ÌâÄ¿ÄÚÈÝ

4£®£¨1£©¹¤ÒµÉÏ£¬ÒÔCO2ºÍNH3ΪԭÁÏÔÚÒ»¶¨Ìõ¼þϺϳÉÄòËصĻ¯Ñ§·½³ÌʽΪ£º
2NH3£¨g£©+CO2£¨g£©?CO£¨NH2£©2£¨s£©+H2O£¨g£©
д³ö¸Ã·´Ó¦Æ½ºâ³£ÊýµÄ±í´ïʽ$\frac{c£¨{H}_{2}O£©}{{c}^{2}£¨N{H}_{3}£©c£¨C{O}_{2}£©}$£»ÆäËûÌõ¼þ²»±äʱÔö´óѹǿ£¬»ìºÏÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿ÎÞ·¨È·¶¨£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±¡¢¡°²»±ä¡±»ò¡°ÎÞ·¨È·¶¨¡±£©£®
£¨2£©ºÏ³É°±µÄÔ­ÁÏ֮һΪÇâÆø£¬ÒÔÌìÈ»ÆøΪȼÁϺϳÉÇâÆøµÄÔ­ÀíÈçÏ£º
¢ÙCH4£¨g£©+H2O£¨g£©?CO£¨g£©+3H2£¨g£©¡÷H=+206.4 kJ•mol-1
¢ÚCO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£©¡÷H=-41.2 kJ•mol-1
Ôò¼×ÍéÓëË®ÕôÆû·´Ó¦Éú³É¶þÑõ»¯Ì¼ºÍÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽΪCH4£¨g£©+2H2O£¨g£©?CO2£¨g£©+4H2£¨g£©¡÷H=+165.2kJ•mol-1£®
£¨3£©¶ÔÓÚ£¨2£©Öеķ´Ó¦¢Ù£¬ÏÂÁдëÊ©Ò»¶¨ÄÜʹƽºâÌåϵÖеÄH2Ìå»ý·ÖÊýÔö´óµÄÊÇAD
A£®Éý¸ßζÈ
B£®Ôö´óË®ÕôÆûŨ¶È
C£®¼ÓÈë¸ßЧ´ß»¯¼Á
D£®¼õСѹǿ£®

·ÖÎö £¨1£©»¯Ñ§·´Ó¦µÄƽºâ³£ÊýK=$\frac{Éú³ÉÎïƽºâŨ¶ÈÃݴη½³Ë»ý}{·´Ó¦ÎïƽºâŨ¶ÈÃݴη½³Ë»ý}$£¬ÆäËûÌõ¼þ²»±äʱÔö´óѹǿ£¬Æ½ºâÏòÆøÌåÌå»ý¼õСµÄ·½Ïò½øÐУ¬¸Ã·´Ó¦ÊÇÆøÌåÌå»ý½øÐеķ´Ó¦ÆøÌåÖÊÁ¿Ò²¼õС£¬ÔòM=$\frac{m}{n}$±ÈÖµÎÞ·¨È·¶¨£»
£¨2£©ÒÀ¾ÝÈÈ»¯Ñ§·½³ÌʽºÍ¸Ç˹¶¨ÂɼÆËãµÃµ½ËùÐèÈÈ»¯Ñ§·½³Ìʽ£¬¢Ù+¢ÚµÃµ½£»
£¨3£©¢ÙCH4£¨g£©+H2O£¨g£©?CO£¨g£©+3H2£¨g£©¡÷H=+206.4 kJ•mol-1
·´Ó¦ÊÇÆøÌåÌå»ýÔö´óµÄÎüÈÈ·´Ó¦£¬Ò»¶¨ÄÜʹƽºâÌåϵÖеÄH2Ìå»ý·ÖÊýÔö´ó£¬Æ½ºâÕýÏò½øÐУ¬ÇÒ²»ÄÜÔö´óÇâÆø±¾ÉíµÄÁ¿£¬½áºÏƽºâÒƶ¯Ô­Àí·ÖÎöÅжÏÑ¡Ï

