ÌâÄ¿ÄÚÈÝ
ÓйØÄÜÁ¿µÄÅжϻò±íʾ·½·¨ÕýÈ·µÄÊÇ £¨ £©
A£®ÓÉ£ºH++OH¡ª¡úH2O+57.3KJ£¬¿ÉÖª£ºº¬0.5mol H2SO4µÄŨÈÜÒºÓ뺬1molNaOHµÄÈÜÒº»ìºÏ£¬·ÅÈÈÁ¿´óÓÚ57.5KJ
B£®´ÓC£¨Ê¯Ä«£©¡úC (½ð¸Õʯ)-119KJ£¬¿ÉÖª£º½ð¸Õʯ±Èʯī¸üÎȶ¨
C£®µÈÖÊÁ¿µÄÁòÕôÆøºÍÁò¹ÌÌå·Ö±ðÍêȫȼÉÕ£¬ºóÕ߷ųöÈÈÁ¿¸ü¶à
D£®2gH2ÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö285.8KJÈÈÁ¿£¬ÔòÇâÆøÈ¼ÉÕµÄÈÈ»¯Ñ§·½³ÌʽΪ£º
2H2£¨g£©+O2(g) ¡ú2H2O(1)+285.8KJ
A
½âÎö:
±¾Ì⿼²éÈÈ»¯Ñ§·½³Ìʽ¡£Ê¯Ä«µÄÄÜÁ¿µÍÓÚ½ð¸Õʯ£¬ËµÃ÷ʯī±È½ð¸ÕʯÎȶ¨£¬BÑ¡Ïî´íÎó£»ÁòÕôÆø×ª»¯ÎªÁò¹ÌÌåÊÇ·ÅÈȹý³Ì£¬ËùÒÔǰÕß·ÅÈȸü¶à£¬CÑ¡Ïî´íÎó£»2gH2ÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö285.8KJÈÈÁ¿£¬Ôò4gH2ÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö571.6KJÈÈÁ¿£¬ÕýÈ·µÄÈÈ»¯Ñ§·½³ÌʽΪ£º2H2£¨g£©+O2(g) ¡ú2H2O(1)+ 571.6KJ£¬DÑ¡Ïî´íÎó¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