ÌâÄ¿ÄÚÈÝ

5£®ÑÇÁòËáÑÎÊÇÒ»ÖÖ³£¼ûµÄʳƷÌí¼Ó¼Á£®ÓÃÈçͼʵÑé¿É¼ìÑéijʳƷÖÐÑÇÁòËáÑκ¬Á¿£¨º¬Á¿Í¨³£ÒÔ1kgÑùÆ·Öк¬SO2µÄÖÊÁ¿¼Æ£»Ëù¼ÓÊÔ¼Á¾ù×ãÁ¿£©£¬ÏÂÁÐ˵·¨²»ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®ÑÇÁòËáÑÎ×÷ΪʳƷÌí¼Ó¼Á×÷ÓÃÊÇ·À¸¯¡¢¿¹Ñõ»¯
B£®·´Ó¦¢ÙÖÐͨÈëN2µÄ×÷ÓÃÊǽ«Éú³ÉµÄÆøÌåÈ«²¿¸Ï³ö
C£®²â¶¨ÑùÆ·ÖÊÁ¿¼°¢ÛÖкļîÁ¿£¬¿É²â¶¨ÑùÆ·ÖÐÑÇÁòËáÑκ¬Á¿
D£®Èô½ö½«¢ÚÖеÄÑõ»¯¼Á¡°H2O2ÈÜÒº¡±Ì滻ΪµâË®£¬¶Ô²â¶¨½á¹ûÎÞÓ°Ïì

·ÖÎö ÑùÆ·ÖмÓÏ¡ÁòËáÉú³É¶þÑõ»¯Áò£¬Í¨µªÆø½«Éú³ÉµÄ¶þÑõ»¯Áò´ÓÈÜÒºÖÐÈ«²¿¸Ï³ö£¬µÃµ½ÆøÌåAΪµªÆøºÍ¶þÑõ»¯ÁòµÄ»ìºÏÆøÌ壬ͨÈëË«ÑõË®ÖУ¬¶þÑõ»¯Áò±»Ñõ»¯³ÉÁòËᣬÔÙÓÃÇâÑõ»¯ÄÆÖк͵ÃÖкÍÒºº¬ÓÐÁòËáÄÆ£¬
A£®ÑÇÁòËáÑÎÓл¹Ô­ÐÔ£¬¿ÉÒÔ·ÀֹʳƷ±»¿ÕÆøÖÐÑõÆøÑõ»¯£»
B£®¸ù¾ÝÉÏÃæµÄ·ÖÎö¿ÉÖª£¬·´Ó¦¢ÙÖÐͨÈëN2µÄ×÷ÓÃÊǽ«Éú³ÉµÄÆøÌåÈ«²¿¸Ï³ö£»
C£®¸ù¾Ý¢ÛÖкļîÁ¿¿ÉÒÔ¼ÆËã³öÉú³ÉµÄÁòËáµÄÎïÖʵÄÁ¿£¬ÀûÓÃÁòÔªËØÊØºã¿ÉÖª¶þÑõ»¯ÁòµÄÖÊÁ¿£¬½áºÏÑùÆ·µÄÖÊÁ¿¿ÉÇóµÃÑùÆ·ÖÐÑÇÁòËáÑκ¬Á¿£»
D£®H2O2ÈÜÒºÌæ»»ÎªµâË®£¬Èç¹ûµâ¹ýÁ¿£¬µâÓëÇâÑõ»¯ÄÆ·´Ó¦£¬ËùÒÔÎÞ·¨¸ù¾ÝÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿È·¶¨Éú³ÉµÄÁòËáµÄÎïÖʵÄÁ¿£®

½â´ð ½â£ºÑùÆ·ÖмÓÏ¡ÁòËáÉú³É¶þÑõ»¯Áò£¬Í¨µªÆø½«Éú³ÉµÄ¶þÑõ»¯Áò´ÓÈÜÒºÖÐÈ«²¿¸Ï³ö£¬µÃµ½ÆøÌåAΪµªÆøºÍ¶þÑõ»¯ÁòµÄ»ìºÏÆøÌ壬ͨÈëË«ÑõË®ÖУ¬¶þÑõ»¯Áò±»Ñõ»¯³ÉÁòËᣬÔÙÓÃÇâÑõ»¯ÄÆÖк͵ÃÖкÍÒºº¬ÓÐÁòËáÄÆ£¬
A£®ÑÇÁòËáÑÎÓл¹Ô­ÐÔ£¬¿ÉÒÔ·ÀֹʳƷ±»¿ÕÆøÖÐÑõÆøÑõ»¯£¬Æðµ½ÁË·À¸¯¡¢¿¹Ñõ»¯×÷Ó㬹ÊAÕýÈ·£»
B£®¸ù¾ÝÉÏÃæµÄ·ÖÎö¿ÉÖª£¬·´Ó¦¢ÙÖÐͨÈëN2µÄ×÷ÓÃÊǽ«Éú³ÉµÄÆøÌåÈ«²¿¸Ï³ö£¬¹ÊBÕýÈ·£»
C£®¸ù¾Ý¢ÛÖкļîÁ¿¿ÉÒÔ¼ÆËã³öÉú³ÉµÄÁòËáµÄÎïÖʵÄÁ¿£¬ÀûÓÃÁòÔªËØÊØºã¿ÉÖª¶þÑõ»¯ÁòµÄÖÊÁ¿£¬½áºÏÑùÆ·µÄÖÊÁ¿¿ÉÇóµÃÑùÆ·ÖÐÑÇÁòËáÑκ¬Á¿£¬¹ÊCÕýÈ·£»
D£®H2O2ÈÜÒºÌæ»»ÎªµâË®£¬Èç¹ûµâ¹ýÁ¿£¬µâÓëÇâÑõ»¯ÄÆ·´Ó¦£¬ËùÒÔÎÞ·¨¸ù¾ÝÇâÑõ»¯ÄƵÄÎïÖʵÄÁ¿È·¶¨Éú³ÉµÄÁòËáµÄÎïÖʵÄÁ¿£¬¹ÊD´íÎó£»
¹ÊÑ¡D£®

µãÆÀ ±¾Ì⿼²éÁËʵÑé·½°¸µÄÉè¼ÆÓëÆÀ¼Û£¬ÌâÄ¿ÄѶÈÖеȣ¬¸ù¾ÝʵÑéÁ÷³ÌÃ÷ȷʵÑéÔ­ÀíΪ½â´ð¹Ø¼ü£¬×¢ÒâÕÆÎÕÑÇÏõËáÑεÄÐÔÖʼ°ÐÔÖÊ·½°¸Éè¼ÆÓëÆÀ¼ÛÔ­Ôò£¬ÊÔÌâÅàÑøÁËѧÉúµÄ·ÖÎöÄÜÁ¦¼°»¯Ñ§ÊµÑéÄÜÁ¦£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
13£®¿ª·¢ÐÂÐÍ´¢Çâ²ÄÁÏÊÇÇâÄÜÀûÓõÄÖØÒªÑо¿·½Ïò£®
£¨1£©Ti£¨BH4£©3ÊÇÒ»ÖÖ´¢Çâ²ÄÁÏ£¬¿ÉÓÉTiCl4ºÍLiBH4·´Ó¦ÖƵã®
¢Ù»ù̬ԭ×ÓTiÓÐ7ÖÖÄÜÁ¿²»Í¬µÄµç×Ó£¬»ù̬Ti3+µÄδ³É¶Ôµç×ÓÓÐ1¸ö£®
¢ÚLiBH4ÓÉLi+ºÍBH4-¹¹³É£¬BH4-µÄÁ¢Ìå¹¹ÐÍÊÇsp3ÔÓ»¯£¬LiBH4Öв»´æÔÚµÄ×÷ÓÃÁ¦
ÓÐc£¨Ìî×Öĸ£©£®a£®Àë×Ó¼ü¡¡¡¡b£®¹²¼Û¼ü¡¡¡¡c£®½ðÊô¼ü¡¡¡¡d£®Åäλ¼ü
¢ÛLi¡¢B¡¢HÔªËØµÄµç¸ºÐÔÓÉ´óµ½Ð¡ÅÅÁÐ˳ÐòΪH£¾B£¾Li£®
£¨2£©½ðÊôÇ⻯ÎïÊǾßÓÐÁ¼ºÃ·¢Õ¹Ç°¾°µÄ´¢Çâ²ÄÁÏ£®
¢ÙLiHÖУ¬Àë×Ó°ë¾¶Li+£¼H-£¨Ìî¡°£¾¡±¡¢¡°=¡±»ò¡°£¼¡±£©£®
¢Úij´¢Çâ²ÄÁÏÊǶÌÖÜÆÚ½ðÊôÔªËØMµÄÇ⻯ÎMµÄ²¿·ÖµçÀëÄÜÈçϱíËùʾ£º
I1I2I3I4I5
I/kJ•mol-17381 4517 73310 54013 630
MÊÇMg£¨ÌîÔªËØ·ûºÅ£©£®
£¨3£©°±ÅðÍ飨NH3BH3£©ÓëïçÄøºÏ½ð£¨LaNix£©¶¼ÊÇÓÅÁ¼µÄ´¢Çâ²ÄÁÏ£®
¢ÙH3BNH3µÄµÈµç×ÓÌåµÄ»¯Ñ§Ê½ÎªC2H6£®
¢ÚïçÄøºÏ½ðµÄ¾§°û½á¹¹Ê¾ÒâͼÈçͼËùʾ£¨Ö»ÓÐ1¸öÔ­×ÓλÓÚ¾§°ûÄÚ²¿£©£¬Ôòx=5£®°±ÅðÍéÔÚ¸ßÎÂÏÂÊÍ·ÅÇâºóÉú³ÉµÄÁ¢·½µª»¯Åð¾§Ì壬¾ßÓÐÀàËÆ½ð¸ÕʯµÄ½á¹¹£¬Ó²¶ÈÂÔСÓÚ½ð¸Õʯ£®ÔòÔÚÏÂÁи÷ÏîÖУ¬Á¢·½µª»¯Åð¾§Ìå²»¿ÉÓÃ×÷c£¨Ìî×Öĸ£©£®
a£®ÄÍÄ¥²ÄÁÏ¡¡    b£®ÇÐÏ÷¹¤¾ß¡¡   c£®µ¼µç²ÄÁÏ¡¡    d£®×ê̽×êÍ·£®
17£®Èçͼ1±íʾµª¼°Æä»¯ºÏÎïÔÚÒ»¶¨Ìõ¼þϵÄת»¯¹ØÏµ£º

£¨1£©·´Ó¦I£ºN2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H=-92kJ•mol-1
ͼ2ÊÇ·´Ó¦IÖÐÆ½ºâ»ìºÏÆøÖÐNH3µÄÌå»ý·ÖÊýËæÎ¶Ȼòѹǿ±ä»¯µÄÇúÏߣ¬Í¼ÖÐL£¨L1¡¢L2£©¡¢X·Ö±ð´ú±íζȻòѹǿ£®ÆäÖÐX´ú±íµÄÊÇѹǿ£¨Ìζȡ±»ò¡°Ñ¹Ç¿¡±£©£¬ÅжÏL1¡¢L2µÄ´óС¹ØÏµ²¢ËµÃ÷ÀíÓÉL1£¼L2 ºÏ³É°±µÄ·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬Ñ¹Ç¿Ïàͬʱ£¬Î¶ÈÉý¸ß£¬Æ½ºâÄæÏòÒÆ¶¯£¬°±µÄÌå»ý·ÖÊý¼õС£®
£¨2£©¢Ù·´Ó¦IIµÄ»¯Ñ§·½³ÌʽÊÇ8NH3+3Cl2=N2+6NH4Cl£®
¢Ú·´Ó¦IIÆäÖÐÒ»²½·´Ó¦Îª
2NH3£¨g£©+3Cl2£¨g£©¨TN2£¨g£©+6HCl£¨g£©¡÷H=-462kJ•mol-1
ÒÑÖª£ºN2£¨g£©$\stackrel{945kJ•mol-1}{¡ú}$2N£¨g£©    Cl2£¨g£©$\stackrel{243kJ•mol-1}{¡ú}$2Cl£¨g£©
¶Ï¿ª1mol H-N¼üÓë¶Ï¿ª1mol H-Cl¼üËùÐèÄÜÁ¿Ïà²îԼΪ41 kJ£®
£¨3£©·´Ó¦IIIÊÇÀûÓÃͼ3ËùʾװÖõç½âÖÆ±¸NCl3£¨ÂȵϝºÏ¼ÛΪ+1£©£¬ÆäÔ­ÀíÊÇ£º
NH4Cl+2HCl$\frac{\underline{\;µç½â\;}}{\;}$NCl3+3H2¡ü£®
¢Ùb½ÓµçÔ´µÄ¸º¼«£¨Ìî¡°Õý¡±»ò¡°¸º¡±£©£®
¢ÚÑô¼«·´Ó¦Ê½ÊÇ3Cl--6e-+NH4+=NCl3+4H+£®
£¨4£©·´Ó¦IIIµÃµ½µÄNCl3¿ÉÒÔºÍNaClO2ÖÆ±¸ ClO2£¬Í¬Ê±Éú³ÉNH3£¬¸Ã·´Ó¦µÄÀë×Ó·½³ÌʽÊÇNCl3+6ClO2-+3H2O=6ClO2+NH3+3Cl-+3OH-£®

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

¾«Ó¢¼Ò½ÌÍø