Question

6£®Ñо¿ÈËԱͨ¹ý¶Ô±±¾©µØÇøPM2.5µÄ»¯Ñ§×é³ÉÑо¿·¢ÏÖ£¬Æû³µÎ²ÆøºÍȼúÎÛȾ·Ö±ðÕ¼4%¡¢18%£®
£¨1£©ÓÃÏ¡ÍÁµÈ´ß»¯¼ÁÄܽ«Æû³µÎ²ÆøÖеÄCO¡¢NO_x̼Ç⻯ºÏÎïת»¯³ÉÎÞ¶¾ÎïÖÊ£¬´Ó¶ø¼õÉÙÆû³µÎ²Æø´«È¾£®ÒÑÖª£º
N₂£¨g£©+O₂£¨g£©=2NO£¨g£©¡÷H=+180.5kJ•mol^-1
2C£¨s£©+O₂£¨g£©=2CO£¨g£©¡÷H=-221.0kJ•mol^-1
C£¨s£©+O₂£¨g£©=CO₂£¨g£©¡÷H=-393.5kJ•mol^-1
д³öNO£¨g£©ÓëCO£¨g£©´ß»¯×ª»¯³ÉN₂£¨g£©ºÍCO₂£¨g£©µÄÈÈ»¯Ñ§·½³Ìʽ£º2NO £¨g£©+2CO£¨g£©=N₂£¨g£©+2CO₂£¨g£©¡÷H=-746.5 kJ/mol£®
£¨2£©»¹¿ÉÒÔÓûîÐÔÌ¿»¹Ô­·¨´¦ÀíµªÑõ»¯Î·´Ó¦ÎªC£¨s£©+2NO£¨g£©?N₂£¨g£©+CO₂£¨g£©H£®ÏòijÃܱÕÈÝÆ÷¼ÓÈëÒ»¶¨Á¿µÄ»îÐÔÌ¿ºÍNO£¬ºãΣ¨T₁¡æ£©Ìõ¼þÏ·´Ó¦£¬·´Ó¦½øÐе½²»Í¬Ê±¼ä²âµÃ¸÷ÎïÖʵÄŨ¶ÈÈçÏ£º

¡¡¡¡¡¡Å¨¶È/mol•L-1 ʱ¼ä/min	NO	N2	CO2
0	0.100	0	0
10	0.058	0.021	0.021
20	0.050	0.025	0.025
30	0.050	0.025	0.025
40	0.036	0.032	0.010
50	0.036	0.032	0.010

¢ÙT₁¡æʱ£¬¸Ã·´Ó¦µÄƽºâ³£ÊýK=0.25£¨±£ÁôÁ½Î»Ð¡Êý£©£®
¢ÚÇ°10minÄÚÓÃv£¨NO£©±íʾµÄ»¯Ñ§·´Ó¦ËÙÂÊΪ0.0042mol/£¨L£®min£©£»30minºó£¬¸Ä±äijһÌõ¼þ£¬·´Ó¦ÖØдﵽƽºâ£¬Ôò¸Ä±äµÄÌõ¼þ¿ÉÄÜÊǽµµÍCO₂Ũ¶È£®
¢ÛÈô30minºóÉý¸ßζÈÖÁT₂¡æ£¬´ïµ½Æ½ºâʱ£¬ÈÝÆ÷ÖÐNO¡¢N₂¡¢CO₂Ũ¶ÈÖ®±ÈΪ3£º1£º1£¬Ôò¸Ã·´Ó¦µÄ¡÷H£¼0£¨Ìî¡°£¾¡±¡°=¡±»ò¡°£¼¡±£©£®
£¨3£©¢ÙÓÃÆøÌå´«¸ÐÆ÷¿ÉÒÔ¼ì²âÆû³µÎ²ÆøÖÐCOµÄº¬Á¿£®´«¸ÐÆ÷ÊÇÒÔȼÁϵç³ØΪ¹¤×÷Ô­Àí£¬Æä×°ÖÃÈçͼËùʾ£¬¸Ãµç³ØÖеç½âÖÊΪÑõ»¯îÆ-Ñõ»¯ÄÆ£¬ÆäÖÐO^2-¿ÉÒÔÔÚ¹ÌÌå½éÖÊNASICONÖÐ×ÔÓÉÒƶ¯£®aΪ¸º¼«£¬µç¼«·´Ó¦Ê½ÎªCO-2e^-+O^2-¨TCO₂£®µç³Ø¹¤×÷¹ý³ÌÖУ¬O^2-ÓÉb¼«£¨Ìî¡°a¡±»ò¡°b¡±£¬ÏÂͬ£©ÒÆÏòa¼«£®
¢ÚÅ·IVÐÍÆû³µÈÏÖ¤ºÍÉú²úÒ»ÖÂÐÔÅÅ·ÅÏÞÖµ£ºÆûÓÍ»úCO1.00g/km£®Ä³²âÊÔ³µ¼ì²â¹ý³ÌÖУ¬Ã¿ÐÐÊ»1km´«¸ÐÆ÷ÖÐͨ¹ýµç×ÓΪ0.08mol£¬Ôò¸Ã³µ²»·ûºÏ£¨Ìî¡°·ûºÏ¡±»ò¡°²»·ûºÏ¡±£©Å·IVÅŷűê×¼£®

Answer 1

·ÖÎö £¨1£©¸ù¾Ý·½³ÌʽµÄ¼Ó¼õµÃ³öNOÓëCO´ß»¯×ª»¯³ÉN₂ºÍCO₂µÄ»¯Ñ§·½³Ìʽ£¬ìʱäÏàÓ¦µÄ¼Ó¼õ£¬´Ó¶øµÃ³öÆäÈÈ»¯Ñ§·´Ó¦·½³Ìʽ£»
£¨2£©¢Ù20minʱ·´Ó¦µÃµ½Æ½ºâ£¬½«Æ½ºâŨ¶È´úÈëƽºâ³£ÊýK=$\frac{c£¨{N}_{2}£©¡Ác£¨C{O}_{2}£©}{{c}^{2}£¨NO£©}$¼ÆË㣻
¢Ú¸ù¾Ýv=$\frac{¡÷c}{¡÷t}$¼ÆËãv£¨NO£©£»
30min¸Ä±äÌõ¼þºóÔÚ40minÓÖµ½´ïƽºâ£¬NO¡¢CO₂Ũ¶È¼õС£¬N₂µÄŨ¶ÈŨ¶ÈÔö´ó£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬ÇÒ±íÖÐNO¡¢N₂Ũ¶È±ä»¯Á¿Ö®±ÈΪ£¨0.05-0.036£©£º£¨0.032-0.025£©=2£º1£¬µÈÓÚ»¯Ñ§¼ÆÁ¿ÊýÖ®±È£¬¹ÊÓ¦ÊǽµµÍCO₂Ũ¶È£»
¢Û30min´¦ÓÚƽºâ״̬£¬NO¡¢N₂¡¢CO₂Ũ¶ÈÖ®±ÈΪ2£º1£º1£¬Èô30minºóÉý¸ßζÈÖÁT₂¡æ£¬´ïµ½Æ½ºâʱ£¬ÈÝÆ÷ÖÐNO¡¢N₂¡¢CO₂Ũ¶ÈÖ®±ÈΪ3£º1£º1£¬ËµÃ÷ƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£»
£¨3£©¢Ù±¾ÖÊÊÇCOÓëÑõÆø·´Ó¦Éú³ÉCO₂£¬COÔÚa¼«·¢ÉúÑõ»¯·´Ó¦£¬¹ÊaΪ¸º¼«£¬bΪÕý¼«£¬ÒõÀë×ÓÏò¸º¼«Òƶ¯£¬O^2-Óɵ缫bÁ÷Ïòµç¼«a£»
¢Ú¸ù¾ÝתÒƵç×Ó¼ÆËãÐÐÊ»1KmʱÅÅ·ÅCOµÄÖÊÁ¿£¬½ø¶øÅжÏÊÇ·ñ·ûºÏÅŷűê×¼£®

