25 ¡æÊ±£¬ºÏ³É°±·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£º
N2 (g) +3H2 (g) 2NH3(g) ¦¤H£½£92.4 kJ/mol
ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ
A£®»ìºÏÆøÌåµÄÖÊÁ¿²»±äʱ£¬ËµÃ÷¸Ã·´Ó¦Ò»¶¨´ïµ½Æ½ºâ״̬
B£®½«ÈÝÆ÷µÄÌå»ýÀ©´óµ½ÔÀ´µÄ¶þ±¶£¬¦Ô(Õý)¼õС£¬¦Ô(Äæ)Ôö´ó£¬Æ½ºâÏò×óÒÆ¶¯
C£®´ß»¯¼Á¼ÈÄÜÌá¸ßN2µÄת»¯ÂÊ£¬ÓÖÄÜËõ¶Ì´ïµ½Æ½ºâËùÐèʱ¼ä£¬Ìá¸ßÉú²úÐ§Òæ
D£®ÔÚÃܱÕÈÝÆ÷ÖзÅÈë1 mol N2ºÍ3 mol H2½øÐз´Ó¦£¬²âµÃ·´Ó¦·Å³öµÄÈÈÁ¿Ð¡ÓÚ92.4 kJ
D?
½âÎö:A£®²»¹ÜÊÇ·´Ó¦ÎﻹÊÇÉú³ÉÎ¶¼ÊÇÆøÌ壬ËùÒÔ˵£¬ÎÞÂÛÊÇ·ñ´ïµ½Æ½ºâ£¬»ìºÏÆøÌåµÄÖÊÁ¿¶¼²»±ä¡£
B£®½«ÈÝÆ÷µÄÌå»ýÀ©´óµ½ÔÀ´µÄ¶þ±¶£¬·´Ó¦Îï¡¢Éú³ÉÎïŨ¶È¾ù¼õС£¬Òò´Ë¦Ô(Õý)¡¢¦Ô(Äæ)¾ù½ÏС£¬Æ½ºâÏò×óÒÆ¶¯
C£®´ß»¯¼Á¼È²»ÄÜÌá¸ßN2µÄת»¯ÂÊ,µ«ÄÜËõ¶Ì´ïµ½Æ½ºâËùÐèʱ¼ä£¬Ìá¸ßÉú²úÐ§Òæ
D£®ÓÉÓÚ´Ë·´Ó¦Ê±¿ÉÄæ·´Ó¦£¬ÔÚÃܱÕÈÝÆ÷ÖзÅÈë1 mol N2ºÍ3 mol H2½øÐз´Ó¦£¬²âµÃ·´Ó¦·Å³öµÄÈÈÁ¿Ð¡ÓÚ92.4 kJ
| Äê¼¶ | ¸ßÖÐ¿Î³Ì | Äê¼¶ | ³õÖÐ¿Î³Ì |
| ¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
| ¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍÆ¼ö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
| 1 |
| 3 |
| 10 | 20 | 30 | 60 | |
| 300 | 52.0 | 64.2 | 71.0 | 84.2 |
| 400 | 25.1 | 38.2 | 47.0 | 65.2 |
| 500 | 10.6 | 19.1 | 26.4 | 42.0 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013-2014ѧÄêÉÂÎ÷Ê¡µÈÎåУ¸ßÈýµÚ¶þ´ÎÁª¿¼Àí×Û»¯Ñ§ÊÔ¾í£¨½âÎö°æ£© ÌâÐÍ£ºÌî¿ÕÌâ
£¨1£©ÒÑÖª£ºO2 (g)= O2£« (g)+e£ ¡÷H1= +1175.7 kJ¡¤mol£1
PtF6(g)+ e£= PtF6£(g)???? ¡÷H2= - 771.1 kJ¡¤mol£1
O2+PtF6£(s)=O2+(g)+PtF6£ (g)?? ¡÷H3=+482.2 kJ¡¤mol£1
Ôò·´Ó¦£ºO2£¨g£©+ PtF6 (g) = O2+PtF6(s)µÄ¡÷H=_____ kJ¡¤mol-1¡£
ÈçͼΪºÏ³É°±·´Ó¦ÔÚʹÓÃÏàͬµÄ´ß»¯¼Á£¬²»Í¬Î¶ȺÍѹǿÌõ¼þϽøÐз´ Ó¦£¬³õʼʱN2ºÍH2µÄÌå»ý±ÈΪ1:3ʱµÄƽºâ»ìºÏÎïÖа±µÄÌå»ý·ÖÊý£º
![]()
¢Ù ÔÚÒ»¶¨µÄζÈÏ£¬ÏòÌå»ý²»±äµÄÃܱÕÈÝÆ÷ÖгäÈëµªÆøºÍÇâÆø·¢ÉúÉÏÊö·´Ó¦£¬ÏÂÁÐÄÜ˵Ã÷·´Ó¦´ïµ½Æ½ºâ״̬µÄÊÇ??????? ¡£
a£®ÌåϵµÄѹǿ±£³Ö²»±ä?? ???? b£®»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
c£®N2ºÍH2µÄÌå»ý±ÈΪ1:3????? d£®»ìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿²»±ä
¢Ú·Ö±ðÓÃvA£¨NH3£©ºÍvB£¨NH3£©±íʾ´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ״̬A¡¢BʱµÄ·´Ó¦ËÙÂÊ£¬ÔòvA£¨NH3£©??? vB£¨NH3£©£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©£¬¸Ã·´Ó¦µÄµÄƽºâ³£ÊýkA ??? kB£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©£¬ÔÚ250 ¡æ¡¢1.0¡Á104kPaÏ´ﵽƽºâ£¬H2µÄת»¯ÂÊΪ????? %£¨¼ÆËã½á¹û±£ÁôСÊýµãºóһ룩£»
£¨3£©25¡æÊ±£¬½«a mol NH4NO3ÈÜÓÚË®£¬ÈÜÒº³ÊËáÐÔ£¬ÔÒò???????????????????????? ?????????? £¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©¡£Ïò¸ÃÈÜÒºÖмÓÈëbL°±Ë®ºóÈÜÒº³ÊÖÐÐÔ£¬ÔòËù¼Ó°±Ë®µÄŨ¶ÈΪ?????????? mol/L£¨Óú¬a¡¢bµÄ´úÊýʽ±íʾ£¬NH3¡¤H2OµÄµçÀëÆ½ºâ³£ÊýΪKb=2¡Á10-5£©
£¨4£©ÈçͼËùʾ£¬×°ÖâñΪ¼×ÍéȼÁÏµç³Ø£¨µç½âÖÊÈÜҺΪKOHÈÜÒº£©£¬Í¨¹ý×°ÖâòʵÏÖÌú°ôÉ϶ÆÍ¡£µç¶ÆÒ»¶Îʱ¼äºó£¬×°ÖâñÖÐÈÜÒºµÄpH ???? £¨Ìî¡°±ä´ó¡±¡¢¡°±äС¡±»ò¡°²»±ä¡±£©£¬a¼«µç¼«·´Ó¦·½³ÌʽΪ????????????????? £»Èôµç¶Æ½áÊøºó£¬·¢ÏÖ×°ÖâòÖÐÒõ¼«ÖÊÁ¿±ä»¯ÁË25.6g£¨ÈÜÒºÖÐÁòËáÍÓÐÊ£Óࣩ£¬Ôò×°ÖâñÖÐÀíÂÛÉÏÏûºÄ¼×Íé????? L£¨±ê×¼×´¿öÏ£©¡£
![]()
²é¿´´ð°¸ºÍ½âÎö>>
¹ú¼ÊѧУÓÅÑ¡ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com