ÌâÄ¿ÄÚÈÝ
ÏÂÁÐ˵·¨»ò±íʾ·½·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A¡¢µÈÎïÖʵÄÁ¿µÄÁòÕôÆøºÍÁò¹ÌÌå·Ö±ðÍêȫȼÉÕ£¬ºóÕ߷ųöÈÈÁ¿¶à | B¡¢ÓÉH+£¨aq£©+OH-£¨aq£©¨TH2O£¨l£©¡÷H=-57.3kJ?mol-1¿ÉÖª£¬Èô½«º¬1 mol CH3COOHµÄÏ¡ÈÜÒºÓ뺬1 mol NaOHµÄÏ¡ÈÜÒº»ìºÏ£¬·Å³öµÄÈÈÁ¿Ð¡ÓÚ57.3 kJ | C¡¢300¡æ¡¢30MPaÏ£¬½«0.5molN2£¨g£©ºÍ1.5mol H2£¨g£©ÖÃÓÚÃܱÕÈÝÆ÷Öгä·Ö·´Ó¦Éú³ÉNH3£¨g£©£¬·ÅÈÈ19.3kJ£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪ£ºN2£¨g£©+3H2£¨g£©=2NH3£¨g£©¡÷H=-38.6kJ?mol-1 | D¡¢ÓÉC£¨Ê¯Ä«£©¨TC£¨½ð¸Õʯ£©¡÷H=+1.90 kJ?mol-1¿ÉÖª£¬½ð¸Õʯ±ÈʯīÎȶ¨ |
·ÖÎö£ºA¡¢ÁòÕôÆø±ä»¯ÎªÁò¹ÌÌåΪ·ÅÈȹý³Ì£»
B¡¢CH3COOHΪÈõËᣬµçÀëʱÎüÈÈ£»
C¡¢ºÏ³É°±·´Ó¦Îª¿ÉÄæ·´Ó¦£»
D¡¢ÒÀ¾Ý·´Ó¦ÈÈÁ¿±ä»¯ÅжÏÎïÖÊÄÜÁ¿´óС£¬ÎïÖÊÄÜÁ¿Ô½¸ßÔ½Ô½»îÆã®
B¡¢CH3COOHΪÈõËᣬµçÀëʱÎüÈÈ£»
C¡¢ºÏ³É°±·´Ó¦Îª¿ÉÄæ·´Ó¦£»
D¡¢ÒÀ¾Ý·´Ó¦ÈÈÁ¿±ä»¯ÅжÏÎïÖÊÄÜÁ¿´óС£¬ÎïÖÊÄÜÁ¿Ô½¸ßÔ½Ô½»îÆã®
½â´ð£º½â£ºA¡¢ÁòÕôÆø±ä»¯ÎªÁò¹ÌÌåΪ·ÅÈȹý³Ì£¬ÔòµÈÁ¿µÄÁòÕôÆøºÍÁò¹ÌÌåÔÚÑõÆøÖзֱðÍêȫȼÉÕ£¬·Å³öÈÈÁ¿ÁòÕôÆø¶à£¬¹ÊA´íÎó£»
B¡¢´×ËáΪÈõµç½âÖÊ£¬µçÀëÎüÈÈ£¬ËùÒÔ½«º¬1 mol CH3COOHµÄÏ¡ÈÜÒºÓ뺬1 mol NaOHµÄÏ¡ÈÜÒº»ìºÏ£¬·Å³öµÄÈÈÁ¿Ð¡ÓÚ57.3kJ£¬¹ÊBÕýÈ·£»
C¡¢ÓÉÓںϳɰ±·´Ó¦Îª¿ÉÄæ·´Ó¦£¬·´Ó¦Îï²»ÄÜÈ«²¿×ª»¯ÎªÉú³ÉÎËùÒÔ0.5molN2£¨g£©ºÍ1.5mol H2£¨g£©ÍêÈ«·´Ó¦·Å³öµÄÈÈÁ¿´óÓÚ19.3kJ£¬¹ÊC´íÎó£»
D¡¢ÓÉC£¨Ê¯Ä«£©¡úC£¨½ð¸Õʯ£©£¬¡÷H=+19kJ/mol£¬·´Ó¦ÎüÈÈ£¬½ð¸ÕʯÄÜÁ¿¸ßÓÚʯī£¬¿É֪ʯī±È½ð¸ÕʯÎȶ¨£¬¹ÊD´íÎó£»
¹ÊÑ¡B£®
B¡¢´×ËáΪÈõµç½âÖÊ£¬µçÀëÎüÈÈ£¬ËùÒÔ½«º¬1 mol CH3COOHµÄÏ¡ÈÜÒºÓ뺬1 mol NaOHµÄÏ¡ÈÜÒº»ìºÏ£¬·Å³öµÄÈÈÁ¿Ð¡ÓÚ57.3kJ£¬¹ÊBÕýÈ·£»
C¡¢ÓÉÓںϳɰ±·´Ó¦Îª¿ÉÄæ·´Ó¦£¬·´Ó¦Îï²»ÄÜÈ«²¿×ª»¯ÎªÉú³ÉÎËùÒÔ0.5molN2£¨g£©ºÍ1.5mol H2£¨g£©ÍêÈ«·´Ó¦·Å³öµÄÈÈÁ¿´óÓÚ19.3kJ£¬¹ÊC´íÎó£»
D¡¢ÓÉC£¨Ê¯Ä«£©¡úC£¨½ð¸Õʯ£©£¬¡÷H=+19kJ/mol£¬·´Ó¦ÎüÈÈ£¬½ð¸ÕʯÄÜÁ¿¸ßÓÚʯī£¬¿É֪ʯī±È½ð¸ÕʯÎȶ¨£¬¹ÊD´íÎó£»
¹ÊÑ¡B£®
µãÆÀ£º±¾Ì⿼²éÁË·´Ó¦ÈÈÁ¿±ä»¯£¬ÈÈ»¯Ñ§·½³ÌʽµÄÕýÎóÅжϣ¬ÎïÖÊÄÜÁ¿ÓëÎȶ¨ÐԵıȽϷ½·¨£¬ÌâÄ¿ÄѶȲ»´ó£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