½â´ð ½â£º£¨1£©ÒÑÖª£º¢Ù£®N₂£¨g£©+O₂£¨g£©=2NO£¨g£©¡÷H=+180.5kJ/mol¢Ù
¢Ú£®2C£¨s£©+O₂£¨g£©=2CO£¨g£©¡÷H=-221.0kJ/mol¢Ú
¢Û£®C£¨s£©+O₂£¨g£©=CO₂£¨g£©¡÷H=-393.5kJ/mol ¢Û
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Û¡Á2-¢Ú-¢Ù¿ÉµÃ£ºNO £¨g£©+2CO£¨g£©=N₂£¨g£©+2CO₂£¨g£©£¬ËùÒÔ¡÷H=£¨-393.5kJ/mol£©¡Á2-£¨-221.0kJ/mol£©-£¨+180.5kJ/mol£©=-746.5 kJ/mol£¬
¹Ê´ð°¸Îª£º2NO £¨g£©+2CO£¨g£©=N₂£¨g£©+2CO₂£¨g£©¡÷H=-746.5 kJ/mol£»
£¨2£©¢Ù20minʱ·´Ó¦µÃµ½Æ½ºâ£¬½«Æ½ºâŨ¶È´úÈë¼ÆË㣬¿ÉÖªK=$\frac{c£¨{N}_{2}£©¡Ác£¨C{O}_{2}£©}{{c}^{2}£¨NO£©}$=$\frac{0.025¡Á0.025}{0.0{5}^{2}}$=0.25£¬
¹Ê´ð°¸Îª£º0.25£»
¢Ú¸ù¾Ýv=$\frac{¡÷c}{¡÷t}$¿ÉÖª£¬v£¨NO£©=$\frac{0.1mol/L-0.058mol/L}{10min}$=0.0042mol/£¨L£®min£©£»
30min¸Ä±äÌõ¼þºóÔÚ40minÓÖµ½´ïƽºâ£¬NO¡¢CO₂Ũ¶È¼õС£¬N₂µÄŨ¶ÈŨ¶ÈÔö´ó£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬ÇÒ±íÖÐNO¡¢N₂Ũ¶È±ä»¯Á¿Ö®±ÈΪ£¨0.05-0.036£©£º£¨0.032-0.025£©=2£º1£¬µÈÓÚ»¯Ñ§¼ÆÁ¿ÊýÖ®±È£¬¹ÊÓ¦ÊǽµµÍCO₂Ũ¶È£¬
¹Ê´ð°¸Îª£º0.0042mol/£¨L£®min£©£»½µµÍCO₂Ũ¶È£»
¢Û30min´¦ÓÚƽºâ״̬£¬NO¡¢N₂¡¢CO₂Ũ¶ÈÖ®±ÈΪ2£º1£º1£¬Èô30minºóÉý¸ßζÈÖÁT₂¡æ£¬´ïµ½Æ½ºâʱ£¬ÈÝÆ÷ÖÐNO¡¢N₂¡¢CO₂Ũ¶ÈÖ®±ÈΪ3£º1£º1£¬ËµÃ÷ƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬Ôò¸Ã·´Ó¦µÄ¡÷H£¼0£¬
¹Ê´ð°¸Îª£º£¼£»
£¨3£©¢Ù±¾ÖÊÊÇCOÓëÑõÆø·´Ó¦Éú³ÉCO₂£¬COÔÚa¼«·¢ÉúÑõ»¯·´Ó¦£¬¹ÊaΪ¸º¼«£¬bΪÕý¼«£¬ÒõÀë×ÓÏò¸º¼«Òƶ¯£¬O^2-Óɵ缫bÁ÷Ïòµç¼«a£¬¸º¼«µç¼«·´Ó¦Ê½Îª£ºCO-2e^-+O^2-¨TCO₂£¬
¹Ê´ð°¸Îª£º¸º£»CO-2e^-+O^2-¨TCO₂£»b£»a£»
¢ÚÿÐÐÊ»1km´«¸ÐÆ÷ÖÐͨ¹ýµç×ÓΪ0.08mol£¬ÓÉCO-2e^-+O^2-¨TCO₂¿ÉÖª£¬ÅÅ·ÅCOµÄÖÊÁ¿Îª0.08mol¡Á$\frac{1}{2}$¡Á28g/mol=1.12g£¾1.00g£¬²»·ûºÏÅŷűê×¼£¬
¹Ê´ð°¸Îª£º²»·ûºÏ£®

µãÆÀ ±¾Ì⿼²éÈÈ»¯Ñ§·½³ÌʽÊéд¡¢Æ½ºâ³£Êý¼ÆËã¡¢·´Ó¦ËÙÂʼÆËã¡¢»¯Ñ§Æ½ºâÒƶ¯¡¢Ô­µç³Ø¼°Æä¼ÆËãµÈ£¬ÊôÓÚÆ´ºÏÐÍÌâÄ¿£¬ÊǶÔѧÉú×ÛºÏÄÜÁ¦µÄ¿¼²é£¬ÄѶÈÖеȣ®