Question

11£®ÒÔÏÂÊÇ´¦ÓÚÑо¿½×¶ÎµÄ¡°È˹¤¹Ìµª¡±µÄз½·¨£®N₂ÔÚ´ß»¯¼Á±íÃæÓëË®·¢Éú·´Ó¦£º2N₂£¨g£©+6H₂O£¨l£©?4NH₃£¨g£©+3O₂£¨g£©¡÷H=+1530.4kJ/mol£»Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©ÉÏÊö·´Ó¦ÔÚ¸ßΣ¨Ìî¸ßΡ¢µÍΡ¢ÈκÎζȣ©ÏÂÄÜ×Ô·¢½øÐУ®
£¨2£©¸Ã·´Ó¦Æ½ºâ³£ÊýKµÄ±í´ïʽK=$\frac{{c}^{4}£¨N{H}_{3}£©•{c}^{3}£¨{O}_{2}£©}{{c}^{2}£¨{N}_{2}£©}$£®
£¨3£©ÉÏÊö·´Ó¦´ïµ½Æ½ºâºó£¬±£³ÖÆäËûÌõ¼þ²»±ä£¬Éý¸ßζȣ¬ÖØдﵽƽºâ£¬Ôòac£®
a£®Æ½ºâ³£ÊýKÔö´ó  b£®H₂OµÄŨ¶È¼õС  c£®ÈÝÆ÷ÄÚµÄѹǿÔö´ó  d£®v_Ä棨O₂£©¼õС
£¨4£©Ñо¿Ð¡×é·Ö±ðÔÚËĸöÈÝ»ýΪ2ÉýµÄÃܱÕÈÝÆ÷ÖУ¬³äÈëN₂ 1mol¡¢H₂O 3mol£¬ÔÚ´ß»¯¼ÁÌõ¼þϽøÐз´Ó¦3Сʱ£®ÊµÑéÊý¾Ý¼û±í£º

ÐòºÅ	µÚÒ»×é	µÚ¶þ×é	µÚÈý×é	µÚËÄ×é
t/¡æ	30	40	50	80
NH3Éú³ÉÁ¿/£¨10-6mol£©	4.8	5.9	6.0	2.0

µÚËÄ×éʵÑéÖÐÒÔNH₃±íʾ·´Ó¦µÄËÙÂÊÊÇ33¡Á10^-7mol/L•h£¬ÓëÇ°Èý×éÏà±È£¬NH₃ Éú³ÉÁ¿×îСµÄÔ­Òò¿ÉÄÜÊÇ´ß»¯¼ÁÔÚ80¡æ»îÐÔ¼õС£¬·´Ó¦ËÙÂÊ·´¶ø¼õÂý£®
£¨5£©°±Ë®ÊÇʵÑéÊÒ³£ÓõÄÈõ¼î£®
¢ÙÍùCaCl₂ÈÜÒºÖÐͨÈëCO₂ÖÁ±¥ºÍ£¬ÎÞÃ÷ÏÔÏÖÏó£®ÔÙͨÈëÒ»¶¨Á¿µÄNH₃ºó²úÉú°×É«³Áµí£¬´ËʱÈÜÒºÖÐÒ»¶¨ÓеÄÈÜÖÊÊÇNH₄Cl£®ÇëÓõçÀëƽºâÀíÂÛ½âÊÍÉÏÊöʵÑéÏÖÏ󱥺ÍH₂CO₃ÈÜÒºÖеçÀë²úÉúµÄCO₃^2-ºÜÉÙ£¬Òò´ËûÓгÁµí£¬¼ÓÈ백ˮºó£¬´Ù½øH₂CO₃µÄµçÀ룬CO₃^2-Àë×ÓŨ¶ÈÔö´ó£¬ÓгÁµí²úÉú£®
¢ÚÏòÑÎËáÖеμӰ±Ë®¹ýÁ¿ÖÁÈÜÒº³ÊÖÐÐÔ£¬ÔòÀë×ÓŨ¶ÈµÄ¹ØϵÊÇc£¨NH₄⁺£©£¾c£¨OH^-£©£¾c£¨Cl^-£©£¾c£¨H⁺£©£®

Answer 1

·ÖÎö £¨1£©¸ù¾Ý¡÷G=¡÷H-T•¡÷S½øÐÐÅжϣ¬Èç¡÷G£¼0£¬Ôò·´Ó¦ÄÜ×Ô·¢½øÐУ¬Èç¡÷G£¾0£¬Ôò²»ÄÜ×Ô·¢½øÐУ»
£¨2£©¸ù¾Ýƽºâ³£Êý±í´ïµÄ¶¨Ò壬Éú³ÉÎïŨ¶ÈÃÝÖ®»ý±ÈÉÏ·´Ó¦ÎïŨ¶ÈÃÝÖ®»ý£»
£¨3£©a£®Õý·´Ó¦ÎªÎüÈÈ·´Ó¦£¬Éý¸ßζȣ¬Æ½ºâ³£ÊýKÔö´ó£»
b£®Éý¸ßζȣ¬H₂OµÄŨ¶È²»±ä£»
c£®Éý¸ßζȣ¬Æ½ºâÕýÏòÒƶ¯£¬ÈÝÆ÷ÄÚµÄѹǿÔö´ó£»
d£®Éý¸ßζÈÕýÄæ·´Ó¦ËÙÂʶ¼Ôö¼Ó£»
£¨4£©3СʱÄÚÒÔNH₃Ũ¶ÈµÄ±ä»¯Á¿À´±íʾ£»µÚËÄ×éζÈ×î¸ßµ«´ïƽºâʱ£¬°±ÆøµÄÎïÖʵÄÁ¿×îС£¬¿ÉÄÜÊÇζȶԴ߻¯¼ÁµÄ»îÐÔµÄÓ°Ï죻
£¨5£©¢ÙͨÈëÒ»¶¨Á¿µÄNH₃ºóʹÈÜÒº³Ê¼îÐÔ£¬Äܹ»Éú³ÉNH₄Cl£»
¢Ú¸ù¾ÝµçºÉÊغ㣬ÈçÈÜÒº³ÊÖÐÐÔ£¬c£¨H⁺£©=c£¨OH^-£©£¬ÔòÓÐc £¨C1^-£©=c£¨NH₄⁺£©£¬ÌåϵΪNH₄ClÈÜÒººÍNH₃£®H₂O£¬°±Ë®¹ýÁ¿½Ï¶àʱ£¬ÈÜÒº³Ê¼îÐÔ£ºc£¨NH₄⁺£©£¾c£¨OH^-£©£¾c£¨Cl^-£©£¾c£¨H⁺£©£®

½â´ð ½â£º£¨1£©2N₂£¨g£©+6H₂O£¨l£©?4NH₃£¨g£©+3O₂£¨g£©¡÷H=+1530.4kJ/mol£¬·´Ó¦ÖС÷H£¾0£¬¡÷S£¾0£¬ËùÒÔ±ØÐëÔÚ¸ßεÄÌõ¼þϲÅÄÜ×Ô·¢½øÐУ¬¹Ê´ð°¸Îª£º¸ßΣ»
£¨2£©·´Ó¦2N₂£¨g£©+6H₂O£¨l£©?4NH₃£¨g£©+3O₂£¨g£©µÄƽºâ³£Êý±í´ïʽΪ£ºK=$\frac{{c}^{4}£¨N{H}_{3}£©•{c}^{3}£¨{O}_{2}£©}{{c}^{2}£¨{N}_{2}£©}$£¬¹Ê´ð°¸Îª£ºK=$\frac{{c}^{4}£¨N{H}_{3}£©•{c}^{3}£¨{O}_{2}£©}{{c}^{2}£¨{N}_{2}£©}$£»
£¨3£©a£®Õý·´Ó¦ÎªÎüÈÈ·´Ó¦£¬Éý¸ßζȣ¬Æ½ºâ³£ÊýKÔö´ó£¬¹ÊÑ¡£»
b£®Éý¸ßζȣ¬H₂OµÄŨ¶È²»±ä£¬¹Ê²»Ñ¡£»
c£®Éý¸ßζȣ¬Æ½ºâÕýÏòÒƶ¯£¬ÈÝÆ÷ÄÚµÄѹǿÔö´ó£¬¹ÊÑ¡£»
d£®Éý¸ßζÈÕýÄæ·´Ó¦ËÙÂʶ¼Ôö¼Ó£¬¹Ê²»Ñ¡£»
¹Ê´ð°¸Îª£ºac£»
£¨4£©3СʱÄÚÒÔNH₃Ũ¶ÈµÄ±ä»¯Á¿£¬v£¨NH₃£©=$\frac{\frac{2¡Á1{0}^{-6}mol-0}{2L}}{3h}$=33¡Á10^-7mol/L•h£¬µÚËÄ×éζÈ×î¸ßµ«´ïƽºâʱ£¬°±ÆøµÄÎïÖʵÄÁ¿×îС£¬¿ÉÄÜÊÇ´ß»¯¼ÁÔÚ80¡æ»îÐÔ¼õС£¬·´Ó¦ËÙÂÊ·´¶ø¼õÂý£¬¹Ê´ð°¸Îª£º33¡Á10^-7mol/L•h£»´ß»¯¼ÁÔÚ80¡æ»îÐÔ¼õС£¬·´Ó¦ËÙÂÊ·´¶ø¼õÂý£»
£¨5£©¢ÙͨÈëÒ»¶¨Á¿µÄNH₃ºóʹÈÜÒº³Ê¼îÐÔ£¬Äܹ»Éú³ÉNH₄Cl£¬±¥ºÍH₂CO₃ÈÜÒºÖеçÀë²úÉúµÄCO₃^2-ºÜÉÙ£¬Òò´ËûÓгÁµí£®¼ÓÈ백ˮºó£¬´Ù½øH₂CO₃µÄµçÀ룬CO₃^2-Àë×ÓŨ¶ÈÔö´ó£¬ÓгÁµí²úÉú£¬¹Ê´ð°¸Îª£ºNH₄Cl£»±¥ºÍH₂CO₃ÈÜÒºÖеçÀë²úÉúµÄCO₃^2-ºÜÉÙ£¬Òò´ËûÓгÁµí£¬¼ÓÈ백ˮºó£¬´Ù½øH₂CO₃µÄµçÀ룬CO₃^2-Àë×ÓŨ¶ÈÔö´ó£¬ÓгÁµí²úÉú£»
¢ÚÌåϵΪNH₄ClÈÜÒººÍNH₃£®H₂O£¬°±Ë®¹ýÁ¿½Ï¶àʱ£¬ÈÜÒº³Ê¼îÐÔ£ºc£¨NH₄⁺£©£¾c£¨OH^-£©£¾c£¨Cl^-£©£¾c£¨H⁺£©£¬
¹Ê´ð°¸Îª£ºc£¨NH₄⁺£©£¾c£¨OH^-£©£¾c£¨Cl^-£©£¾c£¨H⁺£©£®

µãÆÀ ±¾Ì⿼²é»¯Ñ§Æ½ºâ³£Êý±í´ïʽµÄÊéд£¬ÒÔ¼°Ó°Ï컯ѧËÙÂʵÄÒòËغÍÈõµç½âÖʵÄʵÑé̽¾¿£¬ÀûÓüÙÉè·¨À´·ÖÎö½â´ð¼´¿É£¬ÄѶȲ»´ó£®