½â´ð ½â£º£¨1£©2NH3£¨g£©+CO2£¨g£©?CO£¨NH2£©2£¨s£©+H2O£¨g£©
»¯Ñ§·´Ó¦µÄƽºâ³£ÊýK=$\frac{Éú³ÉÎïƽºâŨ¶ÈÃݴη½³Ë»ý}{·´Ó¦ÎïƽºâŨ¶ÈÃݴη½³Ë»ý}$=$\frac{c£¨{H}_{2}O£©}{{c}^{2}£¨N{H}_{3}£©c£¨C{O}_{2}£©}$£¬ÆäËûÌõ¼þ²»±äʱÔö´óѹǿ£¬Æ½ºâÏòÆøÌåÌå»ý¼õСµÄ·½Ïò½øÐУ¬¸Ã·´Ó¦ÊÇÆøÌåÌå»ý½øÐеķ´Ó¦ÆøÌåÖÊÁ¿Ò²¼õС£¬ÔòM=$\frac{m}{n}$±ÈÖµÎÞ·¨È·¶¨£¬
¹Ê´ð°¸Îª£º$\frac{c£¨{H}_{2}O£©}{{c}^{2}£¨N{H}_{3}£©c£¨C{O}_{2}£©}$£¬ÎÞ·¨È·¶¨£»
£¨2£©¢ÙCH4£¨g£©+H2O£¨g£©?CO£¨g£©+3H2£¨g£©¡÷H=+206.4 kJ•mol-1
¢ÚCO£¨g£©+H2O£¨g£©?CO2£¨g£©+H2£¨g£©¡÷H=-41.2 kJ•mol-1
Ôò¼×ÍéÓëË®ÕôÆû·´Ó¦Éú³É¶þÑõ»¯Ì¼ºÍÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽÒÀ¾ÝÈÈ»¯Ñ§·½³ÌʽºÍ¸Ç˹¶¨ÂɼÆËã¢Ù+¢ÚµÃµ½ËùÐèÈÈ»¯Ñ§·½³Ìʽ£¬CH4£¨g£©+2H2O£¨g£©?CO2£¨g£©+4H2£¨g£©¡÷H=+165.2kJ•mol-1£¬
¹Ê´ð°¸Îª£ºCH4£¨g£©+2H2O£¨g£©?CO2£¨g£©+4H2£¨g£©¡÷H=+165.2kJ•mol-1£»
£¨3£©¢ÙCH4£¨g£©+H2O£¨g£©?CO£¨g£©+3H2£¨g£©¡÷H=+206.4 kJ•mol-1
·´Ó¦ÊÇÆøÌåÌå»ýÔö´óµÄÎüÈÈ·´Ó¦£¬Ò»¶¨ÄÜʹƽºâÌåϵÖеÄH2Ìå»ý·ÖÊýÔö´ó£¬Æ½ºâÕýÏò½øÐУ¬ÇÒ²»ÄÜÔö´óÇâÆø±¾ÉíµÄÁ¿£¬½áºÏƽºâÒƶ¯Ô­Àí·ÖÎöÅжÏÑ¡Ï
A£®·´Ó¦ÊÇÎüÈÈ·´Ó¦£¬Éý¸ßζÈƽºâÕýÏò½øÐУ¬H2Ìå»ý·ÖÊýÔö´ó£¬¹ÊAÕýÈ·£»
B£®Ôö´óË®ÕôÆûŨ¶È£¬»áÌá¸ß¼×Íéת»¯ÂÊ£¬µ«ÇâÆøµÄÌå»ý·ÖÊý²»Ò»¶¨Ôö´ó£¬¹ÊB´íÎó£»
C£®¼ÓÈë¸ßЧ´ß»¯¼Á¸Ä±ä·´Ó¦ËÙÂÊ£¬²»¸Ä±ä»¯Ñ§Æ½ºâ£¬ÇâÆøº¬Á¿²»±ä£¬¹ÊC´íÎó£»
D£®¼õСѹǿ£¬Æ½ºâÕýÏò½øÐУ¬ÇâÆøÌå»ý·ÖÊýÔö´ó£¬¹ÊDÕýÈ·£»
¹ÊÑ¡AD£®

µãÆÀ ±¾Ì⿼²éÁËÈÈ»¯Ñ§·½³ÌʽÊéд·½·¨£¬»¯Ñ§Æ½ºâÓ°ÏìÒòËØ·ÖÎö£¬Æ½ºâ³£Êý¸ÅÄîµÄ¼ÆËãÓ¦Óã¬ÕÆÎÕ»ù´¡Êǹؼü£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø