Question

4£®£¨1£©0.3molµÄÆø̬¸ßÄÜȼÁÏÒÒÅðÍ飨B₂H₆£©ÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³öÈÈÁ¿649.5 ÖÐkJ£¬ÆäÈÈ»¯Ñ§·½³ÌΪB₂H₆£¨g£©+3O₂£¨g£©¨TB₂O₃£¨s£©+3H₂O£¨l£©¡÷H=-2165kJ/mol£®
ÓÖÒÑÖª£ºH₂O£¨l£©¨TH₂O£¨g£©£»¡÷H=+44.0kJ/mol£®Ôò11.2L£¨±ê×¼×´¿ö£©ÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆø̬ˮʱ·Å³öµÄÈÈÁ¿Îª1016.5KJ£®ºãΡ¢ºãѹ£¨1.01¡Á10⁵Pa£©Ï£¬½«2.0molSO₂ÓëamolO₂µÄ»ìºÏÆøÌåͨÈëÒ»¸öÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖУ¬·¢ÉúÈçÏ·´Ó¦£º2SO₂£¨g£©+O₂£¨g£©?2SO₃£¨g£©£»¡÷H£¼0
£¨2£©T₁¡æÏ·´Ó¦´ïƽºâʱ£¬²âµÃÈÝÆ÷ÄÚn£¨SO₃£©=1.3mol£¬n£¨O₂£©=3.0mol£¬Ôòa=3.65£»Æ½ºâʱSO₂µÄת»¯ÂÊΪ65%£®T₂¡æÏ·´Ó¦´ïƽºâʱ£¬»ìºÏÆøÌåµÄÎïÖʵÄÁ¿¹²Îª4.8mol£®T₁ÓëT₂µÄ¹ØϵÊÇ¢Ù£¨Ñ¡ÌîÐòºÅ£©£®
¢ÙT₁£¾T₂ ¢ÚT₁£¼T₂    ¢ÛT₁=T₂£®

Answer 1

·ÖÎö £¨1£©0.3molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5KJµÄÈÈÁ¿£¬Ôò1molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö2165KJµÄÈÈÁ¿£¬¸ù¾Ý¸Ç˹¶¨ÂÉд³öÆäÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£¬ºó¸ù¾ÝÎïÖʵÄÎïÖʵÄÁ¿Óë·´Ó¦·Å³öµÄÈÈÁ¿³ÉÕý±ÈÀ´½â´ð£»
£¨2£©T₁¡æÏ·´Ó¦´ïƽºâʱ£¬²âµÃÈÝÆ÷ÄÚn£¨SO₃£©=1.3mol£¬n£¨O₂£©=3.0mol£¬Ôò£º
           2SO₂ £¨g£©+O₂£¨g£©?2SO₃£¨g£©
ÆðʼÁ¿£¨mol£©£º2        a         0
ת»¯Á¿£¨mol£©£º1.3      0.65      1.3
ƽºâÁ¿£¨mol£©£º0.7      3         1.3
ÑõÆøƽºâÎïÖʵÄÁ¿=ÑõÆøÆðʼÎïÖʵÄÁ¿-ÑõÆøÎïÖʵÄÁ¿±ä»¯Á¿£¬Æ½ºâʱSO₂ µÄת»¯ÂÊ=$\frac{ÏûºÄ¶þÑõ»¯ÁòÎïÖʵÄÁ¿}{¶þÑõ»¯ÁòÆðʼÎïÖʵÄÁ¿}$¡Á100%£»T₁¡æÏ·´Ó¦´ïƽºâʱ£¬»ìºÏÆøÌå×ÜÎïÖʵÄÁ¿Îª5mol£¬T₂¡æÏ·´Ó¦´ïƽºâʱ£¬»ìºÏÆøÌåµÄÎïÖʵÄÁ¿¹²Îª4.8mol£¬ÓëT₁¡æÏ·´Ó¦´ïƽºâÏà±ÈƽºâÕýÏòÒƶ¯£¬Õý·´Ó¦Îª·ÅÈÈ·´Ó¦£¬½µµÍζÈƽºâÕýÏòÒƶ¯£®

