ÌâÄ¿ÄÚÈÝ

ÔÚÒ»¸öÈÝ»ý¹Ì¶¨Îª2LµÄÃܱÕÈÝÆ÷ÖУ¬·¢Éú·´Ó¦£ºaA£¨g£©+bB£¨g£©¨TpC£¨g£©¡÷H=£¿·´Ó¦Çé¿ö¼Ç¼ÈçÏÂ±í£º
ʱ¼ä c£¨A£©/mol L-1 c£¨B£©/mol L-1 c£¨C£©/mol L-1
0min 1 3 0
µÚ2min 0.8 2.6 0.4
µÚ4min 0.4 1.8 1.2
µÚ6min 0.4 1.8 1.2
µÚ8min 0.1 2.0 1.8
µÚ9min 0.05 1.9 0.3
Çë×Ðϸ·ÖÎö¸ù¾Ý±íÖÐÊý¾Ý£¬»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©µÚ2minµ½µÚ4minÄÚAµÄƽ¾ù·´Ó¦ËÙÂÊv£¨A£©=
 
mol?L-1?min-1
£¨2£©ÓɱíÖÐÊý¾Ý¿ÉÖª·´Ó¦ÔÚµÚ4minµ½µÚ6minʱ´¦ÓÚÆ½ºâ״̬£¬ÈôÔÚµÚ2min¡¢µÚ6min¡¢µÚ8minʱ·Ö±ð¸Ä±äÁËijһ·´Ó¦Ìõ¼þ£¬Ôò¸Ä±äµÄÌõ¼þ·Ö±ð¿ÉÄÜÊÇ£º
¢ÙµÚ2min
 
»ò
 

¢ÚµÚ6min
 
£»
¢ÛµÚ8min
 
£»
£¨3£©Èô´Ó¿ªÊ¼µ½µÚ4min½¨Á¢Æ½ºâʱ·´Ó¦·Å³öµÄÈÈÁ¿Îª235.92kJ£¬Ôò¸Ã·´Ó¦µÄ¡÷H=
 
£®
·ÖÎö£º£¨1£©AÎïÖʵĻ¯Ñ§·´Ó¦ËÙÂÊv£¨A£©=
¡÷c(A)
¡÷t
£¬´úÈëÊý¾Ý½øÐмÆËã¼´¿É£»
£¨2£©Ó°Ï컯ѧ·´Ó¦ËÙÂʵÄÒòËØÓУºÊ¹Óô߻¯¼Á¡¢¸Ä±äζȡ¢Å¨¶È¡¢¸Ä±äÎïÖʵıíÃæ»ýµÈ£»
£¨3£©ÔÚ¿ªÊ¼µ½µÚ4min½¨Á¢Æ½ºâʱ£¬AÎïÖʵÄÏûºÄÁ¿Îª0.6mol/£¨l?min£©£¬¸ù¾Ý»¯Ñ§·½³Ìʽ¼ÆËã¼´¿É£®
½â´ð£º½â£º£¨1£©AÎïÖʵĻ¯Ñ§·´Ó¦ËÙÂÊv£¨A£©=
¡÷c(A)
¡÷t
=
0.8mol/L-0.4mol/L
2min
¨T0.2mol/£¨l?min£©£¬¹Ê´ð°¸Îª£º0.2£»
£¨2£©¢ÙÔÚµÚ0¡«2minÄÚ£¬AµÄŨ¶È¼õСÁË0.2mol/l£¬ÔÚ2¡«4minÄÚAµÄŨ¶ÈÔÚÔ­»ù´¡ÉϼõСÁË0.4mol/l£¬»¯Ñ§·´Ó¦ËÙÂʼӿìÁË£¬¿ÉÒÔÊÇʹÓô߻¯¼Á»òÉý¸ßζȣ¬¹Ê´ð°¸Îª£ºÊ¹Óô߻¯¼Á£»Éý¸ßζȣ»
¢ÚµÚ6min µ½µÚ8min£¬×÷Ϊ·´Ó¦ÎŨ¶ÈÓ¦¸ÃÊǼõСµÄÇ÷ÊÆ£¬µ«ÊÇBµÄŨ¶È´Ó1.8mol/lÔö¼Óµ½ÁË2.0mol/l£¬ËùÒÔ¿ÉÖªÒ»¶¨ÊǼÓÈëÁËBÎïÖÊ£¬¹Ê´ð°¸Îª£ºÔö¼ÓBµÄŨ¶È£»
¢ÛµÚ8minµ½µÚ9minʱ¼ä¶ÎÄÚ£¬C×÷ΪÉú³ÉÎËüµÄŨ¶ÈÓ¦ÊÇÔö´óµÄÇ÷ÊÆ£¬µ«ÊÇÊý¾Ý±íÃ÷£¬ÆäŨ¶È¼õСÁË£¬Ò»¶¨ÊǼõÉÙÁËCµÄŨ¶È£¬¹Ê´ð°¸Îª£º¼õСCµÄŨ¶È£®
£¨3£©¸ù¾ÝÈÈ»¯Ñ§·½³Ìʽ£¬É裺·´Ó¦µÄìʱäֵΪX
A£¨g£©+2B£¨g£©¾«Ó¢¼Ò½ÌÍø2C£¨g£©¡÷H=-XkJ/mol
1mol                           XkJ
£¨1mol/l-0.4mol/l£©¡Á2L       235.92kJ
ÔòÓУº
1mol
1.2mol
=
XkJ
235.9kJ
£¬½âµÃ£ºX=-196.6 kJ?mol-1
 ¹Ê´ð°¸Îª£º-196.6 kJ?mol-1£®
µãÆÀ£º±¾Ì⿼²éÁËѧÉúÓ°Ï컯ѧ·´Ó¦ËÙÂʵÄÒòËØºÍÓйط´Ó¦ËÙÂʵļÆË㣬ÓÐÒ»¶¨µÄ×ÛºÏÐÔ£®
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´¦ÓÚÆ½ºâ״̬£¬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