ÌâÄ¿ÄÚÈÝ

15£®£¨1£©ÖкÍÈȵIJⶨʵÑéµÄ¹Ø¼üÊÇÒª±È½Ï׼ȷµØÅäÖÆÒ»¶¨µÄÎïÖʵÄÁ¿Å¨¶ÈµÄÈÜÒº£¬Á¿ÈÈÆ÷Òª¾¡Á¿×öµ½¾øÈÈ£»ÔÚÁ¿ÈȵĹý³ÌÖÐÒª¾¡Á¿±ÜÃâÈÈÁ¿µÄɢʧ£¬ÒªÇó±È½Ï׼ȷµØ²âÁ¿³ö·´Ó¦Ç°ºóÈÜҺζȵı仯£®»Ø´ðÏÂÁÐÎÊÌ⣺
¢ÙÖÐѧ»¯Ñ§ÊµÑéÖеÄÖкÍÈȵIJⶨËùÐèµÄ²£Á§ÒÇÆ÷ÓУº´óÉÕ±­¡¢Ð¡ÉÕ±­¡¢Á¿Í²¡¢Î¶ȼơ¢»·Ðβ£Á§½Á°è°ô£¬ÔÚ´óСÉÕ±­Ö®¼äÌîÂúËéÅÝÄ­£¨»òÖ½Ìõ£©Æä×÷ÓÃÊDZ£Î¡¢¸ôÈÈ¡¢¼õÉÙʵÑé¹ý³ÌÖÐÈÈÁ¿µÄËðʧ£®
¢Ú¸ÃʵÑé¿ÉÓÃ0.50mol•L-1HClºÍ0.55mol•L-1µÄNaOHÈÜÒº¸÷50mL£®NaOHµÄŨ¶È´óÓÚHClµÄŨ¶È×÷ÓÃÊDZ£Ö¤ÑÎËáÍêÈ«±»Öкͣ®µ±ÊÒεÍÓÚ10¡æÊ±½øÐУ¬¶ÔʵÑé½á¹û»áÔì³É½Ï´óµÄÎó²îÆäÔ­ÒòÊÇÉ¢ÈÈÌ«¿ì£®
£¨2£©Ô̲ØÔÚº£µ×µÄ¡°¿Éȼ±ù¡±ÊǸßѹÏÂÐγɵÄÍâ¹ÛÏó±ùµÄ¼×ÍéË®ºÏÎï¹ÌÌ壮¼×Í鯸ÌåȼÉÕºÍË®Æû»¯µÄÈÈ»¯Ñ§·½³Ìʽ·Ö±ðΪ£ºCH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O£¨g£©£»¡÷H=-802.3kJ/mol£¬
H2O£¨l£©=H2O£¨g£©£»¡÷H=+44kJ/mol£»
Ôò¼×ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪCH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-890.3kJ•mol-1£®

·ÖÎö £¨1£©¢Ù¸ù¾ÝÁ¿ÈȼƵĹ¹ÔìºÍʵÑé²½ÖèÀ´È·¶¨ÊµÑéµÄÒÇÆ÷£»ÖкÍÈȲⶨʵÑé³É°ÜµÄ¹Ø¼üÊDZ£Î¹¤×÷£»
¢ÚΪÁ˱£Ö¤HClÍêÈ«·´Ó¦£¬¼îÓ¦¹ýÁ¿£»Î¶ȹýµÍ£¬É¢Èȼӿ죬¶ÔʵÑé½á¹ýÔì³ÉµÄÎó²î½Ï´ó£»
£¨2£©È¼ÉÕÈÈÖ¸1mol¿ÉȼÎïÍêȫȼÉÕÉú³ÉÎȶ¨µÄ»¯ºÏÎïʱËù·Å³öµÄÈÈÁ¿£¬·´Ó¦ÎïÖÐC¡úCO2£¨Æø£©£¬H¡úH2O£¨Òº£©£¬S¡úSO2£¨Æø£©µÈ£¬ÀûÓøÇ˹¶¨ÂÉÀ´½â´ð£»

½â´ð ½â£º£¨1£©¢ÙÖÐѧ»¯Ñ§ÊµÑéÖеÄÖкÍÈȵIJⶨ£¬ÐèÒªÁ¿Í²Á¿È¡Ëá¼îµÄÌå»ý¡¢ÐèÒª»·Ðβ£Á§½Á°è°ô½Á°èÈÜҺʹ֮»ìºÏ¾ùÔÈ£¬Î¶ȼƲâÁ¿·´Ó¦Ç°Óë·´Ó¦ºóµÄ×î¸ßζȣ»´óСÉÕ±­Ö®¼äÌîÂúËéÅÝÄ­ËÜÁϵÄ×÷ÓÃÊÇ£º±£Î¡¢¸ôÈÈ¡¢¼õÉÙʵÑé¹ý³ÌÖеÄÈÈÁ¿É¢Ê§£¬
¹Ê´ð°¸Îª£ºÎ¶ȼơ¢»·Ðβ£Á§½Á°è°ô£»±£Î¡¢¸ôÈÈ¡¢¼õÉÙʵÑé¹ý³ÌÖÐÈÈÁ¿µÄËðʧ£»
¢ÚNaOHµÄŨ¶È´óÓÚHClµÄŨ¶È£¬Ê¹ÑÎËáÍêÈ«±»Öкͣ»µ±ÊÒεÍÓÚ10¡æÊ±½øÐУ¬É¢Èȼӿ죬¶ÔʵÑé½á¹ýÔì³ÉµÄÎó²î½Ï´ó£»
¹Ê´ð°¸Îª£º±£Ö¤ÑÎËáÍêÈ«±»Öкͣ»É¢ÈÈÌ«¿ì£»
£¨2£©¢ÙCH4£¨g£©+2O2£¨g£©=CO2£¨g£©+2H2O£¨g£©¡÷H=-802.3kJ/mol£¬
    ¢ÚH2O£¨l£©=H2O£¨g£©¡÷H=+44kJ/mol£»
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬ÓÉ¢Ù-¢Ú¡Á2µÃ£¬CH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-890.3 kJ•mol-1£»
¹Ê´ð°¸Îª£ºCH4£¨g£©+2O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-890.3 kJ•mol-1£»

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éѧÉúÓйØÖкÍÈȵIJⶨ¡¢ÈÈ»¯Ñ§·½³ÌʽµÄÊéд¡¢¸Ç˹¶¨ÂɵÄÔËÓõÈÎÊÌ⣬²àÖØÖкÍÈȵIJⶨ¼°¸Ç˹¶¨ÂÉÓ¦ÓõĿ¼²é£¬ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø