ÌâÄ¿ÄÚÈÝ

8£®£¨1£©N2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H=-94.4kJ/mol£®ºãÈÝʱ£¬ÌåϵÖи÷ÎïÖÊŨ¶ÈËæÊ±Ïò±ä»¯µÄÇúÏßÈçͼËùʾ£ºa£®ÔÚ1LÈÝÆ÷Öз¢Éú·´Ó¦£¬Ç°20minÄÚv£¨NH3£©=0.050 mol•£¨L•min£©-1£¬·Å³öµÄÈÈÁ¿Îª47.2 kJ£®
b.25minʱ²ÉÈ¡µÄ´ëÊ©Êǽ«NH3´Ó·´Ó¦ÌåϵÖзÖÀë³öÈ¥£®c£®Ê±¶ÎIIIÌõ¼þÏ·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ$\frac{0£®{5}^{2}}{0.7{5}^{3}•0.25}$£¨ÓÃͼÖÐÊý¾Ý±íʾ£¬²»¼ÆËã½á¹û£©£®

£¨2£©ÏõËṤҵµÄ»ù´¡Êǰ±µÄ´ß»¯Ñõ»¯£¬ÔÚ´ß»¯¼ÁµÄ×÷ÓÃÏ·¢ÉúÈçÏ·´Ó¦£º
¢Ù4NH2£¨g£©+5O2£¨g£©+6H2O£¨g£©¡÷H=-905kJ/molÖ÷·´Ó¦
¢Ú4NH2£¨g£©+3O2?2N2£¨g£©+6HO£¨g£©¡÷H=-1268kJ/mol¸±·´Ó¦
a£®ÓÉ·´Ó¦¢Ù¢Ú¿ÉÖª·´Ó¦N2£¨g£©+O2£¨g£©?2NO£¨g£©µÄ·´Ó¦ÈÈ¡÷H=+181.5 kJ/mol
b£®·´Ó¦¢ÙÔÚÒ»¶¨Ìõ¼þÏÂÄÜ£¨Ìî¡°ÄÜ¡±»ò¡°²»ÄÜ¡±£©×Ô·¢·´Ó¦£¬ÀíÓÉÊÇ¡÷H£¼0ÇÒÊÇ»ìÂÒ¶ÈÔö´óµÄ·´Ó¦£®
£¨3£©ÒÀ¾Ý·´Ó¦¢Ú¿ÉÒÔÉè¼Æ³ÉÖ±½Ó¹©°±Ê½¼îÐÔȼÁÏµç³ØÈçͼËùʾ£®ÔòͼÖÐAΪ¸º¼«£¨Ìî¡°Õý¼«¡±»ò¡°¸º¼«¡±£©£¬µç¼«·´Ó¦·½³ÌʽΪ2NH3-6e-+6OH-=N2+6H2O£®
£¨4£©NH3ÓëN2H4¶¼¾ßÓл¹Ô­ÐÔ£¬¿ÉÒÔÓëÆäËûÇ¿Ñõ»¯¼Á·´Ó¦£¬ÀýÈçÔÚÒ»¶¨Ìõ¼þÏ£¬°±¿ÉÒÔ±»Ë«ÑõË®Ñõ»¯ÎªÓÎÀë̬µª£¬Ð´³ö¸Ã·´Ó¦µÄ»¯Ñ§·½³Ìʽ2NH3+3H2O2=N2+6H2O£®

·ÖÎö £¨1£©a¡¢»¯Ñ§·´Ó¦ËÙÂÊv=$\frac{¡÷c}{¡÷t}$½áºÏͼÏó´úÈëÊý¾ÝÀ´¼ÆËã¼´¿É£»
b¡¢¸ù¾Ýͼʾ£¬25¡æÊ±°±ÆøµÄŨ¶ÈѸËÙ¼õС£¬¾Ý´Ë»Ø´ð£»
c¡¢¸ù¾Ýƽºâ³£ÊýKµÈÓÚ¸÷Éú³ÉÎïÆ½ºâŨ¶ÈϵÊý´Î·½µÄ³Ë»ýºÍ¸÷·´Ó¦ÎïÆ½ºâŨ¶ÈϵÊý´Î·½Ö®»ýµÄ±ÈÖµÀ´»Ø´ð£»
£¨2£©a¡¢4NH3£¨g£©+5O2?4NO£¨g£©+6H2O£¨g£©£»¡÷H=-905kJ/mol¢Ù
4NH3£¨g£©+3O2£¨g£©?2N2£¨g£©+6H2O£¨g£©£»¡÷H=-1268kJ/mol¢Ú£»
ÒÀ¾Ý¸Ç˹¶¨ÂÉ¢Ù-¢ÚµÃµ½£¬2N2£¨g£©+2O2£¨g£©=4NO£¨g£©£¬ÒÀ¾ÝÈÈ»¯Ñ§·½³ÌʽºÍ¸Ç˹¶¨ÂɼÆËã·ÖÎöÅжϣ»
b¡¢¸ù¾Ý·´Ó¦ÊÇ·ñ×Ô·¢µÄÅоݡ÷H-T¡÷S£¼0À´ÅжϷ´Ó¦ÊÇ·ñ×Ô·¢£»
£¨3£©ÓÉͼ¿ÉÖª£¬A¼«Í¨ÈëµÄΪ°±Æø£¬·¢ÉúÑõ»¯·´Ó¦£¬Îª¸º¼«£¬°±ÆøÔÚ¼îÐÔÌõ¼þÏ·ŵçÉú³ÉµªÆøÓëË®£»
£¨4£©°±¿ÉÒÔ±»Ë«ÑõË®Ñõ»¯ÎªÓÎÀë̬µª£¬¸ù¾Ýµç×ÓÊØºãºÍÔ­×ÓÊØºãÊéдÅ䯽·½³Ìʽ¼´¿É£®

½â´ð ½â£º£¨1£©a£®»¯Ñ§·´Ó¦ËÙÂÊv=$\frac{¡÷c}{¡÷t}$=$\frac{1.0mol/L}{20min}$=0.050 mol•£¨L•min£©-1£¬¸ù¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÒâÒ壬Éú³É2mol°±Æø·Å³öÈÈÁ¿94.4kJ£¬ËùÒÔǰ20min£¬Éú³É1mol/LµÄ°±Æø£¬·Å³öµÄÈÈÁ¿ÊÇ47.2 kJ£¬¹Ê´ð°¸Îª£º0.050 mol•£¨L•min£©-1£»47.2 kJ£»
b£®¾Ýͼʾ£¬25¡æÊ±°±ÆøµÄŨ¶ÈѸËÙ¼õС£¬25minʱ²ÉÈ¡µÄ´ëÊ©ÊÇ£º½«NH3´Ó·´Ó¦ÌåϵÖзÖÀë³öÈ¥£¬¹Ê´ð°¸Îª£º½«NH3´Ó·´Ó¦ÌåϵÖзÖÀë³öÈ¥£»
c£®Æ½ºâ³£ÊýKµÈÓÚ¸÷Éú³ÉÎïÆ½ºâŨ¶ÈϵÊý´Î·½µÄ³Ë»ýºÍ¸÷·´Ó¦ÎïÆ½ºâŨ¶ÈϵÊý´Î·½Ö®»ýµÄ±ÈÖµ£¬·´Ó¦µÄƽºâ³£Êý±í´ïʽΪK=$\frac{{c}^{2}£¨N{H}_{3}£©}{c£¨{N}_{2}£©•{c}^{3}£¨{H}_{2}£©}$£¬Ê±¶ÎIIIÌõ¼þÏ´øÈëÏàÓ¦Êý¾Ý£¬·´Ó¦µÄƽºâ³£Êý±í´ïʽΪK=$\frac{0£®{5}^{2}}{0.7{5}^{3}•0.25}$£»¹Ê´ð°¸Îª£º$\frac{0£®{5}^{2}}{0.7{5}^{3}•0.25}$£»
£¨2£©a¡¢4NH3£¨g£©+5O2?4NO£¨g£©+6H2O£¨g£©£»¡÷H=-905kJ/mol¢Ù
4NH3£¨g£©+3O2£¨g£©?2N2£¨g£©+6H2O£¨g£©£»¡÷H=-1268kJ/mol¢Ú£»
ÒÀ¾Ý¸Ç˹¶¨ÂÉ¢Ù-¢ÚµÃµ½£¬2N2£¨g£©+2O2£¨g£©=4NO£¨g£©¡÷H=+363KJ/mol£¬ÔòµªÆøÑõ»¯ÎªNOµÄÈÈ»¯Ñ§·½³ÌʽΪ£ºN2£¨g£©+O2£¨g£©?2NO£¨g£©£»¡÷H=+181.5 kJ/mol£¬¹Ê´ð°¸Îª£º+181.5 kJ/mol£»
 b£®·´Ó¦¢Ù4NH2£¨g£©+5O2£¨g£©+6H2O£¨g£©¡÷H=-905kJ/mol£¬£©£»¡÷H£¼0ÇÒÊÇ»ìÂÒ¶ÈÔö´óµÄ·´Ó¦£¬ËùÒÔÈκÎÌõ¼þÏ£¬·´Ó¦¶¼×Ô·¢½øÐУ¬¹Ê´ð°¸Îª£ºÄÜ£»¡÷H£¼0ÇÒÊÇ»ìÂÒ¶ÈÔö´óµÄ·´Ó¦£»
£¨3£©ÓÉͼ¿ÉÖª£¬A¼«Í¨ÈëµÄΪ°±Æø£¬·¢ÉúÑõ»¯·´Ó¦£¬Îª¸º¼«£¬°±ÆøÔÚ¼îÐÔÌõ¼þÏ·ŵçÉú³ÉµªÆøÓëË®£¬µç¼«·´Ó¦Ê½Îª£º2NH3-6e-+6OH-=N2+6H2O£¬
¹Ê´ð°¸Îª£º¸º¼«£»2NH3-6e-+6OH-=N2+6H2O£»
£¨4£©°±±»Ë«ÑõË®Ñõ»¯ÎªÓÎÀë̬µª£¬Ã¿mol°±Ê§3molµç×Ó£¬H2O2µÃµç×ÓÉú³ÉH2O£¬Ã¿molµÃ2molµç×Ó£¬ËùÒÔ°±·Ö×Ӻ͹ýÑõ»¯Çâ·Ö×Ó»¯Ñ§¼ÆÁ¿ÊýÖ®±ÈΪ2£º3£¬ÔÙ¾ÝÔ­×ÓÊØºãÊéд»¯Ñ§·½³ÌʽΪ£º2NH3+3H2O2=N2+6H2O£¬¹Ê´ð°¸Îª£º2NH3+3H2O2=N2+6H2O£®

µãÆÀ ±¾Ì⿼²é·´Ó¦ÈȵļÆËã¡¢»¯Ñ§Æ½ºâÌõ¼þ¿ØÖÆ¡¢Ô­µç³Ø¹¤×÷Ô­ÀíµÈ£¬ÌâÄ¿½ÏΪ×ۺϣ¬ÄѶȲ»´ó£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
3£®ÏÖ´ú´«¸ÐÐÅÏ¢¼¼ÊõÔÚ»¯Ñ§ÊµÑéÖÐÓй㷺µÄÓ¦Óã®

ijС×éÓô«¸Ð¼¼Êõ²â¶¨ÅçȪʵÑéÖеÄѹǿ±ä»¯À´ÈÏʶÅçȪʵÑéµÄÔ­Àí£¨Í¼1£©£®
£¨1£©ÖÆÈ¡°±Æø£®ÉÕÆ¿ÖÐÖÆÈ¡NH3µÄ»¯Ñ§·½³ÌʽΪNH3•H2O+CaO=Ca£¨OH£©2+NH3¡ü£¬¼ìÑéÈý¾±Æ¿¼¯ÂúNH3µÄ·½·¨Êǽ«ÊªÈóµÄºìɫʯÈïÊÔÖ½¿¿½üÆ¿¿Úc£¬ÊÔÖ½±äÀ¶É«£¬Ö¤Ã÷NH3ÒÑÊÕÂú£»»ò£º½«ÕºÓÐŨÑÎËáµÄ²£Á§°ô¿¿½üÆ¿¿Úc£¬Óа×ÑÌÉú³É£¬Ö¤Ã÷NH3ÒÑÊÕÂú£®
£¨2£©¹Ø±Õa£¬½«ÎüÓÐ2mLË®µÄ½ºÍ·µÎ¹ÜÈû½ô¾±¿Úc£¬´ò¿ªb£¬Íê³ÉÅçȪʵÑ飬µçÄÔ»æÖÆÈý¾±Æ¿ÄÚÆøÑ¹±ä»¯ÇúÏߣ¨Í¼2£©£®Í¼2ÖÐDµãʱÅçȪ×î¾çÁÒ£®
²â¶¨NH3•H2OµÄŨ¶È¼°µçÀëÆ½ºâ³£ÊýKb
£¨3£©´ÓÈý¾±Æ¿ÖÐÓüîʽµÎ¶¨¹Ü£¨»òÒÆÒº¹Ü£©£¨ÌîÒÇÆ÷Ãû³Æ£©Á¿È¡25.00mL°±Ë®ÖÁ×¶ÐÎÆ¿ÖУ¬ÓÃ0.0500mol•L-1HClµÎ¶¨£®ÓÃpH¼Æ²É¼¯Êý¾Ý¡¢µçÄÔ»æÖƵζ¨ÇúÏßÈçÏÂͼ£®
£¨4£©¾Ýͼ£¬¼ÆË㰱ˮµÄŨ¶ÈΪ0.045mol•L-1£»Ð´³öNH3•H2OµçÀëÆ½ºâ³£ÊýKbµÄ±í´ïʽ£¬Kb=$\frac{c£¨O{H}^{-}£©£®c£¨N{{H}_{4}}^{+}£©}{c£¨N{H}_{3}£®{H}_{2}O£©}$£¬µ±pH=11.0ʱ¼ÆËãKbµÄ½üËÆÖµ£¬Kb¡Ö2.2¡Á10-5£®
£¨5£©¹ØÓڸõζ¨ÊµÑéµÄ˵·¨ÖУ¬ÕýÈ·µÄÊÇAC£®
A£®×¶ÐÎÆ¿ÖÐÓÐÉÙÁ¿ÕôÁóË®²»Ó°Ïì²â¶¨½á¹û
B£®Î´µÎ¼ÓËá¼îָʾ¼Á£¬ÊµÑé½á¹û²»¿ÆÑ§
C£®ËáʽµÎ¶¨¹ÜδÓÃÑÎËáÈóÏ´»áµ¼Ö²âµÃ°±Ë®µÄŨ¶ÈÆ«¸ß
D£®µÎ¶¨ÖÕµãʱ¸©ÊÓ¶ÁÊý»áµ¼Ö²âµÃ°±Ë®µÄŨ¶ÈÆ«¸ß£®

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø