Question

11£®£¨1£©¢ÙCaCO₃£¨s£©=CaO£¨s£©+CO₂£¨g£©£»¡÷H=177.7kJ/mol
¢ÚC£¨s£©+H₂O£¨g£©=CO£¨g£©+H₂£¨g£©£»¡÷H=-131.3kJ/mol
¢Û$\frac{1}{2}$ H₂SO₄£¨l£©+NaOH£¨l£©=$\frac{1}{2}$Na₂SO₄£¨l£©+H₂O£¨l£©£»¡÷H=-57.3kJ/mol
¢ÜC£¨s£©+O₂£¨g£©=CO₂£¨g£©£»¡÷H=-393.5kJ/mol
¢ÝCO£¨g£©+$\frac{1}{2}$ O₂£¨g£©=CO₂£¨g£©£»¡÷H=-283kJ/mol
¢ÞHNO₃ £¨aq£©+NaOH£¨aq£©=NaNO₃£¨aq£©+H₂O£¨l£©£»¡÷H=-57.3kJ/mol
¢ß2H₂£¨g£©+O₂£¨g£©=2H₂O£¨l£©£»¡÷H=-517.6kJ/mol
£¨a£©ÉÏÊöÈÈ»¯Ñ§·½³ÌʽÖУ¬²»ÕýÈ·µÄÓТ٢ڢÛ
£¨b£©¸ù¾ÝÉÏÊöÐÅÏ¢£¬Ð´³öCת»¯ÎªCOµÄÈÈ»¯Ñ§·½³ÌʽC£¨s£©+1/2 O₂£¨g£©¨TCO₂ £¨g£©¡÷H=-110.5kJ/mol£®
£¨2£©ÒÑÖªÈÈ»¯Ñ§·½³Ìʽ£ºH₂£¨g£©+$\frac{1}{2}$O₂£¨g£©=H₂O£¨g£©¡÷H=-241.8kJ•mol^-1£¬¸Ã·´Ó¦µÄ»î»¯ÄÜΪ167.2kJ•mol^-1£¬ÔòÆäÄæ·´Ó¦µÄ»î»¯ÄÜΪ409.0 kJ•mol^-1
£¨3£©ÓÃCH₄´ß»¯»¹Ô­NO_x¿ÉÒÔÏû³ýµªÑõ»¯ÎïµÄÎÛȾ£®ÀýÈ磺
CH₄£¨g£©+4NO₂£¨g£©¨T4NO£¨g£©+CO₂£¨g£©+2H₂O£¨g£©¡÷H=-574kJ•mol^-1
CH₄£¨g£©+4NO£¨g£©¨T2N₂£¨g£©+CO₂£¨g£©+2H₂O£¨g£©¡÷H=-1 160kJ•mol^-1
ÈôÓñê×¼×´¿öÏÂ4.48L CH₄»¹Ô­NO₂Éú³ÉN₂£¬·´Ó¦ÖÐתÒƵĵç×Ó×ÜÊýΪ1.6 N_A£¨°¢·ü¼ÓµÂÂÞ³£ÊýÓÃN_A±íʾ£©£¬·Å³öµÄÈÈÁ¿Îª173.4kJ£®

Answer 1

·ÖÎö £¨1£©£¨a£©ÈÈ»¯Ñ§·½³ÌʽҪ±êÃ÷ÎïÖʵľۼ¯×´Ì¬£¬²¢ÇÒìʱäµÄ·ûºÅºÍµ¥Î»ÒªÕýÈ·£¬¸ù¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨À´»Ø´ð£»
£¨b£©½áºÏ»¯Ñ§·´Ó¦¢Ü¢Ý£¬ÀûÓøÇ˹¶¨ÂÉÀ´»Ø´ð£»
£¨2£©·´Ó¦µÄ»î»¯ÄÜÊÇʹÆÕͨ·Ö×Ó±ä³É»î»¯·Ö×ÓËùÐèÌṩµÄ×îµÍÏ޶ȵÄÄÜÁ¿£¬ÒÀ¾ÝͼÏóÄÜÁ¿¹Øϵ¿ÉÖª£¬Äæ·´Ó¦µÄ»î»¯ÄÜ=Õý·´Ó¦µÄ»î»¯ÄÜ+·´Ó¦µÄìʱ䣻
£¨3£©ÒÑÖª¢ÙCH₄£¨g£©+4NO₂£¨g£©¨T4NO£¨g£©+CO₂£¨g£©+2H₂O£¨g£©¡÷H=-574kJ•mol^-1£¬¢ÚCH₄£¨g£©+4NO£¨g£©¨T2N₂£¨g£©+CO₂£¨g£©+2H₂O£¨g£©¡÷H=-1 160kJ•mol^-1£¬
ÀûÓøÇ˹¶¨ÂÉ$\frac{1}{2}$£¨¢Ù+¢Ú£©¿ÉµÃ£ºCH₄£¨g£©+2NO₂£¨g£©¨TN₂£¨g£©+CO₂£¨g£©+2H₂O£¨g£©£¬¡÷H=-$\frac{574+1160}{2}$kJ•mol^-1=867kJ•mol^-1£¬¸ù¾Ý»¯ºÏ¼Û±ä»¯¼ÆËã³ö·´Ó¦0.2mol¼×ÍéתÒƵĵç×ÓÊý¼°·Å³öµÄÈÈÁ¿£®

