C(s)+H2O(g)
CO(g)+H2(g)£»DH£¾0
ÒÑ֪ƽºâʱ£¬COΪ0.12mol£®ÊÔÌî¿Õ£º
(1)ÈôÓÃv(H2O)±íʾ¸Ã·´Ó¦Ç°20sÄÚÆ½¾ùËÙÂÊ£¬Ôòv(H2O)=________£®
(2)ÈôÔÚÉÏÊöƽºâ»ìºÏÎïÖмÓÈëÉÙÁ¿Na2O¹ÌÌ壬²¢ÔÚ´ËζÈÏÂÔٴδﵽÐÂµÄÆ½ºâ(µÚ¶þƽºâ)£¬ÔòƽºâʱH2µÄÎïÖʵÄÁ¿________(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)£®ÀíÓÉÊÇ________£®
(3)ÈôÏòÉÏÊöµÚһƽºâµÄ»ìºÏÎïÖÐÔÙ³äÈëamolH2(a£¼0.12)ÔÚÏàͬÌõ¼þÏ´ﵽÐÂµÄÆ½ºâ(µÚÈýƽºâ)£¬´ËʱCOµÄÎïÖʵÄÁ¿nµÄ·¶Î§ÊÇ________£®
| (1)0.0030mol¡¤L-1¡¤s-1
(2)¼õС,Na2OÓëH2O·´Ó¦£¬Ê¹Ë®ÕôÆøÅ¨¶È½µµÍ£¬Æ½ºâÏò×óÒÆ¶¯ (3)(0.12-a)£¼n£¼0.12 ½âÎö£º(1)20sÄÚÓÃCO±íʾ¸Ã·´Ó¦µÄƽ¾ùËÙÂÊv(CO)=(0.12mol¡Â2L)/20 s=0.003mol¡¤L-1¡¤s-1£® ¸ù¾Ý£º¸ù¾Ý»¯Ñ§·´Ó¦ËÙÂÊÖ®±ÈµÈÓÚ»¯Ñ§¼ÆÁ¿ÊýÖ®±È£¬ÓÃH2O±íʾ¸Ã·´Ó¦Ç°20sÄ򵀮½¾ùËÙÂÊv(H2O)=0.003mol¡¤L-1¡¤s-1£® (2)ÓÉÓÚNa2OÓëH2O·´Ó¦£¬Ê¹Ë®ÕôÆøÅ¨¶È½µµÍ£¬Æ½ºâÏò×óÒÆ¶¯£¬Òò´ËµÚ¶þ´Îƽºâʱ£¬H2µÄŨ¶È¼õС£® (3)µÚÒ»´ÎƽºâʱCO¡¢H2ÎïÖʵÄÁ¿¸÷Ϊ0.12mol£¬¼ÓÈëamolH2£¬Æ½ºâÏò×óÒÆ¶¯£¬COµÄÎïÖʵÄÁ¿¼õС£¬´ïµ½Æ½ºâ(µÚÈýƽºâ)ʱ£¬COµÄÎïÖʵÄÁ¿Ð¡ÓÚ0.12mol£®¸ù¾ÝÀÕÏÄÌØÁÐÔÀí£¬¼ÓÈëamolH2ºó»¯Ñ§Æ½ºâÏò¼õÈõÕâÖָıäµÄ·½ÏòÒÆ¶¯(²»ÄܵÖÏûÕâÖָıä)£¬Òò´Ë±ä»¯µÄH2µÄÎïÖʵÄÁ¿Ð¡ÓÚamol£¬µÚÈýƽºâʱCOµÄÎïÖʵÄÁ¿´óÓÚ(0.12-a)mol£®
|
| 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£ºÔĶÁÀí½â
| ||
| ¸ßΠ|
| ||
| ¸ßΠ|
| 4 |
| 7 |
| ||
| ¸ßΠ|
| T/¡æ | 200 | 300 | 400 |
| K | K1 | K2 | 0.5 |
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
£¨1£©ÒÑÖª£ºN2(g)+O2(g)=2NO(g)£»¡÷H= +180.5kJ/mol
4NH3(g)+5O2(g)=4NO(g)+6H2O(g)£»¡÷H=£905kJ/mol
