¢ñ.ÔÚijѹǿºã¶¨µÄÃܱÕÈÝÆ÷ÖмÓÈë2 mol N2ºÍ4 mol H2£¬·¢ÉúÈçÏ·´Ó¦£º

N2(g)+3H2(g)2NH3(g);¦¤H=-92.4 kJ¡¤mol-1¡£´ïµ½Æ½ºâʱ£¬Ìå»ýΪ·´Ó¦Ç°µÄÈý·ÖÖ®¶þ¡£Çó£º

£¨1£©´ïµ½Æ½ºâʱ£¬N2µÄת»¯ÂÊΪ____________¡£

£¨2£©ÈôÏò¸ÃÈÝÆ÷ÖмÓÈëa mol N2¡¢b mol H2¡¢c mol NH3£¬ÇÒa¡¢b¡¢c¾ù´óÓÚ0£¬ÔÚÏàͬÌõ¼þÏ´ﵽƽºâʱ£¬»ìºÏÎïÖи÷×é·ÖµÄÎïÖʵÄÁ¿ÓëÉÏÊöƽºâÏàͬ¡£ÊԱȽϷ´Ó¦·Å³öµÄÄÜÁ¿£º£¨1£©__________£¨2£©£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©¡£

¢ò.Èô½«2 mol N2ºÍ4 mol H2·ÅÈëÆðʼÌå»ýÏàͬµÄºãÈÝÈÝÆ÷ÖУ¬ÔÚÓë¢ñÏàͬµÄζÈÏ´ﵽƽºâ¡£ÊԱȽÏƽºâʱNH3µÄŨ¶È£º¢ñ____________¢ò£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©¡£

¢ñ.£¨1£©50%

£¨2£©£¾

¢ò.£¾

½âÎö£º¢ñ.£¨1£©

              N2(g)+3H2(g)2NH3(g);¦¤H=-92.4 kJ¡¤mol-1¡ª×Ü

Æð(mol):   2        4                    0                                            6

±ä(mol):   x       3x                  2x

ƽ(mol):2-x    4-3x                 2x                                          6-2x

ÒòͬÎÂͬѹ£¬£¬µÃx=1 mol£¬Ôò¦Á(N2)=¡Á100%=50%¡£(2)Èôc£¾2£¬ÄæÏò·´Ó¦´ïµ½Æ½ºâ£¬´Ë¹ý³ÌÎüÊÕÄÜÁ¿£»Èôc=2£¬N2¡¢H2¡¢NH3´¦ÓÚƽºâ״̬£¬²»ÊÍ·ÅÄÜÁ¿£»Èôc£¼2£¬·´Ó¦ÏòÓҴﵽƽºâ£¬´Ë¹ý³Ì½öÉú³ÉNH3(2-c) mol£¬·Å³öµÄÄÜÁ¿±äÉÙ¡£×ÛºÏÒÔÉÏÇé¿ö£¬·Å³öµÄÄÜÁ¿£¨1£©£¾£¨2£©¡£¢ò.ÓÒÏò·´Ó¦ÖÐÆøÌå×ÜÎïÖʵÄÁ¿¼õÉÙ£¬ºãѹÌõ¼þ½¨Á¢Æ½ºâ¹ý³ÌÖÐÈÝ»ý²»¶Ï¼õС£¬´ïµ½Æ½ºâʱÓÒÏò½øÐеij̶ÈÏà¶ÔÓÚºãÈÝ·´Ó¦´ó¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ºÏ³É°±¹¤Òµ¶Ô»¯Ñ§¹¤ÒµºÍ¹ú·À¹¤Òµ¾ßÓÐÖØÒªÒâÒ壮
£¨1£©³£ÎÂÏ°±Æø¼«Ò×ÈÜÓÚË®£¬ÆäË®ÈÜÒº¿ÉÒÔµ¼µç£®
¢ÙÓ÷½³Ìʽ±íʾ°±ÆøÈÜÓÚË®µÄ¹ý³ÌÖдæÔڵĿÉÄæ¹ý³Ì
NH3+H2O?NH3?H2O?NH4++OH-
NH3+H2O?NH3?H2O?NH4++OH-