½â´ð ½â£º£¨1£©£¨a£©¸ù¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨£¬¢Ù¢Ú¢Û¶¼ÊÇÓдíÎóµÄ£¬´íÎóÖ®´¦·Ö±ðÊÇ£º¢ÙÖС÷H=+177.7/mol£¬¢Ú·´Ó¦ÎüÈÈ£¬¡÷H=+131£®KJ/mol£¬¢Û³ýH₂OÍâÆäËüÎïÖʵÄ״̬²»ÊÇ¡°l¡±Ó¦Îª¡°aq¡±£»
¹Ê´ð°¸Îª£º¢Ù¢Ú¢Û£»
£¨b£©½áºÏ»¯Ñ§·´Ó¦¢Ü¢Ý£¬ÀûÓøÇ˹¶¨ÂÉ£¬·´Ó¦C£¨s£©+$\frac{1}{2}$O₂£¨g£©=CO£¨g£©ÊÇ¢Ü-¢ÝµÃÀ´£¬ËùÒÔìʱä¡÷H=£¨-393.5kJ/mol£©-£¨-283kJ/mol£©=-110.5KJ/mol£¬
¹Ê´ð°¸Îª£ºC£¨s£©+$\frac{1}{2}$O₂£¨g£©=CO£¨g£©¡÷H=-110.5KJ/mol£»
£¨2£©·´Ó¦µÄ»î»¯ÄÜÊÇʹÆÕͨ·Ö×Ó±ä³É»î»¯·Ö×ÓËùÐèÌṩµÄ×îµÍÏ޶ȵÄÄÜÁ¿£¬ÒÀ¾ÝͼÏóÄÜÁ¿¹Øϵ¿ÉÖª£¬Äæ·´Ó¦µÄ»î»¯ÄÜ=Õý·´Ó¦µÄ»î»¯ÄÜ+·´Ó¦µÄìʱ䣻H₂£¨g£©+$\frac{1}{2}$O₂£¨g£©¨TH₂O£¨g£©¡÷H=-241.8kJ•mol^-1£¬¸Ã·´Ó¦µÄ»î»¯ÄÜΪ167.2kJ•mol^-1£¬ÔòÆäÄæ·´Ó¦µÄ»î»¯ÄÜ=167.2KJ/mol+241.8KJ/mol=409.0KJ/mol£¬
¹Ê´ð°¸Îª£º409.0£»
£¨3£©ÒÑÖª¢ÙCH₄£¨g£©+4NO₂£¨g£©¨T4NO£¨g£©+CO₂£¨g£©+2H₂O£¨g£©¡÷H=-574kJ•mol^-1£¬¢ÚCH₄£¨g£©+4NO£¨g£©¨T2N₂£¨g£©+CO₂£¨g£©+2H₂O£¨g£©¡÷H=-1 160kJ•mol^-1£¬
ÀûÓøÇ˹¶¨ÂÉ$\frac{1}{2}$£¨¢Ù+¢Ú£©¿ÉµÃ£º¢ÛCH₄£¨g£©+2NO₂£¨g£©¨TN₂£¨g£©+CO₂£¨g£©+2H₂O£¨g£©£¬¡÷H=-$\frac{574+1160}{2}$kJ•mol^-1=-867kJ•mol^-1£¬
±ê¿öÏÂ4.48L¼×ÍéµÄÎïÖʵÄÁ¿Îª£ºn£¨CH₄£©=$\frac{4.48L}{22.4L/mol}$=0.2mol£¬¼×ÍéÖÐ̼ԪËصĻ¯ºÏ¼ÛΪ-4¼Û£¬·´Ó¦²úÎï¶þÑõ»¯Ì¼ÖÐCÔªËØ»¯ºÏ¼ÛΪ+4¼Û£¬»¯ºÏ¼Û±ä»¯Îª8£¬¼´ÍêÈ«·´Ó¦1mol¼×ÍéתÒÆÁË8mol£¨8N_A£©µç×Ó£¬Ôò·´Ó¦0.2mol¼×ÍéתÒƵĵç×Ó×ÜÊýΪ£º0.2mol¡Á8N_A=1.6N_A£»¸ù¾ÝÈÈ»¯Ñ§·½³Ìʽ¢Û¿ÉÖª£¬ÍêÈ«·´Ó¦0.2mol¼×Íé·ÅÈȵÄÈÈÁ¿Îª£º867kJ•mol^-1¡Á0.2mol=173.4kJ£¬
¹Ê´ð°¸Îª£º1.6N_A£»173.4£®

µãÆÀ ±¾Ì⿼²é·´Ó¦ÈÈÒÔ¼°Ñõ»¯»¹Ô­·´Ó¦µÄ¼ÆË㣬ÌâÄ¿ÄѶÈÖеȣ¬×¢ÒâÕÆÎÕ¸Ç˹¶¨Âɵĺ¬Òå¼°ÈÈ»¯Ñ§·½³ÌʽµÄÊéдԭÔò£¬Ã÷È·Ñõ»¯»¹Ô­·´Ó¦Öеç×ÓתÒƵļÆËã·½·¨£¬ÊÔÌâ²àÖØÓÚѧÉúµÄ¼ÆËãÄÜÁ¦ºÍ·ÖÎöÄÜÁ¦µÄ¿¼²é£®