½â´ð ½â£º£¨1£©0.3molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5KJµÄÈÈÁ¿£¬Ôò1molÆø̬¸ßÄÜȼÁÏÒÒÅðÍéÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö2165KJµÄÈÈÁ¿£¬·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ£ºB₂H₆£¨g£©+3O₂£¨g£©¨TB₂O₃£¨s£©+3H₂O£¨l£©¡÷H=-2165kJ/mol£¬
¢ÙB₂H₆£¨g£©+3O₂£¨g£©¨TB₂O₃£¨s£©+3H₂O£¨l£©¡÷H=-2165kJ/mol£¬
¢ÚH₂O£¨l£©¡úH₂O£¨g£©¡÷H=+44kJ/moL£¬
ÓɸÇ˹¶¨ÂÉ¿ÉÖª¢Ù+¢Ú¡Á3µÃ£ºB₂H₆£¨g£©+3O₂£¨g£©¨TB₂O₃£¨s£©+3H₂O£¨g£©¡÷H=-2033kJ/mol£¬11.2L£¨±ê×¼×´¿ö£©¼´0.5molÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆø̬ˮʱ·Å³öµÄÈÈÁ¿ÊÇ2033kJ¡Á0.5=1016.5kJ£¬
¹Ê´ð°¸Îª£ºB₂H₆£¨g£©+3O₂£¨g£©¨TB₂O₃£¨s£©+3H₂O£¨l£©¡÷H=-2165kJ/mol£»1016.5KJ£»
£¨2£©T₁¡æÏ·´Ó¦´ïƽºâʱ£¬²âµÃÈÝÆ÷ÄÚn£¨SO₃£©=1.3mol£¬n£¨O₂£©=3.0mol£¬Ôò£º
            2SO₂ £¨g£©+O₂£¨g£©?2SO₃£¨g£©
ÆðʼÁ¿£¨mol£©£º2         a       0
ת»¯Á¿£¨mol£©£º1.3       0.65   1.3
ƽºâÁ¿£¨mol£©£º0.7       3      1.3
Ôòa=3+0.65=3.65£¬Æ½ºâʱSO₂ µÄת»¯ÂÊ=$\frac{1.3mol}{2mol}$¡Á100%=65%£¬
T₁¡æÏ·´Ó¦´ïƽºâʱ£¬»ìºÏÆøÌå×ÜÎïÖʵÄÁ¿Îª5mol£¬T₂¡æÏ·´Ó¦´ïƽºâʱ£¬»ìºÏÆøÌåµÄÎïÖʵÄÁ¿¹²Îª4.8mol£¬ÓëT₁¡æÏ·´Ó¦´ïƽºâÏà±ÈƽºâÕýÏòÒƶ¯£¬Õý·´Ó¦Îª·ÅÈÈ·´Ó¦£¬½µµÍζÈƽºâÕýÏòÒƶ¯£¬¹ÊζÈT₁£¾T₂£¬
¹Ê´ð°¸Îª£º3.65£»65%£»¢Ù£®

µãÆÀ ±¾Ì⿼²éȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽµÄÊéд¡¢ìʱäÇóËã¡¢»¯Ñ§Æ½ºâ¼ÆËãÓëÓ°ÏìÒòËØ£¬ÌâÄ¿ÄѶȽϴó£¬×¢Òâ±íʾÈÈ»¯Ñ§·½³ÌʽµÄÊéдҪעÃ÷ÎïÖʾۼ¯×´Ì¬ºÍ·´Ó¦ÈȵÄÕý¸ººÅÎÊÌ⣬עÒâÈý¶Îʽ½âÌâ·¨ÔÚ»¯Ñ§Æ½ºâ¼ÆËãÖÐÓ¦Óã®