â´ð£º ½â£º£¨1£©½«0.4molµÄËÄÑõ»¯¶þµªÆøÌå³äÈë2LÒѳé¿ÕµÄ¹Ì¶¨ÈÝ»ýµÄÃܱÕÈÝÆ÷ÖУ¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºN2O4?2NO2£¬
¹Ê´ð°¸Îª£ºN2O4?2NO2£»
£¨2£©¸ôÒ»¶Îʱ¼ä¶Ô¸ÃÈÝÆ÷ÄÚµÄÎïÖʽøÐзÖÎö£¬µÃµ½ÈçÏÂÊý¾Ý£º·´Ó¦½øÐÐÖÁ100sºó½«·´Ó¦»ìºÏÎïµÄζȽµµÍ£¬·¢ÏÖÆøÌåµÄÑÕÉ«±ädz£¬ËµÃ÷·´Ó¦ÄæÏò½øÐУ¬ÄæÏòÊÇ·ÅÈÈ·´Ó¦£¬ÕýÏòÊÇÎüÈÈ·´Ó¦£»ÒÀ¾Ý»¯Ñ§Æ½ºâÈý¶ÎʽÁÐʽ¼ÆËã·ÖÎöÅжϣ»½øÐе½40SºÍ½øÐе½60Sʱ£»
              N2O4?2NO2 
ÆðʼÁ¿£¨mol£©  0.4    0
±ä»¯Á¿£¨mol£© 0.2   0.4
40SÄ©£¨mol£©  0.2    0.4
µÃµ½b=0.2mol/L£»
½øÐе½60SºÍ½øÐе½60Sʱ£»
              N2O4?2NO2 
ÆðʼÁ¿£¨mol£©  0.4    0
±ä»¯Á¿£¨mol£© 0.22    0.44
60SÄ©£¨mol£©  0.18    0.44
c=0.18 mol/L  
¼ÆËã±È½ÏµÃµ½b£¾c£»
¹Ê´ð°¸Îª£º£¾£»
£¨3£©½øÐе½20S£»
              N2O4?2NO2 
ÆðʼÁ¿£¨mol£©  0.4    0
±ä»¯Á¿£¨mol£© 0.12   0.24
20SÄ©£¨mol£©  0.28  0.24
20sʱ£¬N2O4µÄŨ¶È=
0.28mol
2L
=0.14mol/L£¬
¹Ê´ð°¸Îª£º0.14£»
£¨4£©60¡ãC·´Ó¦´ïµ½Æ½ºâ״̬ºÍ80¡ãCÊÇÏàͬµÄƽºâ״̬£¬ÒÀ¾Ý£¨2£©¼ÆËãµÃµ½£¬Æ½ºâŨ¶Èc£¨NO2£©=0.22mol/L£¬c£¨N2O4£©=0.09mol/L£»ÆøÌåÎïÖʵÄÁ¿n£¨NO2£©=0.44mol£¬n£¨N2O4£©=0.18mol£»
ƽºâʱ»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿=
0.44mol¡Á46g/mol+0.18mol¡Á92g/mol
0.44mol+0.18mol
=59g/mol£¬
´ð£º»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿59£»
µãÆÀ£º±¾Ì⿼²éÁË»¯Ñ§Æ½ºâµÄ½¨Á¢Åжϣ¬Æ½ºâ³£ÊýµÄ¼ÆËãÓ¦Óã¬Æ½ºâÒÆ¶¯Ô­ÀíµÄ·ÖÎöÅжϣ¬ÌâÄ¿ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø