(11·Ö)£¨1£©Ä¿Ç°¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2À´Éú²úȼÁϼ״¼¡£ÎªÌ½¾¿·´Ó¦ÔÀí£¬ÏÖ½øÐÐÈçÏÂʵÑ飬ÔÚÌå»ýΪ1 LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1mol CO2ºÍ3mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º
CO2(g)£«3H2(g)CH3OH(g)£«H2O(g) £¬¡÷H£½£49.0kJ/mol£»²âµÃCO2ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ¡£
¢Ù´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬Æ½¾ù·´Ó¦ËÙÂÊv(CO2)£½ mol/(L¡¤min)¡£
¢Ú¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ________________________________________¡£
¢ÛÏÂÁдëÊ©ÖÐÄÜʹn(CH3OH)£¯n(CO2)Ôö´óµÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
A£®Éý¸ßÎÂ¶È | B£®³äÈëHe(g)£¬Ê¹ÌåϵѹǿÔö´ó |
C£®½«H2O(g)´ÓÌåϵÖзÖÀë | D£®ÔÙ³äÈë1mol H2 |
Ä꼶 | ¸ßÖÐ¿Î³Ì | Ä꼶 | ³õÖÐ¿Î³Ì |
¸ßÒ» | ¸ßÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ | ³õÒ» | ³õÒ»Ãâ·Ñ¿Î³ÌÍƼö£¡ |
¸ß¶þ | ¸ß¶þÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õ¶þ | ³õ¶þÃâ·Ñ¿Î³ÌÍƼö£¡ |
¸ßÈý | ¸ßÈýÃâ·Ñ¿Î³ÌÍƼö£¡ | ³õÈý | ³õÈýÃâ·Ñ¿Î³ÌÍƼö£¡ |
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â
£¨11 ·Ö£©¿Æѧ¼ÒÒ»Ö±ÖÂÁ¦Ñо¿³£Î¡¢³£Ñ¹Ï¡°È˹¤¹Ìµª¡±µÄз½·¨¡£ÔøÓÐʵÑ鱨µÀ£ºÔÚ³£Î¡¢³£Ñ¹¡¢¹âÕÕÌõ¼þÏ£¬N2ÔÚ´ß»¯¼Á£¨²ôÓÐÉÙÁ¿Fe2O3µÄTiO2£©±íÃæÓëË®·¢Éú·´Ó¦£¬Éú³ÉµÄÖ÷Òª²úÎïΪNH3£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÈçÏ£ºN2(g)+ 3H2O(l) 2NH3(g)+ O2(g)¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©½øÒ»²½Ñо¿NH3Éú³ÉÁ¿ÓëζȵĹØϵ£¬²¿·ÖʵÑéÊý¾Ý¼ûÏÂ±í£¨¹âÕÕ¡¢N2ѹÁ¦1.0¡Á105Pa¡¢·´Ó¦Ê±¼ä3 h£©£¬Ôò¸Ã·´Ó¦µÄÕý·´Ó¦Îª ·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©
T/K | 303 | 313 | 323 |
NH3Éú³ÉÁ¿/£¨10-6 mol£© | 4.8 | 5.9 | 6.0 |
£¨2£©ÓëÄ¿Ç°¹ã·ºÊ¹ÓõĹ¤ÒµºÏ³É°±·½·¨Ïà±È£¬¸Ã·½·¨Öй̵ª·´Ó¦ËÙÂÊÂý¡£ÇëÌá³ö¿ÉÌá¸ßÆä·´Ó¦ËÙÂÊÇÒÔö´óNH3Éú³ÉÁ¿µÄ½¨Ò飺¡¡¡¡ ¡£
£¨3£©ºÏ³É°±¹¤ÒµÖÐÔÁÏÆøN2¿É´Ó¿ÕÆøÖзÖÀëµÃµ½£¬H2¿ÉÓü×ÍéÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦ÖƵ᣼×ÍéÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦·´Ó¦·½³ÌʽΪ£ºCH4(g)£«H2O(g)£½CO(g)£«3H2(g)¡£²¿·ÖÎïÖʵÄȼÉÕÈÈÊý
¾ÝÈçÏ£º
H2(g) £º¡÷H=£285.8 kJ・mol£1£»
CO(g) £º¡÷H =£283.0 kJ・mol£1£»
CH4(g) £º¡÷H=£890.3 kJ・mol£1 ¡£
ÒÑÖª1mol H2O(g)ת±äΪ1mol H2O(l)ʱ·Å³ö44.0 kJÈÈÁ¿¡£Ð´³öCH4ºÍH2OÔÚ¸ßÎÂÏ·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ__________________________________¡£
£¨4£©ÓÐÈËÉèÏëÑ°ÇóÊʺϵĴ߻¯¼ÁºÍµç¼«²ÄÁÏ£¬ÒÔN2¡¢H2Ϊµç¼«·´Ó¦ÎÒÔHCl¡ª¡ªNH4ClΪµç½âÖÊÈÜÒºÖƳÉÐÂÐÍȼÁϵç³Ø£¬Çëд³ö¸Ãµç¼«µÄÕý¼«·´Ó¦Ê½
£¨5£©Éú³ÉµÄNH3¿ÉÓÃÓÚÖÆï§Ì¬µª·Ê£¬Èç(NH4)2SO4¡¢NH4Cl£¬ÕâЩ·ÊÁÏÏÔ ÐÔ£¬ÔÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©___________________________£¬Ê¹ÓÃʱ±ÜÃâÓë________________ÎïÖʺÏÊ©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(11·Ö) £¨1£©Ä¿Ç°¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2À´Éú²úȼÁϼ״¼¡£ÎªÌ½¾¿·´Ó¦ÔÀí£¬ÏÖ½øÐÐÈçÏÂʵÑ飬ÔÚÌå»ýΪ1 LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1mol CO2ºÍ3mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º
CO2(g)£«3H2(g)CH3OH(g)£«H2O(g) £¬¡÷H£½£49.0kJ/mol£»²âµÃCO2ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ¡£
¢Ù´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬Æ½¾ù·´Ó¦ËÙÂÊv(CO2)£½ mol/(L¡¤min)¡£
¢Ú¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ________________________________________¡£
¢ÛÏÂÁдëÊ©ÖÐÄÜʹn(CH3OH)£¯n(CO2)Ôö´óµÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
A£®Éý¸ßÎÂ¶È B£®³äÈëHe(g)£¬Ê¹ÌåϵѹǿÔö´ó
C£®½«H2O(g)´ÓÌåϵÖзÖÀë D£®ÔÙ³äÈë1mol H2
£¨2£©ÔÚÔØÈ˺½ÌìÆ÷µÄÉú̬ϵͳÖУ¬²»½öÒªÇó·ÖÀëÈ¥³ýCO2£¬»¹ÒªÇóÌṩ³ä×ãµÄO2¡£Ä³Öֵ绯ѧװÖÿÉʵÏÖÈçÏÂת»¯£º2CO2£½2CO£«O2¡¡£¬CO¿ÉÓÃ×÷ȼÁÏ¡£
ÒÑÖª¸Ã·´Ó¦µÄÑô¼«·´Ó¦Îª£º4OH¨D¨D4e¨D£½O2¡ü£«2H2O
ÔòÒõ¼«·´Ó¦Ê½Îª£º ¡£
ÓÐÈËÌá³ö£¬¿ÉÒÔÉè¼Æ·´Ó¦2CO£½2C£«O2£¨¡÷H£¾0¡¢¡÷S£¼0£©À´Ïû³ýCOµÄÎÛȾ¡£ÇëÄãÅжÏÊÇ·ñ¿ÉÐв¢Ëµ³öÀíÓÉ£º¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ £¬¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2013½ìÌì½òÊи߶þÉÏѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
£¨11 ·Ö£©¿Æѧ¼ÒÒ»Ö±ÖÂÁ¦Ñо¿³£Î¡¢³£Ñ¹Ï¡°È˹¤¹Ìµª¡±µÄз½·¨¡£ÔøÓÐʵÑ鱨µÀ£ºÔÚ³£Î¡¢³£Ñ¹¡¢¹âÕÕÌõ¼þÏ£¬N2ÔÚ´ß»¯¼Á£¨²ôÓÐÉÙÁ¿Fe2O3µÄTiO2£©±íÃæÓëË®·¢Éú·´Ó¦£¬Éú³ÉµÄÖ÷Òª²úÎïΪNH3£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÈçÏ£ºN2(g)+ 3H2O(l) 2NH3(g)+ O2(g)¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©½øÒ»²½Ñо¿NH3Éú³ÉÁ¿ÓëζȵĹØϵ£¬²¿·ÖʵÑéÊý¾Ý¼ûÏÂ±í£¨¹âÕÕ¡¢N2ѹÁ¦1.0¡Á105 Pa¡¢·´Ó¦Ê±¼ä3 h£©£¬Ôò¸Ã·´Ó¦µÄÕý·´Ó¦Îª ·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©
T/K |
303 |
313 |
323 |
NH3Éú³ÉÁ¿/£¨10-6 mol£© |
4.8 |
5.9 |
6.0 |
£¨2£©ÓëÄ¿Ç°¹ã·ºÊ¹ÓõĹ¤ÒµºÏ³É°±·½·¨Ïà±È£¬¸Ã·½·¨Öй̵ª·´Ó¦ËÙÂÊÂý¡£ÇëÌá³ö¿ÉÌá¸ßÆä·´Ó¦ËÙÂÊÇÒÔö´óNH3Éú³ÉÁ¿µÄ½¨Ò飺¡¡¡¡ ¡£
