ÌâÄ¿ÄÚÈÝ

19£®ÒÑÖª1molCOÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌå·Å³ö283kJÈÈÁ¿£»1molÇâÆøÍêȫȼÉÕÉú³ÉÆø̬ˮ·Å³ö242kJÈÈÁ¿£»1molCH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍҺ̬ˮ·Å³ö890kJÈÈÁ¿£®1molҺ̬ˮÕô·¢ÎªÆø̬ˮÊÇÒªÎüÊÕ44kJÈÈÁ¿£®
£¨1£©Ð´³öÄܱíʾCOȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽCO£¨g £©+$\frac{1}{2}$ O2£¨g £©=CO2£¨g£©¡÷H=-283 kJ/mol£®
£¨2£©Èô1molCH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍË®ÕôÆø£¬·Å³öÈÈÁ¿Îª£¼890kJ£¨Ìî¡°£¾¡±¡¢¡°£¼¡±¡¢¡°=¡±£©£®
£¨3£©È¼ÉÕ10gH2Éú³ÉҺ̬ˮ£¬·Å³öµÄÈÈÁ¿Îª1430KJ£®
£¨4£©ÇâÆøµÄȼÉÕÈÈΪ286 kJ/mol£®
£¨5£©Ð´³öCH4ÆøÌå²»ÍêȫȼÉÕÉú³ÉÖ»Éú³ÉCOÆøÌåºÍҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽCH4£¨g £©+$\frac{3}{2}$O2£¨g £©=CO£¨g£©+2H2O£¨l£©¡÷H=-607kJ/mol£®

·ÖÎö £¨1£©È¼ÉÕÈÈÊÇ1mol¿ÉȼÎïÍêȫȼÉÕÉú³ÉÎȶ¨Ñõ»¯Îï·Å³öµÄÈÈÁ¿£¬ÒÀ¾ÝÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨Ð´³ö£¬±ê×¢ÎïÖʾۼ¯×´Ì¬ºÍ¶ÔÓ¦ìʱ䣻
£¨2£©¸ù¾Ý²»Í¬¾Û¼¯×´Ì¬µÄË®µÄÄÜÁ¿²»Í¬£¬ÒÔ¼°ÄÜÁ¿¸ßµÍ½øÐÐÅжϣ»
£¨3£©1molÇâÆøÍêȫȼÉÕÉú³ÉÆø̬ˮ·Å³ö242kJÈÈÁ¿£¬1molҺ̬ˮÕô·¢ÎªÆø̬ˮÊÇÒªÎüÊÕ44kJÈÈÁ¿£¬ÊéдÈÈ»¯Ñ§·½³Ìʽ½áºÏ¸Ç˹¶¨ÂɼÆËãµÃµ½£»
£¨4£©È¼ÉÕÈÈÊÇ1mol¿ÉȼÎïÍêȫȼÉÕÉú³ÉÎȶ¨Ñõ»¯Îï·Å³öµÄÈÈÁ¿£¬1molÇâÆøȼÉÕÉú³ÉҺ̬ˮ·Å³öµÄÈÈÁ¿ÎªÈ¼ÉÕÈÈ£»
£¨5£©1molCOÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌå·Å³ö283kJÈÈÁ¿£¬ÊéдCOȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ£¬1molCH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍҺ̬ˮ·Å³ö890kJÈÈÁ¿ÊéдÈÈ»¯Ñ§·½³Ìʽ£¬½áºÏ¸Ç˹¶¨ÂɼÆËãµÃµ½£®

