Question

16£®£¨1£©°ÑAlCl₃ ÈÜÒºÕô¸ÉºóÔÙ×ÆÉÕ£¬×îºóµÃµ½µÄÖ÷Òª¹ÌÌå²úÎïÊÇAl₂O₃£¬ÆäÀíÓÉÊÇ£¨ÓÃÀë×Ó·½³Ìʽ»ò»¯Ñ§·½³Ìʽ±íʾ£©Al³⁺+3H₂O?Al£¨OH£©₃+3H⁺¡¢2Al£¨OH£©₃$\frac{\underline{\;¼ÓÈÈ\;}}{\;}$Al₂O₃+3H₂O
£¨2£©¶ÔÓÚ¿ÉÄæ·´Ó¦aA£¨g£©+bB£¨g£©?cC£¨g£©+dD£¨g£©¡÷H£¾0£¬Æ½ºâ³£ÊýKµÄ±í´ïʽΪ$\frac{{c}^{c}£¨C£©•{c}^{d}£¨D£©}{{c}^{a}£¨A£©•{c}^{b}£¨B£©}$£¬KÖ»ÊÜζÈÓ°Ï죬200¡æʱΪK₁£¬800¡æʱΪK₂£¬ÔòK₁СÓÚK₂£¨Ìî´ó»òС»òµÈ£©£®
£¨3£©ÂÈ»¯ÌúË®½âµÄÀë×Ó·½³ÌʽΪFe³⁺+3H₂O?Fe£¨OH£©₃+3H⁺£¬ÅäÖÆÂÈ»¯ÌúÈÜҺʱµÎ¼ÓÉÙÁ¿ÑÎËáµÄ×÷ÓÃÊÇÒÖÖÆ£¨Ìî¡°´Ù½ø¡±»ò¡°ÒÖÖÆ¡±£©Ë®½â£»ÈôÏòÂÈ»¯ÌúÈÜÒºÖмÓÈë̼Ëá¸Æ·ÛÄ©£¬·¢ÏÖ̼Ëá¸ÆÖð½¥Öð½¥Èܽ⣬²¢²úÉúÎÞÉ«ÆøÌ壬ͬʱÓкìºÖÉ«³ÁµíÉú³ÉµÄÔ­ÒòÊÇ£ºCaCO₃ÓëÉÏÊöË®½âƽºâÖеÄH⁺·´Ó¦£¬Ê¹Ë®½âƽºâÏòÓÒÒƶ¯£¬µ¼ÖÂFe£¨OH£©₃Ôö¶à£¬ÐγɺìºÖÉ«³Áµí£®
£¨4£©Ò»¶¨Ìõ¼þϵĿÉÄæ·´Ó¦2NO₂ £¨g£©¨PN₂O₄£¨g£©¡÷H=-92.4kJ/mol´ïµ½»¯Ñ§Æ½ºâ״̬ÇÒÆäËüÌõ¼þ²»±äʱ£¬
¢ÙÈç¹ûÉý¸ßζȣ¬Æ½ºâ»ìºÏÎïµÄÑÕÉ«±äÉÌî¡°±äÉ»ò¡°±ädz¡±£©£»
¢ÚÈç¹ûÔÚÌå»ý¹Ì¶¨µÄÈÝÆ÷ÖмÓÈëÒ»¶¨Á¿µÄ¶þÑõ»¯µª£¬»¯Ñ§Æ½ºâÏòÕý·´Ó¦  ·½ÏòÒƶ¯£®

Answer 1

·ÖÎö £¨1£©¸ù¾ÝAlCl₃ÊôÓÚÇ¿ËáÈõ¼îÑΣ¬ÔÚÈÜÒºÖдæÔÚË®½âƽºâ£¬´ÓƽºâÒƶ¯µÄ½Ç¶È·ÖÎö²¢½â´ð¸ÃÌ⣻
£¨2£©Æ½ºâ³£ÊýK=$\frac{Éú³ÉÎïƽºâŨ¶ÈµÄÃÝÖ®»ý}{·´Ó¦ÎïƽºâŨ¶ÈÃÝÖ®»ý}$£¬KÖ»ÊÜζȵÄÓ°Ï죬´Ë·´Ó¦ÎüÈÈ£¬Éý¸ßζȣ¬Æ½ºâÓÒÒÆ£¬KÔö´ó£»
£¨3£©ÂÈ»¯ÌúΪǿËáÈõ¼îÑΣ¬Ë®½â³ÊËáÐÔ£»¼ÓËáÒÖÖÆÇ¿ËáÈõ¼îÑεÄË®½â£»ÂÈ»¯ÌúΪǿËáÈõ¼îÑΣ¬Ë®½â³ÊËáÐÔ£¬CaCO₃ÓëÉÏÊöË®½âƽºâÖеÄH⁺·´Ó¦Ë®½âƽºâÓÒÒÆ£»
£¨4£©¢ÙÕý·´Ó¦·ÅÈÈ£¬Éý¸ßζÈƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£»
¢ÚÔö´ó·´Ó¦µÄŨ¶È£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£®

½â´ð ½â£º£¨1£©ÂÈ»¯ÂÁΪǿËáÈõ¼îÑΣ¬Al³⁺·¢ÉúË®½â£¬Ë®½âµÄ·½³ÌʽΪAl³⁺+3H₂O?Al£¨OH£©₃+3H⁺£¬Ë®½âºóÈÜÒº³ÊËáÐÔ£¬Õô¸ÉºÍ×ÆÉÕ¹ý³ÌÖУ¬HCl»Ó·¢£¬Al£¨OH£©₃²»Îȶ¨£¬×ÆÉÕʱ·Ö½âÉú³ÉAl₂O₃£º2Al£¨OH£©₃$\frac{\underline{\;¼ÓÈÈ\;}}{\;}$Al₂O₃+3H₂O£®
´ð£ºAl₂O₃£»Al³⁺+3H₂O?Al£¨OH£©₃+3H⁺£¬2Al£¨OH£©₃$\frac{\underline{\;¼ÓÈÈ\;}}{\;}$Al₂O₃+3H₂O£»
£¨2£©Æ½ºâ³£ÊýK=$\frac{Éú³ÉÎïƽºâŨ¶ÈµÄÃÝÖ®»ý}{·´Ó¦ÎïƽºâŨ¶ÈÃÝÖ®»ý}$=$\frac{{c}^{c}£¨C£©•{c}^{d}£¨D£©}{{c}^{a}£¨A£©•{c}^{b}£¨B£©}$£¬KÖ»ÊÜζȵÄÓ°Ï죬´Ë·´Ó¦ÎüÈÈ£¬Éý¸ßζȣ¬Æ½ºâÓÒÒÆ£¬KÔö´ó£¬¹ÊK₁СÓÚK₂£®¹Ê´ð°¸Îª£º$\frac{{c}^{c}£¨C£©•{c}^{d}£¨D£©}{{c}^{a}£¨A£©•{c}^{b}£¨B£©}$£»Î¶ȣ»Ð¡£»
£¨3£©FeCl₃ΪǿËáÈõ¼îÑΣ¬Fe³⁺Àë×ÓË®½âʹÈÜÒº³ÊËáÐÔ£¬Ë®½â·½³ÌʽΪFe³⁺+3H₂O?Fe£¨OH£©₃+3H⁺£»FeCl₃ΪǿËáÈõ¼îÑΣ¬Ë®½âÏÔËáÐÔ£¬¼ÓÈëÑÎËáÄÜÒÖÖÆÆäË®½â£»
ÂÈ»¯ÌúË®½â³ÊËáÐÔFe³⁺+3H₂O?Fe£¨OH£©₃+3H⁺£¬CaCO₃ÓëÉÏÊöË®½âƽºâÖеÄH⁺·´Ó¦£¬·´Ó¦ÏûºÄH⁺£¬Ê¹c£¨H⁺£©¼õС£¬ÒýÆðË®½âƽºâÏòÓÒÒƶ¯£¬µ¼ÖÂFe£¨OH£©₃Ôö¶à£¬ÐγɺìºÖÉ«³Áµí£®
¹Ê´ð°¸Îª£ºCaCO₃+2H⁺=Ca²⁺+H₂O+CO₂¡ü£» H⁺£¬ÓÒ£»
£¨4£©¢ÙÕý·´Ó¦Îª·ÅÈÈ·´Ó¦£¬Éý¸ßζÈƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬ÔòÑÕÉ«±äÉ¹Ê´ð°¸Îª£º±äÉ
¢ÚÈç¹ûÔÚÌå»ý¹Ì¶¨ÈÝÆ÷ÖмÓÈëÒ»¶¨Á¿µÄ¶þÑõ»¯µª£¬Ôö´ó·´Ó¦µÄŨ¶È£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬¹Ê´ð°¸Îª£ºÕý·´Ó¦£®

µãÆÀ ±¾Ì⿼²éƽºâÒƶ¯µÄÓ°ÏìÒòËØ£¬ÌâÄ¿ÄѶȲ»´ó£¬±¾Ìâ×¢Òâ°ÑÎÕƽºâÒƶ¯Ô­Àí£®