Question

11£®Ð´³öÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ
£¨1£©¼ÒÓÃÒº»¯ÆøÖÐÖ÷Òª³É·ÖÖ®Ò»ÊǶ¡Í飨C₄H₁₀£©£®1g¶¡ÍéÍêȫȼÉÕÉú³ÉCO₂ºÍҺ̬ˮʱ£¬·Å³öÈÈÁ¿50kJ£®ÊÔд³ö±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽC₄H₁₀£¨g£©+$\frac{13}{2}$O₂£¨g£©¨T4CO₂£¨g£©+5H₂O¡÷H=-2900kJ/mol
£¨2£©0.2molµÄÆø̬¸ßÄÜȼÁÏÒÒÅðÍ飨B₂H₆£©ÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö433kJÈÈÁ¿£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪB₂H₆£¨g£©+3O₂£¨g£©¨TB₂O₃£¨s£©+3H₂O£¨l£©¡÷H=-2165.0kJ/mol£»£»
£¨3£©ÓÃCO₂ºÍÇâÆøºÏ³ÉCH₃OCH₃£¨¼×ÃÑ£©Êǽâ¾öÄÜԴΣ»úµÄÑо¿·½ÏòÖ®Ò»£®
ÒÑÖª£ºCO£¨g£©+2H₂£¨g£©¨TCH₃OH£¨g£©¡÷H=-90.7kJ•mol^-1
2CH₃OH£¨g£©¨TCH₃OCH₃£¨g£©+H₂O£¨g£©¡÷H=-23.5kJ•mol^-1
CO£¨g£©+H₂O£¨g£©¨TCO₂£¨g£©+H₂£¨g£©¡÷H=-41.2kJ•mol^-1
C£¨s£©+$\frac{1}{2}$O₂£¨g£©=CO£¨g£©¡÷H=-110.5kJ/mol
ÔòCO₂ºÍÇâÆøºÏ³ÉCH₃OCH₃£¨g£©µÄÈÈ»¯Ñ§·½³ÌʽΪ2CO₂£¨g£©+6H₂£¨g£©=CH₃OCH₃£¨g£©+3H₂O£¨g£©¡÷H=-122.5 kJ•mol^-1£®

Answer 1

·ÖÎö £¨1£©ÒÀ¾ÝȼÉÕÈȵĸÅÄîºÍÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨Ð´³ö£¬×¢ÒâÎïÖʵľۼ¯×´Ì¬£¬ìʱäµÄ¼ÆË㣻
£¨2£©¸ù¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨¿ÉÖª£¬ÎïÖʵÄÎïÖʵÄÁ¿Óë·´Ó¦·Å³öµÄÈÈÁ¿³ÉÕý±È£¬0.2molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö433.0kJµÄÈÈÁ¿£¬Ôò1molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö433.0¡Á5=2165.0kJµÄÈÈÁ¿£¬×¢Òâ±êÃ÷¸÷ÎïÖʵľۼ¯×´Ì¬À´½â´ð£»
£¨3£©Ê×ÏÈд³ö·´Ó¦µÄ»¯Ñ§·½³Ìʽ£¬¼´2CO₂£¨g£©+6H₂£¨g£©=CH₃OCH₃£¨g£©+3H₂O£¨g£©£¬È»ºóÀûÓøÇ˹¶¨ÂɽâÌ⣮

