ÌâÄ¿ÄÚÈÝ
ÔÚ»ð¼ýÍÆ½øÆ÷ÖÐ×°Óл¹Ô¼Á루N2H4£©ºÍÇ¿Ñõ»¯¼ÁH2O2£¬µ±ËüÃÇ»ìºÏʱ£¬¼´²úÉú´óÁ¿µÄµªÆøºÍË®ÕôÆø£¬²¢·Å³ö´óÁ¿µÄÈÈÁ¿¡£ÒÑÖª0.4molҺ̬ëºÍ×ãÁ¿Ë«ÑõË®·´Ó¦Éú³ÉµªÆøºÍË®ÕôÆøÊ±·Å³ö256.65KJµÄÈÈÁ¿¡£
(1).д³öëº͹ýÑõ»¯ÇâµÄ½á¹¹Ê½, ëÂ________________£¬¹ýÑõ»¯Çâ____________________¡£
(2).ÉÏÊö·´Ó¦Ó¦ÓÃÓÚ»ð¼ýÍÆ½ø¼Á£¬³ýÊͷųö´óÁ¿ÈÈÁ¿ºÍ¿ìËÙ²úÉú´óÁ¿ÆøÌåÍ⣬»¹ÓÐÒ»¸öºÜÍ»³öµÄÓŵãÊÇ__________________________________________¡£
(3).д³öëºÍË«ÑõË®·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ_____________________________¡£
(4).ÒÑÖªH2O(l)==H2O(g)£»¡÷H= +44KJ/mol£¬Ôò16gҺ̬ëÂÓë×ãÁ¿Ë«ÑõË®·´Ó¦Éú³ÉµªÆøºÍҺ̬ˮʱ£¬·Å³öµÄÈÈÁ¿ÊÇ___________¡£
(5).ÒÑÖªN2(g) + 2O2(g) == 2NO2(g)£»¡÷H= +67.7KJ/mol£¬
N2H4(g) + O2(g) == N2(g)+2H2O(g)£»¡÷H= -534KJ/mol£¬
ÔòëÂÓëNO2ÍêÈ«·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ_________________________________________¡£
¡¾´ð°¸¡¿¢Å
£¨2·Ö£©
¢Æ ²úÎïΪµªÆøºÍË®£¬ÎÞÎÛȾ £¨2·Ö£©
¢Ç N2H4(l) + 2H2O2(l) =N2(g)+4H2O(g)£»¡÷H= ¡ª641.6KJ/mol£¬
(3·Ö£¬×´Ì¬±ê´í¿Û1·Ö£¬¡÷H¼ÆËã´íÎó¿Û1·Ö£¬Á½Õß¾ù´í²»µÃ·Ö)
¢È 408.8KJ (2·Ö£¬Ã»µ¥Î»¿Û1·Ö)
¢É 2N2H4(g) + 2NO2(g) = 3N2(g)+4H2O(g)£»¡÷H= ¡ª1135.7KJ/mol£¬
(3·Ö£¬×´Ì¬±ê´í¿Û1·Ö£¬¡÷H¼ÆËã´íÎó¿Û1·Ö£¬Á½Õß¾ù´í²»µÃ·Ö)
¡¾½âÎö¡¿
£¨1£©![]()
£¨2£©ëÂÓëH2O2·´Ó¦µÄ²úÎïΪµªÆøºÍË®£¬ÎÞÎÛȾ¡£
£¨3£©0.4molҺ̬ëºÍ×ãÁ¿Ë«ÑõË®·´Ó¦Éú³ÉµªÆøºÍË®ÕôÆøÊ±·Å³ö256.65KJµÄÈÈÁ¿£¬Ôò1molҺ̬ëºÍ×ãÁ¿Ë«ÑõË®·´Ó¦Éú³ÉµªÆøºÍË®ÕôÆøÊ±·Å³ö641.6KJµÄÈÈÁ¿¡£
£¨4£©16gҺ̬ëÂÓë×ãÁ¿Ë«ÑõË®·´Ó¦Éú³ÉµªÆøºÍÆøÌ¬Ë®Ê±£¬·Å³ö320.8KJÈÈÁ¿£¬Éú³ÉҺ̬ˮʱ£¬¶à·Å³ö88 KJÈÈÁ¿£¬¼´×ܹ²·Å³ö408.8KJÈÈÁ¿¡£
¢É 2N2H4(g) + 2NO2(g) = 3N2(g)+4H2O(g)£»¡÷H= ¡ª1135.7KJ/mol