2H2(g)+O2(g)=2H2O(g)£»¡÷H=£483.6kJ/mol
ÔòN2(g)+3H2(g)=2NH3(g)µÄ¡÷H= ¡£
£¨2£©¹¤ÒµºÏ³É°±µÄ·´Ó¦ÎªN2(g)+3H2(g)
2NH3(g)¡£ÔÚÒ»¶¨Î¶ÈÏ£¬½«Ò»¶¨Á¿µÄN2ºÍH2ͨÈë¹Ì¶¨Ìå»ýΪ1LµÄÃܱÕÈÝÆ÷Öдﵽƽºâºó£¬¸Ä±äÏÂÁÐÌõ¼þ£¬ÄÜʹƽºâÏòÕý·´Ó¦·½ÏòÒÆ¶¯µÄÊÇ ¡£
¢ÙÔö´óѹǿ ¢ÚͨÈëHe
¢ÛʹÓô߻¯¼Á ¢Ü½µµÍζÈ
£¨3£©¹¤ÒµºÏ³É°±µÄ·´Ó¦ÎªN2(g)+3H2(g)
2NH3(g)¡£ÉèÔÚÈÝ»ýΪ2.0LµÄÃܱÕÈÝÆ÷ÖгäÈë0.60molN2(g)ºÍ1.60molH2(g)£¬·´Ó¦ÔÚÒ»¶¨Ìõ¼þÏ´ﵽƽºâʱ£¬NH3µÄÎïÖʵÄÁ¿·ÖÊý£¨NH3µÄÎïÖʵÄÁ¿Óë·´Ó¦ÌåϵÖÐ×ܵÄÎïÖʵÄÁ¿Ö®±È£©Îª
¡£¼ÆËã¸ÃÌõ¼þÏ´ﵽƽºâʱN2ת»¯ÂÊΪ £»
£¨4£©ºÏ³É°±µÄÔÁÏÇâÆøÊÇÒ»ÖÖÐÂÐ͵ÄÂÌÉ«ÄÜÔ´£¬¾ßÓйãÀ«µÄ·¢Õ¹Ç°¾°¡£ÏÖÓÃÇâÑõȼÁÏµç³Ø½øÐÐÓÒͼËùʾʵÑ飺£¨ÆäÖÐc¡¢d¾ùΪ̼°ô£¬NaClÈÜÒºµÄÌå»ýΪ500ml£©
![]()
¢Ùb¼«Îª ¼«£¬µç¼«·´Ó¦Ê½ £»
c¼«Îª ¼«£¬µç¼«·´Ó¦Ê½
¢ÚÓÒͼװÖÃÖУ¬µ±b¼«ÉÏÏûºÄµÄO2ÔÚ±ê×¼×´¿öϵÄÌå»ýΪ280mlʱ£¬ÔòÒÒ³ØÈÜÒºµÄPHΪ £¨¼ÙÉ跴ӦǰºóÈÜÒºÌå»ý²»±ä£¬ÇÒNaClÈÜÒº×ãÁ¿£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
£¨1£©ÒÑÖª£ºN2(g)+O2(g)=2NO(g)£»¡÷H= +180.5kJ/mol
4NH3(g)+5O2(g)=4NO(g)+6H2O(g)£»¡÷H=£905kJ/mol
2H2(g)+O2(g)=2H2O(g)£»¡÷H=£483.6kJ/mol
ÔòN2(g)+3H2(g)=2NH3(g)µÄ¡÷H= ¡£
£¨2£©¹¤ÒµºÏ³É°±µÄ·´Ó¦ÎªN2(g)+3H2(g) 2NH3(g)¡£ÔÚÒ»¶¨Î¶ÈÏ£¬½«Ò»¶¨Á¿µÄN2ºÍH2ͨÈë¹Ì¶¨Ìå»ýΪ1LµÄÃܱÕÈÝÆ÷Öдﵽƽºâºó£¬¸Ä±äÏÂÁÐÌõ¼þ£¬ÄÜʹƽºâÏòÕý·´Ó¦·½ÏòÒÆ¶¯µÄÊÇ ¡£
¢ÙÔö´óѹǿ ¢ÚͨÈëHe
¢ÛʹÓô߻¯¼Á ¢Ü½µµÍζÈ
£¨3£©¹¤ÒµºÏ³É°±µÄ·´Ó¦ÎªN2(g)+3H2(g) 2NH3(g)¡£ÉèÔÚÈÝ»ýΪ2.0LµÄÃܱÕÈÝÆ÷ÖгäÈë0.60molN2(g)ºÍ1.60molH2(g)£¬·´Ó¦ÔÚÒ»¶¨Ìõ¼þÏ´ﵽƽºâʱ£¬NH3µÄÎïÖʵÄÁ¿·ÖÊý£¨NH3µÄÎïÖʵÄÁ¿Óë·´Ó¦ÌåϵÖÐ×ܵÄÎïÖʵÄÁ¿Ö®±È£©Îª