£¨3£©ºÏ³É°±¹¤ÒµÖÐÔÁÏÆøN2¿É´Ó¿ÕÆøÖзÖÀëµÃµ½£¬H2¿ÉÓü×ÍéÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦ÖƵ᣼×ÍéÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦·´Ó¦·½³ÌʽΪ£ºCH4(g)£«H2O(g)£½CO(g)£«3H2(g)¡£²¿·ÖÎïÖʵÄȼÉÕÈÈÊý
¾ÝÈçÏ£º
H2(g) £º¡÷H =£285.8 kJ・mol£1£»
CO(g) £º ¡÷H =£283.0 kJ・mol£1£»
CH4(g) £º¡÷H =£890.3 kJ・mol£1 ¡£
ÒÑÖª1mol H2O(g)ת±äΪ1mol H2O(l)ʱ·Å³ö44.0 kJÈÈÁ¿¡£Ð´³öCH4ºÍH2OÔÚ¸ßÎÂÏ·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ__________________________________¡£
£¨4£©ÓÐÈËÉèÏëÑ°ÇóÊʺϵĴ߻¯¼ÁºÍµç¼«²ÄÁÏ£¬ÒÔN2¡¢H2Ϊµç¼«·´Ó¦ÎÒÔHCl¡ª¡ªNH4ClΪµç½âÖÊÈÜÒºÖƳÉÐÂÐÍȼÁϵç³Ø£¬Çëд³ö¸Ãµç¼«µÄÕý¼«·´Ó¦Ê½
£¨5£©Éú³ÉµÄNH3¿ÉÓÃÓÚÖÆï§Ì¬µª·Ê£¬Èç(NH4)2SO4¡¢NH4Cl£¬ÕâЩ·ÊÁÏÏÔ ÐÔ£¬ÔÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©___________________________£¬Ê¹ÓÃʱ±ÜÃâÓë________________ÎïÖʺÏÊ©¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010Äê¹ã¶«Ê¡¶«Ý¸Êи߶þµÚ¶þѧÆÚÆÚÄ©¿¼ÊÔ»¯Ñ§B¾í ÌâÐÍ£ºÊµÑéÌâ
(11·Ö) £¨1£©Ä¿Ç°¹¤ÒµÉÏÓÐÒ»ÖÖ·½·¨ÊÇÓÃCO2À´Éú²úȼÁϼ״¼¡£ÎªÌ½¾¿·´Ó¦ÔÀí£¬ÏÖ½øÐÐÈçÏÂʵÑ飬ÔÚÌå»ýΪ1 LµÄÃܱÕÈÝÆ÷ÖУ¬³äÈë1mol CO2ºÍ3mol H2£¬Ò»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º
CO2(g)£«3H2(g)CH3OH(g)£«H2O(g) £¬¡÷H£½£49.0kJ/mol£»²âµÃCO2ºÍCH3OH(g)µÄŨ¶ÈËæʱ¼ä±ä»¯ÈçͼËùʾ¡£
¢Ù´Ó·´Ó¦¿ªÊ¼µ½Æ½ºâ£¬Æ½¾ù·´Ó¦ËÙÂÊv(CO2)£½ mol/(L¡¤min)¡£
¢Ú¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽΪ________________________________________¡£
¢ÛÏÂÁдëÊ©ÖÐÄÜʹn(CH3OH)£¯n(CO2)Ôö´óµÄÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
A£®Éý¸ßÎÂ¶È B£®³äÈëHe(g)£¬Ê¹ÌåϵѹǿÔö´ó
C£®½«H2O(g)´ÓÌåϵÖзÖÀë D£®ÔÙ³äÈë1mol H2
£¨2£©ÔÚÔØÈ˺½ÌìÆ÷µÄÉú̬ϵͳÖУ¬²»½öÒªÇó·ÖÀëÈ¥³ýCO2£¬»¹ÒªÇóÌṩ³ä×ãµÄO2¡£Ä³Öֵ绯ѧװÖÿÉʵÏÖÈçÏÂת»¯£º2CO2£½2CO£«O2¡¡£¬CO¿ÉÓÃ×÷ȼÁÏ¡£
ÒÑÖª¸Ã·´Ó¦µÄÑô¼«·´Ó¦Îª£º4OH¨D¨D4e¨D£½O2¡ü£«2H2O
ÔòÒõ¼«·´Ó¦Ê½Îª£º ¡£
ÓÐÈËÌá³ö£¬¿ÉÒÔÉè¼Æ·´Ó¦2CO£½2C£«O2£¨¡÷H£¾0¡¢¡÷S£¼0£©À´Ïû³ýCOµÄÎÛȾ¡£ÇëÄãÅжÏÊÇ·ñ¿ÉÐв¢Ëµ³öÀíÓÉ£º¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡ £¬¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡£
²é¿´´ð°¸ºÍ½âÎö>>
°Ù¶ÈÖÂÐÅ - Á·Ï°²áÁбí - ÊÔÌâÁбí
ºþ±±Ê¡»¥ÁªÍøÎ¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨Æ½Ì¨ | ÍøÉÏÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | µçÐÅթƾٱ¨×¨Çø | ÉæÀúÊ·ÐéÎÞÖ÷ÒåÓк¦ÐÅÏ¢¾Ù±¨×¨Çø | ÉæÆóÇÖȨ¾Ù±¨×¨Çø
Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com