½â´ð ½â£º£¨1£©È¼ÉÕÈÈÊÇ1mol¿ÉȼÎïÍêȫȼÉÕÉú³ÉÎȶ¨Ñõ»¯Îï·Å³öµÄÈÈÁ¿£¬1molCOÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌå·Å³ö283kJÈÈÁ¿£¬±íʾCOȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ£ºCO£¨g £©+$\frac{1}{2}$ O2£¨g £©=CO2£¨g£©¡÷H=-283 kJ/mol£»
¹Ê´ð°¸Îª£ºCO£¨g £©+$\frac{1}{2}$ O2£¨g £©=CO2£¨g£©¡÷H=-283 kJ/mol£»
£¨2£©Ë®ÓÉÆø̬±äΪҺ̬ʱҪ·ÅÈÈ£¬1mol CH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍҺ̬ˮ·Å³ö890KJÈÈÁ¿£¬Ôò1mol CH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍË®ÕôÆø£¬·Å³öµÄÈÈÁ¿Ð¡ÓÚ890kJ£¬¹Ê´ð°¸Îª£º£¼£»
£¨3£©H2£¨g£©+$\frac{1}{2}$O2£¨g£©¨TH2O£¨g£©¡÷H=-242kJ/mol£¬H2O£¨l£©=H2O£¨g£©¡÷H=+44kJ/mol£¬È¼ÉÕ1molÇâÆøÉú³ÉҺ̬ˮµÄÈÈ»¯Ñ§·½³ÌʽΪ£ºH2£¨g£©+$\frac{1}{2}$O2£¨g£©¨TH2O£¨l£©¡÷H=-286kJ/mol£¬È¼ÉÕ10gH2Éú³ÉҺ̬ˮ·ÅÈÈ$\frac{10g}{2g/mol}$¡Á286KJ/mol=1430KJ£¬
¹Ê´ð°¸Îª£º1430 kJ£»      
£¨4£©H2£¨g£©+$\frac{1}{2}$O2£¨g£©¨TH2O£¨g£©¡÷H=-242kJ/mol£¬H2O£¨l£©=H2O£¨g£©¡÷H=+44kJ/mol£¬1molÇâÆøÍêȫȼÉÕÉú³ÉҺ̬ˮ·Å³ö286kJÈÈÁ¿£¬ÇâÆøȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ£ºH2£¨g£©+$\frac{1}{2}$O2£¨g£©¨TH2O£¨l£©¡÷H=-286kJ/mol£»
¹Ê´ð°¸Îª£º286 kJ/mol£»
£¨5£©1molCOÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌå·Å³ö283kJÈÈÁ¿£¬ÊéдCOȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ£º¢ÙCO£¨g £©+$\frac{1}{2}$ O2£¨g £©=CO2£¨g£©£»¡÷H=-283 kJ/mol£¬1molCH4ÆøÌåÍêȫȼÉÕÉú³ÉCO2ÆøÌåºÍҺ̬ˮ·Å³ö890kJÈÈÁ¿£¬¢ÚCH4£¨g £©+2O2£¨g £©=CO2£¨g£©+2H2O£¨l£©¡÷H=-890 kJ/mol£¬ÒÀ¾Ý¸Ç˹¶¨ÂÉ¢Ù+¢ÚµÃµ½£¬CH4£¨g £©+$\frac{3}{2}$O2£¨g £©=CO£¨g£©+2H2O£¨l£©¡÷H=-607kJ/mol£¬
¹Ê´ð°¸Îª£ºCH4£¨g £©+$\frac{3}{2}$O2£¨g £©=CO£¨g£©+2H2O£¨l£©¡÷H=-607kJ/mol£®

µãÆÀ ±¾Ì⿼²éÁËÈÈ»¯Ñ§·½³ÌʽÊéд·½·¨£¬¸Ç˹¶¨ÂɵļÆËãÓ¦Óã¬Ö÷ÒªÊÇÎïÖʾۼ¯×´Ì¬ºÍȼÉÕÈȸÅÄîµÄÀí½âÓ¦Óã¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
7£®ÔªËØÖÜÆÚ±í½ÒʾÁËÐí¶àÔªËصÄÏàËÆÐԺ͵ݱä¹æÂÉ£¬Í¬Ò»ÖÜÆÚ¹¹³ÉµÄijЩ΢Á£ÍùÍù¾ßÓÐÏàͬµÄµç×ÓÊý£¬ÖÜÆÚ±íÖÐÖ÷×åÔªËØ¿ÉÒÔ¹¹³ÉÐí¶àµç×ÓÊýΪ10»ò8µÄ΢Á££¬ÈçÏÂÁÐÖÜÆÚ±íËùʾµÄһЩ·Ö×Ó»òÀë×Ó£®

£¨1£©Ð´³ö¢áÔªËØÔÚÔªËØÖÜÆÚ±íÖеÄλÖ㺵ÚËÄÖÜÆÚµÚ¢õ¢ó×壻
£¨2£©Ð´³ö¢Þ¡¢¢ß¡¢¢àÈýÖÖÔªËؼòµ¥Àë×Ӱ뾶ÓÉ´óµ½Ð¡µÄ˳ÐòΪS2-£¾Cl-£¾K+£»
£¨3£©Ð´³ö¢Û¡¢¢Ü¡¢¢Ý¡¢¢àËÄÖÖÔªËصÄ×î¸ß¼ÛÑõ»¯ÎïµÄË®»¯ÎïµÄ¼îÐÔÇ¿Èõ˳Ðò£ºKOH£¾NaOH£¾Mg£¨OH£©2£¾Al£¨OH£©3£»
£¨4£©¢ÙÔªËغ͢ÝÔªËصÄÇâÑõ»¯Îï¾ßÓÐÏàËƵĻ¯Ñ§ÐÔÖÊ£¬Ð´³ö¢ÙÔªËصÄÇâÑõ»¯ÎïÓë×ãÁ¿µÄNaOHÈÜÒº·´Ó¦µÄ»¯Ñ§·½³ÌʽBe£¨OH£©2+2NaOH=Na2BeO2+2H2O£»
£¨5£©Óõç×Óʽ±íʾ¢Ü¡¢¢ßÁ½ÖÖÔªËØÐγɻ¯ºÏÎïµÄ¹ý³Ì£º£»
£¨6£©º¬ÓТÞÔªËØijÖÖ18µç×ÓµÄÀë×ÓÓëH+¼°OH-¾ù¿É·¢Éú·´Ó¦£¬·Ö±ðд³ö·´Ó¦µÄÀë×Ó·½³Ìʽ£ºHS-+H+=H2S¡ü£»HS-+OH-=S2-+H2O£»
£¨7£©¢Ú¡¢¢ÛÁ½ÖÖÔªËصĵ¥ÖÊÔÚ¼ÓÈÈÌõ¼þÏ·¢Éú·´Ó¦Éú³ÉµÄ²úÎïµÄ»¯Ñ§Ê½ÎªNa2O2£¬ÆäÖÐËùº¬µÄ»¯Ñ§¼üÀàÐÍÓзǼ«ÐÔ¼üºÍÀë×Ó¼ü£®
£¨8£©ÔÚ¢Ú¡¢¢ÞÁ½ÖÖÔªËØÐγɵļòµ¥Ç⻯ÎïÖУ¬·Ðµã½Ï¸ßµÄÊÇË®£¬ÆäÖ÷ÒªÔ­ÒòÊÇË®·Ö×ÓÖ®¼äÓÐÇâ¼ü£»
£¨9£©½«¢Ü¡¢¢áÁ½ÖÖÔªËصĵ¥ÖÊ×é³ÉµÄ»ìºÏÎï9.8gÈܽâÔÚ¹ýÁ¿Ä³Å¨¶ÈµÄÏ¡ÏõËáÖУ¬ÍêÈ«·´Ó¦£¬¼ÙÉèÏõËáµÄ»¹Ô­²úÎïÈ«²¿ÎªNO£¬ÆäÎïÖʵÄÁ¿Îª0.2mol£¬ÔòÏò·´Ó¦ºóµÄÈÜÒºÖмÓÈë×ãÁ¿µÄÉÕ¼îÈÜÒº£¬¿ÉÉú³ÉÇâÑõ»¯Îï³ÁµíµÄÖÊÁ¿ÊÇ20g£®

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

¾«Ó¢¼Ò½ÌÍø