ÌâÄ¿ÄÚÈÝ

£¨10·Ö£©.ÏòÒ»¶¨Ìå»ýµÄÃܱÕÈÝÆ÷ÖмÓÈë2 mol A¡¢0.6 mol CºÍÒ»¶¨Á¿µÄBÈýÖÖÆøÌå¡£Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£¬¸÷ÎïÖʵÄÁ¿Å¨¶ÈËæʱ¼ä±ä»¯Èçͼ£¨¢ñ£©Ëùʾ£¬ÆäÖÐt0--t1½×¶Îc£¨B£©Î´»­³ö¡£Í¼£¨¢ò£©Îªt2ʱ¿Ìºó¸Ä±ä·´Ó¦Ìõ¼þ£¬»¯Ñ§·´Ó¦ËÙÂÊËæʱ¼ä±ä»¯µÄÇé¿ö£¬Ëĸö½×¶Î¸Ä±äµÄÌõ¼þ¾ù²»Ïàͬ£¬Ã¿¸ö½×¶ÎÖ»¸Ä±äŨ¶È¡¢Ñ¹Ç¿¡¢Î¶ȡ¢´ß»¯¼ÁÖеÄÒ»¸öÌõ¼þ£¬ÆäÖÐt3---t4½×¶ÎΪʹÓô߻¯¼Á¡£ 
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©Èôt1="15" min£¬Ôòt0---t1½×¶ÎÒÔCÎïÖʵÄŨ¶È±ä»¯±íʾµÄ·´Ó¦ËÙÂÊΪ          mol¡¤ L¡ª1¡¤min¡ª1¡£
£¨2£©t4--t5½×¶Î¸Ä±äµÄÌõ¼þΪ          £¬BµÄÆðʼÎïÖʵÄÁ¿Å¨¶ÈΪ        mol¡¤ L¡ª1¡£
£¨3£©t5----t6½×¶Î±£³ÖÈÝÆ÷ÄÚζȲ»±ä£¬ÈôAµÄÎïÖʵÄÁ¿¹²±ä»¯ÁË0.01mol£¬¶ø´Ë¹ý³ÌÖеÄÈÈЧӦΪa kJÈÈÁ¿£¬Ð´³ö´ËζÈϸ÷´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
                                                             
£¨1£©0.02
£¨2£©¼õСѹǿ  0.5
£¨3£©2A(g)+B(g)="3C(g)"
¡÷H="+200a " KJ/mol
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨15·Ö£©¢ñ.ÔÚÒ»¸öÈÝ»ý¹Ì¶¨Îª2LµÄÃܱÕÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£ºaA (g) + bB(g) pC(g) ¡÷H=?·´Ó¦Çé¿ö¼Ç¼ÈçÏÂ±í£º
ʱ¼ä/(min)
n(A)/( mol)
n(B)/( mol)
n(C)/( mol)
0
1
3
0
µÚ2 min
0.8
2.6
0.4
µÚ4 min
0.4
1.8
1.2
µÚ6 min
0.4
1.8
1.2
µÚ8 min
0.1
2.0
1.8
µÚ9 min
0.05
1.9
0.3
Çë¸ù¾Ý±íÖÐÊý¾Ý×Ðϸ·ÖÎö£¬»Ø´ðÏÂÁÐÎÊÌ⣺
(1)µÚ2minµ½µÚ4minÄÚAµÄƽ¾ù·´Ó¦ËÙÂÊV(A)=                mol?L£­1? min£­1
(2)ÓɱíÖÐÊý¾Ý¿ÉÖª·´Ó¦ÔÚµÚ4minµ½µÚ6minʱ´¦ÓÚƽºâ״̬£¬ÈôÔÚµÚ2min¡¢µÚ6min¡¢µÚ8 minʱ·Ö±ð¸Ä±äÁËijһ¸ö·´Ó¦Ìõ¼þ£¬Ôò¸Ä±äµÄÌõ¼þ·Ö±ð¿ÉÄÜÊÇ£º
¢ÙµÚ2min                            »ò                                  £»
¢ÚµÚ6min                                      £»
¢ÛµÚ8 min                                     ¡£
(3)Èô´Ó¿ªÊ¼µ½µÚ4 min½¨Á¢Æ½ºâʱ·´Ó¦·Å³öµÄÈÈÁ¿Îª235.92kJÔò¸Ã·´Ó¦µÄ¡÷H=       ¡£
(4)·´Ó¦ÔÚµÚ4 min½¨Á¢Æ½ºâ£¬´ËζÈϸ÷´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýK=              .
¢ò.ÒÑÖª£º³£ÎÂÏ£¬AËáµÄÈÜÒºpH£½a, B¼îµÄÈÜÒºpH£½b
(1)ÈôAΪÑÎËᣬBΪÇâÑõ»¯±µ£¬ÇÒa=3,b=11,Á½ÕßµÈÌå»ý»ìºÏ£¬ÈÜÒºµÄpHΪ              ¡£
A£®´óÓÚ7              B.µÈÓÚ7            C. СÓÚ7
(2)ÈôAΪ´×ËᣬBΪÇâÑõ»¯ÄÆ£¬ÇÒa=4,b=12£¬ÄÇôAÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ
            mol?L£­1£¬BÈÜÒºÖÐË®µçÀë³öµÄÇâÀë×ÓŨ¶ÈΪ             mol?L£­1¡£
(3)ÈôAµÄ»¯Ñ§Ê½ÎªHR£¬BµÄ»¯Ñ§Ê½ÎªMOH£¬ÇÒa+b=14,Á½ÕßµÈÌå»ý»ìºÏºóÈÜÒºÏÔ¼îÐÔ¡£Ôò»ìºÏÈÜÒºÖбض¨ÓÐÒ»ÖÖÀë×ÓÄÜ·¢ÉúË®½â£¬¸ÃË®½â·´Ó¦µÄÀë×Ó·½³ÌʽΪ£º
                                                             ¡£

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

¾«Ó¢¼Ò½ÌÍø