ÌâÄ¿ÄÚÈÝ

£¨1£©ÊµÑé²âµÃ£¬5g¼×´¼£¨CH3OH£©ÔÚÑõÆøÖгä·ÖȼÉÕÉú³É¶þÑõ»¯Ì¼ÆøÌåºÍҺ̬ˮʱÊͷųö113.5kJµÄÈÈÁ¿£¬ÊÔд³ö¼×´¼È¼ÉÕµÄÈÈ»¯Ñ§·½³Ìʽ
 
£®
£¨2£©ÒÑÖªH2£¨g£©¡¢C2H4£¨g£©ºÍC2H5OH£¨1£©µÄȼÉÕÈÈ¡÷H·Ö±ðÊÇ-285.8kJ?mlo-1¡¢-1411.0kJ?mlo-1ºÍ-1366.8kJ£¬ÔòÓÉC2H4£¨g£©ºÍH2O£¨I£©·´Ó¦Éú³ÉC2H5OH£¨I£©µÄÈÈ»¯Ñ§·½³Ìʽ
 
£®
¿¼µã£ºÓйط´Ó¦ÈȵļÆËã,ÈÈ»¯Ñ§·½³Ìʽ
רÌ⣺»¯Ñ§·´Ó¦ÖеÄÄÜÁ¿±ä»¯
·ÖÎö£º£¨1£©¸ù¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨¿ÉÖª£¬»¯Ñ§¼ÆÁ¿ÊýÓë·´Ó¦ÈȳÉÕý±È£¬²¢×¢Òâ±êÃ÷ÎïÖʵľۼ¯×´Ì¬£¬¸ù¾ÝȼÉÕÈȵĸÅÄî½áºÏÈÈ»¯Ñ§·½³ÌʽµÄÊéдÀ´»Ø´ð£»
£¨2£©ÓÉȼÉÕÈȵĸÅÄîд³ö¸÷·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£¬ÀûÓøÇ˹¶¨ÂɼÆË㣮
½â´ð£º ½â£º£¨1£©5gCH3OHÔÚÑõÆøÖÐȼÉÕÉú³ÉCO2ºÍҺ̬ˮ£¬·Å³ö113.5kJÈÈÁ¿£¬64g¼´2molCH3OHÔÚÑõÆøÖÐȼÉÕÉú³ÉCO2ºÍҺ̬ˮ£¬·Å³ö1452.8kJÈÈÁ¿£¬
ÔòȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCH3OH£¨g£©+
3
2
O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-726.4KJ/mol£¬
¹Ê´ð°¸Îª£ºCH3OH£¨g£©+
3
2
O2£¨g£©¨TCO2£¨g£©+2H2O£¨l£©¡÷H=-726.4KJ/mol£»
£¨3£©ÒÑÖªH2£¨g£©¡¢C2H4£¨g£©ºÍC2H5OH£¨l£©µÄȼÉÕÈÈ·Ö±ðÊÇ-285.8kJ/mol¡¢-1411.0kJ/molºÍ-1366.8kJ/mol£¬
ÔòÓУº¢ÙH2£¨g£©+
1
2
O2£¨g£©=H2O£¨l£©¡÷H=-285.8kJ/mol£»
      ¢ÚC2H4£¨g£©+2O2£¨g£©=2H2O£¨l£©+2CO2£¨g£©¡÷H=-1411.0kJ/mol£»
    ¢ÛC2H5OH£¨l£©+2O2£¨g£©=3H2O£¨l£©+2CO2 £¨g£©¡÷H=-1366.8kJ/mol£»
¸ù¾Ý¸Ç˹¶¨ÂÉ ¢Ú-¢Û¿ÉµÃ£ºC2H4£¨g£©+H2O£¨l£©=C2H5OH£¨l£©¡÷H=-44.2kJ/mol£»
¹Ê´ð°¸Îª£ºC2H4£¨g£©+H2O£¨l£©=C2H5OH£¨l£©¡÷H=-44.2kJ/mol£»
µãÆÀ£º±¾Ì⿼²é·´Ó¦ÈȵļÆË㣬ÈÈ»¯Ñ§·½³ÌʽÊéд£¬È¼ÉÕÈȸÅÄîµÄ¼ÆËãÓ¦Óã¬ÌâÄ¿ÄѶÈÖеȣ¬×¢Òâ¸Ç˹¶¨ÂÉÓ¦ÓÃÓÚ·´Ó¦ÈȵļÆË㣮
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø