ÌâÄ¿ÄÚÈÝ

6£®½ñÄêËÄ¡¢ÎåÔ·ݣ¬ÎÒ¹úºþ±±¡¢½­Î÷µÈµØÔâÓö´ó·¶Î§³ÖÐø¸Éºµ----Û¶Ñôºþ¸ÉºÔµÈ£®È«¹úÈËÃÇÍŽáÒ»Ö£¬¹²Í¬¿¹ºµ£®
£¨1£©ÓÐЩ´åׯ´òÉȡÓõØÏÂË®£®¼ìÑéµØÏÂË®ÊÇӲˮ»¹ÊÇÈíË®£¬¿ÉÓõÄÎïÖÊÊÇ·ÊÔíË®£®
£¨2£©ÓÐЩ´åÃñÈ¡»ë×ǵĿÓË®×öÉú»îÓÃË®£®ÓÐͬѧÀûÓÃËùѧµÄ֪ʶ½«»ë×ǵĿÓË®½áºÏÈçͼËùʾµÄ¼òÒ×¾»Ë®Æ÷½øÐо»»¯£¬ÆäÖÐСÂÑʯºÍʯӢɰµÄ×÷ÓÃÊǹýÂË£®
£¨3£©Èç¹ûµØÏÂˮӲ¶È´ó£¬»òÕß¿ÓË®ÖÐԭ΢ÉúÎï¹ý¶à£¬¶¼¿ÉÒÔ²ÉÈ¡¼ÓÈÈÖó·Ð µÄ·½·¨£¬À´½µµÍË®µÄÓ²¶ÈºÍɱÃð²¡Ô­Î¢ÉúÎ

·ÖÎö £¨1£©Ó²Ë®Óë·ÊÔíË®»ìºÏ²úÉú´óÁ¿µÄ¸¡Ôü£¬ÈíË®Óë·ÊÔíË®»ìºÏ²úÉú½Ï¶àµÄÅÝÄ­£»
£¨2£©ÔÚ¼òÒ×¾»Ë®Æ÷ÖУ¬Ð¡ÂÑʯºÍʯӢɰÄÜ×èÖ¹²»ÈÜÐÔ¹ÌÌå¿ÅÁ£Í¨¹ý£¬Æðµ½Á˹ýÂ˵Ä×÷Ó㬻îÐÔÌ¿µÄ×÷ÓÃÊÇÎü¸½£»
£¨3£©Ó²Ë®Öк¬Óн϶àµÄ¿ÉÈÜÐÔ¸Æþ»¯ºÏÎ¼ÓÈȺóÄÜ·Ö½âÉú³É²»ÈÜÐÔ¸Æþ»¯ºÏÎ

½â´ð ½â£º£¨1£©Ó²Ë®Óë·ÊÔíË®»ìºÏ²úÉú´óÁ¿µÄ¸¡Ôü£¬ÈíË®Óë·ÊÔíË®»ìºÏ²úÉú½Ï¶àµÄÅÝÄ­£¬¼ìÑéµØÏÂË®ÊÇӲˮ»¹ÊÇÈíË®£¬¿ÉÓõÄÎïÖÊÊÇ·ÊÔíË®£¬¹Ê´ð°¸Îª£º·ÊÔíË®£»
£¨2£©ÔÚ¼òÒ×¾»Ë®Æ÷ÖУ¬Ð¡ÂÑʯºÍʯӢɰÄÜ×èÖ¹²»ÈÜÐÔ¹ÌÌå¿ÅÁ£Í¨¹ý£¬Æðµ½Á˹ýÂ˵Ä×÷Ó㬹ʴð°¸Îª£º¹ýÂË£»
£¨3£©Ó²Ë®Öк¬Óн϶àµÄ¿ÉÈÜÐÔ¸Æþ»¯ºÏÎ¼ÓÈȺóÄÜ·Ö½âÉú³É²»ÈÜÐÔ¸Æþ»¯ºÏÎ¼ÓÈÈ¿ÉÒÔɱËÀϸ¾ú΢ÉúÎ¹ÊÈç¹ûµØÏÂˮӲ¶È´ó£¬»òÕß¿ÓË®Öв¡Ô­Î¢ÉúÎï¹ý¶à£¬¶¼¿ÉÒÔ²ÉÈ¡¼ÓÈÈÖó·Ð·½·¨£¬¹Ê´ð°¸Îª£º¼ÓÈÈÖó·Ð£®

µãÆÀ ±¾Ì⿼²éË®µÄ¾»»¯£¬ÄѶȲ»´ó£¬×¢ÒâӲˮÖк¬Óн϶àµÄ¿ÉÈÜÐÔ¸Æþ»¯ºÏÎ¼ÓÈȺóÄÜ·Ö½âÉú³É²»ÈÜÐÔ¸Æþ»¯ºÏÎ

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
17£®ÌìÈ»ÆøÊÇÒ»ÖÖÖØÒªµÄÇå½àÄÜÔ´ºÍ»¯¹¤Ô­ÁÏ£¬ÆäÖ÷Òª³É·ÖΪ¼×Í飮
£¨1£©ÒÑÖª£º¢ÙCH4£¨g£©+2O2 =CO2£¨g£©+2H2O£¨g£©¡÷H=-890.3 kJ•mol-1
          ¢ÚCO£¨g£©+H2O£¨g£©=H2£¨g£©+CO2£¨g£©¡÷H+2.8 kJ•mol-1
          ¢Û2CO£¨g£©+O2£¨g£©=2CO2£¨g£©¡÷H=-566.0 kJ•mol-1
  Ôò·´Ó¦CH4£¨g£©+CO2£¨g£©?2CO+2H2£¨g£©¡÷H=+247.3kJ•mol-1
£¨2£©ÌìÈ»ÆøµÄÒ»¸öÖØÒªÓÃ;ÊÇÖÆÈ¡H2£¬ÆäÔ­ÀíΪ£ºCH4£¨g£©+CO2£¨g£©?2CO£¨g£©+2H2£¨g£©£®Ò»¶¨Ìõ¼þÏ£¬ÔÚÃܱÕÈÝÆ÷ÖУ¬Í¨ÈëÎïÖʵÄÁ¿Å¨¶È¾ùΪ0.1mol•L-1µÄCH4ÓëCO2£¬ÔÚ·¢Éú·´Ó¦£¬²âµÃCH4µÄƽºâת»¯ÂÊÓëζȼ°Ñ¹Ç¿µÄ¹ØϵÈçͼlËùʾ£¬

¢Ù¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ$\frac{{c}^{2}£¨CO£©•{c}^{2}£¨{H}_{2}£©}{c£¨C{O}_{2}£©•c£¨C{H}_{4}£©}$
¢ÚÔòѹǿP1СÓÚP2¡¡£¨Ìî¡°´óÓÚ¡±»ò¡°Ð¡ÓÚ¡±£©£»Ñ¹Ç¿ÎªP2ʱ£¬ÔÚYµã£ºv£¨Õý£©´óÓÚv£¨Ä棩£¨Ìî¡°´óÓÚ¡°¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡°£©£®
£¨3£©ÌìÈ»ÆøÖеÄÉÙÁ¿H2SÔÓÖʳ£Óð±Ë®ÎüÊÕ£¬²úÎïΪNH4HS£®Ò»¶¨Ìõ¼þÏÂÏòNH4HSÈÜÒºÖÐͨÈë¿ÕÆø£¬µÃµ½µ¥ÖÊÁò²¢Ê¹ÎüÊÕÒºÔÙÉú£¬Ð´³öÔÙÉú·´Ó¦µÄ»¯Ñ§·½³Ìʽ2NH4HS+O2=2NH3•H2O+2S¡ý£®
£¨4£©¹¤ÒµÉÏÓÖ³£½«´ÓÌìÈ»Æø·ÖÀë³öµÄH2S£¬ÔÚ¸ßÎÂÏ·ֽâÉú³ÉÁòÕôÆøºÍH2£¬Èô·´Ó¦ÔÚ²»Í¬Î¶ÈÏ´ﵽƽºâʱ£¬»ìºÏÆøÌåÖи÷×é·ÖµÄÌå»ý·ÖÊýÈçͼ2Ëùʾ£¬H2SÔÚ¸ßÎÂÏ·ֽⷴӦµÄ»¯Ñ§·½³ÌʽΪ2H2S?2H2+S2
£¨5£©¿Æѧ¼ÒÓõª»¯ïزÄÁÏÓëÍ­×é×°Èçͼ3µÄÈ˹¤¹âºÏϵͳ£¬ÀûÓøÃ×°Öóɹ¦µØʵÏÖÁËÒÔCO2ºÍH2OºÏ³ÉCH4£®
¢Ùд³öÍ­µç¼«±íÃæµÄµç¼«·´Ó¦Ê½CO2+8e-+8H+=CH4+2H2O£®
¢ÚΪÌá¸ß¸ÃÈ˹¤¹âºÏϵͳµÄ¹¤×÷ЧÂÊ£¬¿ÉÏò×°ÖÃÖмÓÈëÉÙÁ¿ÁòËᣨѡÌî¡°ÑÎËᡱ»ò¡°ÁòËᡱ£©£®

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

¾«Ó¢¼Ò½ÌÍø