¡£¼ÆËã¸ÃÌõ¼þÏ´ﵽƽºâʱN2ת»¯ÂÊΪ £»
£¨4£©ºÏ³É°±µÄÔÁÏÇâÆøÊÇÒ»ÖÖÐÂÐ͵ÄÂÌÉ«ÄÜÔ´£¬¾ßÓйãÀ«µÄ·¢Õ¹Ç°¾°¡£ÏÖÓÃÇâÑõȼÁÏµç³Ø½øÐÐÓÒͼËùʾʵÑ飺£¨ÆäÖÐc¡¢d¾ùΪ̼°ô£¬NaClÈÜÒºµÄÌå»ýΪ500ml£©
¢Ùb¼«Îª ¼«£¬µç¼«·´Ó¦Ê½ £»
c¼«Îª ¼«£¬µç¼«·´Ó¦Ê½
¢ÚÓÒͼװÖÃÖУ¬µ±b¼«ÉÏÏûºÄµÄO2ÔÚ±ê×¼×´¿öϵÄÌå»ýΪ280mlʱ£¬ÔòÒÒ³ØÈÜÒºµÄPHΪ £¨¼ÙÉ跴ӦǰºóÈÜÒºÌå»ý²»±ä£¬ÇÒNaClÈÜÒº×ãÁ¿£©
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011½ìÔÆÄÏÊ¡ÃÉ×Ը߼¶ÖÐѧ¸ßÈý1ÔÂÔ¿¼£¨Àí×Û£©»¯Ñ§²¿·Ö ÌâÐÍ£ºÌî¿ÕÌâ
£¨1£©ÒÑÖª£ºN2(g)+O2(g)=2NO(g)£»¡÷H= +180.5kJ/mol
4NH3(g)+5O2(g)=4NO(g)+6H2O(g)£»¡÷H=£905kJ/mol
2H2(g)+O2(g)=2H2O(g)£»¡÷H=£483.6kJ/mol
ÔòN2(g)+3H2(g)=2NH3(g)µÄ¡÷H= ¡£
£¨2£©¹¤ÒµºÏ³É°±µÄ·´Ó¦ÎªN2(g)+3H2(g)
2NH3(g)¡£ÔÚÒ»¶¨Î¶ÈÏ£¬½«Ò»¶¨Á¿µÄN2ºÍH2ͨÈë¹Ì¶¨Ìå»ýΪ1LµÄÃܱÕÈÝÆ÷Öдﵽƽºâºó£¬¸Ä±äÏÂÁÐÌõ¼þ£¬ÄÜʹƽºâÏòÕý·´Ó¦·½ÏòÒÆ¶¯µÄÊÇ ¡£
¢ÙÔö´óѹǿ ¢ÚͨÈëHe
¢ÛʹÓô߻¯¼Á ¢Ü½µµÍζÈ
£¨3£©¹¤ÒµºÏ³É°±µÄ·´Ó¦ÎªN2(g)+3H2(g)
2NH3(g)¡£ÉèÔÚÈÝ»ýΪ2.0LµÄÃܱÕÈÝÆ÷ÖгäÈë0.60molN2(g)ºÍ1.60molH2(g)£¬·´Ó¦ÔÚÒ»¶¨Ìõ¼þÏ´ﵽƽºâʱ£¬NH3µÄÎïÖʵÄÁ¿·ÖÊý£¨NH3µÄÎïÖʵÄÁ¿Óë·´Ó¦ÌåϵÖÐ×ܵÄÎïÖʵÄÁ¿Ö®±È£©Îª
¡£¼ÆËã¸ÃÌõ¼þÏ´ﵽƽºâʱN2ת»¯ÂÊΪ £»
£¨4£©ºÏ³É°±µÄÔÁÏÇâÆøÊÇÒ»ÖÖÐÂÐ͵ÄÂÌÉ«ÄÜÔ´£¬¾ßÓйãÀ«µÄ·¢Õ¹Ç°¾°¡£ÏÖÓÃÇâÑõȼÁÏµç³Ø½øÐÐÓÒͼËùʾʵÑ飺£¨ÆäÖÐc¡¢d¾ùΪ̼°ô£¬NaClÈÜÒºµÄÌå»ýΪ500ml£©![]()
¢Ùb¼«Îª ¼«£¬µç¼«·´Ó¦Ê½ £»
c¼«Îª ¼«£¬µç¼«·´Ó¦Ê½
¢ÚÓÒͼװÖÃÖУ¬µ±b¼«ÉÏÏûºÄµÄO2ÔÚ±ê×¼×´¿öϵÄÌå»ýΪ280mlʱ£¬ÔòÒÒ³ØÈÜÒºµÄPHΪ £¨¼ÙÉ跴ӦǰºóÈÜÒºÌå»ý²»±ä£¬ÇÒNaClÈÜÒº×ãÁ¿£©
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