½â´ð ½â£º£¨1£©1g¶¡ÍéÎïÖʵÄÁ¿Îª$\frac{1}{58}$mol£¬ÍêȫȼÉÕ²¢Éú³É¶þÑõ»¯Ì¼ºÍҺ̬ˮʱ£¬·Å³öÈÈÁ¿Îª50kJ£¬ËùÒÔ1mol¶¡ÍéÍêȫȼÉÕÉú³ÉÎȶ¨Ñõ»¯Îïʱ·Å³öµÄÈÈÁ¿Îª50kJ¡Á58=2900kJ£¬È¼ÉÕÈÈÊÇ1mol¿ÉȼÎïÍêȫȼÉÕÉú³ÉÎȶ¨Ñõ»¯Îïʱ·Å³öµÄÈÈÁ¿£¬±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ£ºC₄H₁₀£¨g£©+$\frac{13}{2}$O₂£¨g£©¨T4CO₂£¨g£©+5H₂O¡÷H=-2900kJ/mol£»
¹Ê´ð°¸Îª£ºC₄H₁₀£¨g£©+$\frac{13}{2}$O₂£¨g£©¨T4CO₂£¨g£©+5H₂O¡÷H=-2900kJ/mol£»
£¨2£©0.2molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö433.0kJµÄÈÈÁ¿£¬Ôò1molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö433.0¡Á5=2165.0kJµÄÈÈÁ¿£¬ÈÈ»¯Ñ§·½³ÌʽÖзÅÈÈ¡÷H£¼0£¬ËùÒÔ¡÷H=-2165.0kJ/mol£¬ËùÒԸ÷´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪB₂H₆£¨g£©+3O₂£¨g£©¨TB₂O₃£¨s£©+3H₂O£¨l£©¡÷H=-2165.0kJ/mol£¬
¹Ê´ð°¸Îª£ºB₂H₆£¨g£©+3O₂£¨g£©¨TB₂O₃£¨s£©+3H₂O£¨l£©¡÷H=-2165.0kJ/mol£»
£¨3£©ÀûÓøÇ˹¶¨ÂɽøÐмÆË㣬½«Èý¸ö·½³Ìʽ½øÐÐÐÎʽ±ä»»£¬CO£¨g£©+2H₂£¨g£©¨TCH₃OH£¨g£©¡÷H=-90.7kJ•mol^-1
³ËÒÔ2µÃ£º2CO£¨g£©+4H₂£¨g£©=2CH₃OH£¨g£©¡÷H=-181.4 kJ•mol^-1£»
CO£¨g£©+H₂O£¨g£©¨TCO₂£¨g£©+H₂£¨g£©¡÷H=-41.2kJ•mol^-1£»³ËÒÔ-2µÃ£º2CO₂£¨g£©+2H₂£¨g£©=2CO£¨g£©+2H₂O£¨g£©¡÷H=+82.4 kJ•mol^-1£¬
¢Ù2CO£¨g£©+4H₂£¨g£©=2CH₃OH£¨g£©¡÷H=-181.4 kJ•mol^-1
¢Ú2CH₃OH£¨g=CH₃OCH₃£¨g£©+H₂O£¨g£©¡÷H=-23.5 kJ•mol^-1
¢Û2CO₂£¨g£©+2H₂£¨g£©=2CO£¨g£©+2H₂O£¨g£©¡÷H=+82.4 kJ•mol^-1
¢Ù¢Ú¢ÛÈýʽÏà¼ÓµÃ£º2CO₂£¨g£©+6H₂£¨g£©=CH₃OCH₃£¨g£©+3H₂O£¨g£©¡÷H=-122.5 kJ•mol^-1
¹Ê´ð°¸£º2CO₂£¨g£©+6H₂£¨g£©=CH₃OCH₃£¨g£©+3H₂O£¨g£©¡÷H=-122.5 kJ•mol^-1£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éÁËÈÈ»¯Ñ§·½³ÌʽµÄÊéд£¬ÐèҪעÒâµÄÓУºÎïÖʵÄ״̬¡¢·´Ó¦ÈȵÄÊýÖµÓ뵥룬·´Ó¦ÈȵÄÊýÖµÓ뻯ѧ·½³ÌʽǰÃæµÄϵÊý³ÉÕý±È£¬×¢Òâ¸Ç˹¶¨ÂɵÄÔËÓã¬È¼ÉÕÈȵĸÅÄÌâÄ¿ÄѶÈÖеȣ®