¢Ú°±Ë®ÖÐË®µçÀë³öµÄc£¨OH-£©
£¼
£¼
10-7mol/L£¨Ìîд¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©
¢Û½«ÏàͬÌå»ý¡¢ÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄ°±Ë®ºÍÑÎËá»ìºÏºó£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½ÒÀ´ÎΪ
c£¨Cl-£©£¾c£¨NH4+£©£¾c£¨H+£©£¾c£¨OH-£©
c£¨Cl-£©£¾c£¨NH4+£©£¾c£¨H+£©£¾c£¨OH-£©
£®
£¨2£©°±Æø¾ßÓл¹Ô­ÐÔ£¬ÔÚÍ­µÄ´ß»¯×÷ÓÃÏ£¬°±ÆøºÍ·úÆø·´Ó¦Éú³ÉAºÍB£®AΪï§ÑΣ¬BÔÚ±ê×¼×´¿öÏÂΪÆø̬£®ÔÚ´Ë·´Ó¦ÖУ¬Èôÿ·´Ó¦1Ìå»ý°±Æø£¬Í¬Ê±·´Ó¦0.75Ìå»ý·úÆø£»Èôÿ·´Ó¦8.96L°±Æø£¨±ê×¼×´¿ö£©£¬Í¬Ê±Éú³É0.3molA£®
¢Ùд³ö°±ÆøºÍ·úÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ
4NH3+3F2
 Cu 
.
 
NF3+3NH4F£¨´ß»¯¼ÁҪעÃ÷ÊÇÍ­£©
4NH3+3F2
 Cu 
.
 
NF3+3NH4F£¨´ß»¯¼ÁҪעÃ÷ÊÇÍ­£©
£»
¢ÚÔÚ±ê×¼×´¿öÏ£¬Ã¿Éú³É1mol B£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿Îª
6
6
mol£®
£¨3£©¢ñÔÚijѹǿºã¶¨µÄÃܱÕÈÝÆ÷ÖмÓÈë2mol N2ºÍ4mol H2£¬·¢ÉúÈçÏ·´Ó¦£º
N2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H=-92.4kJ/mol´ïµ½Æ½ºâʱ£¬Ìå»ýΪ·´Ó¦Ç°µÄÈý·ÖÖ®¶þ£®Çó£º
¢Ù´ïµ½Æ½ºâʱ£¬N2µÄת»¯ÂÊΪ
50%
50%
£®
¢ÚÈôÏò¸ÃÈÝÆ÷ÖмÓÈëa mol N2¡¢b mol H2¡¢c mol NH3£¬ÇÒa¡¢b¡¢c¾ù£¾0£¬ÔÚÏàͬÌõ¼þÏ´ﵽƽºâʱ£¬»ìºÏÎïÖи÷×é·ÖµÄÎïÖʵÄÁ¿ÓëÉÏÊöƽºâÏàͬ£®ÊԱȽϷ´Ó¦·Å³öµÄÄÜÁ¿£º¢Ù
£¾
£¾
¢Ú£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©
¢òÈô½«2mol N2ºÍ4mol H2·ÅÈëÆðʼÌå»ýÏàͬµÄºãÈÝÈÝÆ÷ÖУ¬ÔÚÓë¢ñÏàͬµÄζÈÏ´ﵽƽºâ£®
¢ÛÊԱȽÏƽºâʱNH3µÄŨ¶È£º¢ñ
£¾
£¾
¢ò£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¢ñ.ÒÑ֪ijÑÎNaHAµÄË®ÈÜÒºÖнöº¬Ë®·Ö×Ó¡¢Á½ÖÖÑôÀë×Ó¡¢ÈýÖÖÒõÀë×Ó£¬»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©H2AÔÚË®ÖеĵçÀë·½³Ìʽ£º___________________________________¡£

£¨2£©Èô0.1 mol¡¤L-1µÄNaHAÈÜÒºpH=3£¬Ôò¸ÃÈÜÒº¸÷Àë×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ________________£»¸ÃÈÜÒºÓÉË®µçÀë²úÉúµÄc£¨H+£©=_____________mol¡¤L-1¡£

£¨3£©Íù10 mL pH=3µÄ 0.1 mol¡¤L-1µÄNaHAÈÜÒºÖмÓË®ÖÁ1 000 mL£¬ÈÜÒºpH½«___________¡£

A.µÈÓÚ5             B.´óÓÚ5               C.СÓÚ5

£¨4£©ÔÚ0.1 mol¡¤L-1µÄNaHAÈÜÒºÖÐÀë×ÓŨ¶È¹ØϵÕýÈ·µÄÊÇ_____________¡£

A.c(HA-)+c(H2A)+c(A2-)=0.1 mol¡¤L-1          B.c(Na+)+c(H+)=c(OH-)+c(HA-)+c(A2-)

C.c(Na+)£¾c(HA-)£¾c(H+)£¾c(A2-)                   D.c(OH-) =c(H+)-c(A2-)

¢ò.ÔÚijѹǿºã¶¨µÄÃܱÕÈÝÆ÷ÖмÓÈë2 mol N2ºÍ4 mol H2£¬·¢ÉúÈçÏ·´Ó¦£º

N2£¨g£©+3H2£¨g£©2NH3£¨g£©£»¦¤H=-92.4 kJ¡¤mol-1¡£´ïµ½Æ½ºâʱ£¬Ìå»ýΪ·´Ó¦Ç°µÄÈý·ÖÖ®¶þ¡£Çó£º

£¨1£©´ïµ½Æ½ºâʱ£¬N2µÄת»¯ÂÊΪ_____________¡£

£¨2£©ÈôÏò¸ÃÈÝÆ÷ÖмÓÈëa mol N2¡¢b mol H2¡¢c mol NH3£¬ÇÒa¡¢b¡¢c¾ù´óÓÚ0,ÔÚÏàͬÌõ¼þÏ´ﵽƽºâʱ£¬»ìºÏÎïÖи÷×é·ÖµÄÎïÖʵÄÁ¿ÓëÉÏÊöƽºâÏàͬ£¬ÊԱȽϷ´Ó¦·Å³öµÄÄÜÁ¿£º

£¨1£©_____________£¨2£©£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©¡£

(3)Èô½«2 mol N2ºÍ4 mol H2·ÅÈëÆðʼÌå»ýÏàͬµÄºãÈÝÈÝÆ÷ÖУ¨ÏàͬζÈÏ£©£¬´ïµ½Æ½ºâ¡£ÊԱȽÏƽºâʱNH3µÄŨ¶È£º£¨1£©_____________(3)£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ºÏ³É°±¹¤Òµ¶Ô»¯Ñ§¹¤ÒµºÍ¹ú·À¹¤Òµ¾ßÓÐÖØÒªÒâÒå¡£

£¨1£©³£ÎÂÏ°±Æø¼«Ò×ÈÜÓÚË®£¬ÆäË®ÈÜÒº¿ÉÒÔµ¼µç¡£

¢ÙÓ÷½³Ìʽ±íʾ°±ÆøÈÜÓÚË®µÄ¹ý³ÌÖдæÔڵĿÉÄæ¹ý³Ì

_______________________________________________________________

¢Ú°±Ë®ÖÐË®µçÀë³öµÄ£¨Ìîд¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©

¢Û½«ÏàͬÌå»ý¡¢ÏàͬÎïÖʵÄÁ¿Å¨¶ÈµÄ°±Ë®ºÍÑÎËá»ìºÏºó£¬ÈÜÒºÖÐÀë×ÓŨ¶ÈÓÉ´óµ½Ð¡ÒÀ´ÎΪ__________________¡£

£¨2£©°±Æø¾ßÓл¹Ô­ÐÔ£¬ÔÚÍ­µÄ´ß»¯×÷ÓÃÏ£¬°±ÆøºÍ·úÆø·´Ó¦Éú³ÉAºÍB¡£AΪï§ÑΣ¬BÔÚ±ê×¼×´¿öÏÂΪÆø̬¡£ÔÚ´Ë·´Ó¦ÖУ¬Èôÿ·´Ó¦1Ìå»ý°±Æø£¬Í¬Ê±·´Ó¦0.75Ìå»ý·úÆø£»Èôÿ·´Ó¦8.96L°±Æø£¨±ê×¼×´¿ö£©£¬Í¬Ê±Éú³É0.3molA¡£

¢Ùд³ö°±ÆøºÍ·úÆø·´Ó¦µÄ»¯Ñ§·½³Ìʽ___________________________£»

¢ÚÔÚ±ê×¼×´¿öÏ£¬Ã¿Éú³É1mol B£¬×ªÒƵç×ÓµÄÎïÖʵÄÁ¿Îª_________mol¡£

£¨3£©I  ÔÚijѹǿºã¶¨µÄÃܱÕÈÝÆ÷ÖмÓÈë2mol N2ºÍ4 mol H2£¬·¢ÉúÈçÏ·´Ó¦£º

N2£¨g£©+3H2 (g) NH3(g)

´ïµ½Æ½ºâʱ£¬Ìå»ýΪ·´Ó¦Ç°µÄÈý·ÖÖ®¶þ¡£Çó£º

¢Ù´ïµ½Æ½ºâʱ£¬N2µÄת»¯ÂÊΪ_________¡£

¢ÚÈôÏò¸ÃÈÝÆ÷ÖмÓÈëamol N2¡¢b mol H2¡¢c mol NH3£¬ÇÒa¡¢b¡¢c¾ù>0£¬ÔÚÏàͬÌõ¼þÏ´ﵽƽºâʱ£¬»ìºÏÎïÖи÷×é·ÖµÄÎïÖʵÄÁ¿ÓëÉÏÊöƽºâÏàͬ¡£ÊԱȽϷ´Ó¦·Å³öµÄÄÜÁ¿£º

¢Ù_________¢Ú£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£

¢ò  Èô½«2 mol N2ºÍ4 mol H2·ÅÈëÆðʼÌå»ýÏàͬµÄºãÈÝÈÝÆ÷ÖУ¬ÔÚÓëIÏàͬµÄζÈÏ´ﵽƽºâ¡£

¢ÛÊԱȽÏƽºâʱNH3µÄŨ¶ÈI_________¢ò  £¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

¢ñ£®ÔÚijѹǿºã¶¨µÄÃܱÕÈÝÆ÷ÖмÓÈë2mol N2ºÍ4mol H2£¬·¢ÉúÈçÏ·´Ó¦£º

N2£¨g£©+3H2£¨g£©2NH3£¨g£©  ¡÷H=£­92£®4 kJ£®mol-1´ïµ½Æ½ºâʱ£¬Ìå»ýΪ·´Ó¦Ç°µÄ2/3¡£Çó£º

¢Ù´ïµ½Æ½ºâʱ£¬N2µÄת»¯ÂÊΪ                  ¡£

¢ÚÈôÏò¸ÃÈÝÆ÷ÖмÓÈëa mol N2¡¢b mol H2¡¢c mol NH3£¬ÇÒa¡¢b¡¢c¾ù´óÓÚ0£¬ÔÚÏàͬÌõ¼þÏ´ﵽƽºâʱ£¬»ìºÏÎïÖи÷×é·ÖµÄÎïÖʵÄÁ¿ÓëÉÏÊöƽºâÏàͬ¡£ÊԱȽϷ´Ó¦·Å³öµÄÄÜÁ¿£º

¢Ù       ¢Ú£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©

¢ò£®Èô½«2mol N2ºÍ4mol H2·ÅÈëÆðʼÌå»ýÏàͬµÄºãÈÝÈÝÆ÷ÖУ¬ÔÚÓë¢ñÏàͬµÄζÈÏ´ﵽƽºâ¡£

¢ÛÊԱȽÏƽºâʱNH3µÄŨ¶È£º¢ñ        ¢ò£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