(6·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º
(1)ÒÑ֪ϡÈÜÒºÖУ¬1 mol H2SO4ÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Éú³ÉÕýÑÎʱ£¬·Å³ö114.6 kJÈÈÁ¿£¬Ð´³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ_____________________¡£
(2)25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ ¡£
(3)ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l) ¦¤H1£½£870.3 kJ/mol
¢ÚC(s)£«O2(g)===CO2(g) ¡¡ ¦¤H2£½£393.5 kJ/mol
¢ÛH2(g)£«O2(g)===H2O(l ) ¦¤H3£½£285.8 kJ/mol
д³öÓÉC(s)¡¢H2(g)ºÍO2(g)»¯ºÏÉú³ÉCH3COOH(l)µÄÈÈ»¯Ñ§·½³Ìʽ ¡£
(1)H2SO4(aq)£«NaOH(aq)===Na2SO4(aq)£«H2O(l)¡¡ ¦¤H£½£57.3 kJ/mol(»¯Ñ§¼ÆÁ¿Êý¡¢¦¤H¿É³É±ÈÀý±ä»¯)
(2)C4H10(g)£«O2(g)===4CO2(g)£«5H2O(l) ¦¤H£½£16Q kJ/mol
(3)2C(s)£«2H2(g)£«O2(g)===CH3COOH(l) ¦¤H£½£488.3 kJ/mol
¡¾½âÎö¡¿¿¼²éÈÈ»¯Ñ§·½³ÌʽµÄÊéд¡£
£¨1£©Öкͷ´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬¡÷HСÓÚ0£¬·´Ó¦Ê½ÎªH2SO4(aq)£«NaOH(aq)===Na2SO4(aq)£«H2O(l)¡¡ ¦¤H£½£57.3 kJ/mol¡£
£¨2£©25 g°×É«³ÁµíÊÇ̼Ëá¸Æ£¬ÆäÎïÖʵÄÁ¿ÊÇ£¬¸ù¾Ý̼Ô×ÓÊغã¿ÉÖª£¬Éú³ÉµÄCO2ÊÇ0.25mol£¬ÔòÉú³É1molCO2·Å³öµÄÈÈÁ¿ÊÇ4QkJ¡£È¼ÉÕÈÈÊÇÖ¸1mol¿ÉȼÎïÍêȫȼÉÕÉú³ÉÎȶ¨µÄÑõ»¯ÎïʱËù·Å³öµÄÈÈÁ¿£¬ËùÒÔ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³ÌʽΪC4H10(g)£«O2(g)===4CO2(g)£«5H2O(l) ¦¤H£½£16Q kJ/mol¡£
£¨3£©¿¼²é¸Ç˹¶¨ÂɵÄÓ¦Óúͷ´Ó¦ÈȵļÆËã¡£½«¢Ú¡Á2£«¢Û¡Á2£¢Ù£¬¼´µÃµ½2C(s)£«2H2(g)£«O2(g)===CH3COOH(l)£¬ËùÒÔ·´Ó¦ÈÈÊÇ£393.5kJ/mol¡Á2£285.8 kJ/mol¡Á2£«870.3 kJ/mol£½£488.3 kJ/mol¡£
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£º
(6·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º
(1)ÒÑ֪ϡÈÜÒºÖУ¬1mol H2SO4ÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Éú³ÉÕýÑÎʱ£¬·Å³ö114.6 kJÈÈÁ¿£¬Ð´³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ_____________________¡£
(2)25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ ¡£
(3)ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l) ¦¤H1£½£870.3 kJ/mol
¢ÚC(s)£«O2(g)===CO2(g) ¡¡ ¦¤H2£½£393.5 kJ/mol
¢ÛH2(g)£«1/2O2(g)===H2O(l) ¦¤H3£½£285.8 kJ/mol
д³öÓÉC(s)¡¢H2(g)ºÍO2(g)»¯ºÏÉú³ÉCH3COOH(l)µÄÈÈ»¯Ñ§·½³Ìʽ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(6·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º
(1)ÒÑ֪ϡÈÜÒºÖУ¬1mol H2SO4ÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Éú³ÉÕýÑÎʱ£¬·Å³ö114.6 kJÈÈÁ¿£¬Ð´³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ_____________________¡£
(2)25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ ¡£
(3)ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l) ¦¤H1£½£870.3 kJ/mol
¢ÚC(s)£«O2(g)===CO2(g) ¡¡ ¦¤H2£½£393.5 kJ/mol
¢ÛH2(g)£«1/2O2(g)===H2O(l) ¦¤H3£½£285.8 kJ/mol
д³öÓÉC(s)¡¢H2(g)ºÍO2(g)»¯ºÏÉú³ÉCH3COOH(l)µÄÈÈ»¯Ñ§·½³Ìʽ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º
(6·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º
(1)ÒÑ֪ϡÈÜÒºÖУ¬1mol H2SO4ÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Éú³ÉÕýÑÎʱ£¬·Å³ö114.6 kJÈÈÁ¿£¬Ð´³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ_____________________¡£
(2)25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ ¡£
(3)ÒÑÖªÏÂÁÐÈÈ»¯Ñ§·½³Ìʽ£º
¢ÙCH3COOH(l)£«2O2(g)===2CO2(g)£«2H2O(l) ¦¤H1£½£870.3 kJ/mol
¢ÚC(s)£«O2(g)===CO2(g) ¡¡ ¦¤H2£½£393.5 kJ/mol
¢ÛH2(g)£«1/2O2(g)===H2O(l) ¦¤H3£½£285.8 kJ/mol
д³öÓÉC(s)¡¢H2(g)ºÍO2(g)»¯ºÏÉú³ÉCH3COOH(l)µÄÈÈ»¯Ñ§·½³Ìʽ ¡£
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011½ì¸£½¨Ê¡¸£ÖݽðÇŸ߼¶ÖÐѧ¸ßÈý12ÔÂÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
(6·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º
(1)ÒÑ֪ϡÈÜÒºÖУ¬1 mol H2SO4ÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Ê±£¬·Å³ö114.6 kJÈÈÁ¿£¬Ð´³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ_______________ ___
(2)25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ_________________________
²é¿´´ð°¸ºÍ½âÎö>>
¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010-2011ѧÄ긣½¨Ê¡¸ßÈý12ÔÂÔ¿¼»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ
(6·Ö)°´ÒªÇóдÈÈ»¯Ñ§·½³Ìʽ£º
(1)ÒÑ֪ϡÈÜÒºÖУ¬1 mol H2SO4ÓëNaOHÈÜҺǡºÃÍêÈ«·´Ó¦Ê±£¬·Å³ö114.6 kJÈÈÁ¿£¬Ð´³ö±íʾH2SO4ÓëNaOH·´Ó¦µÄÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ_______________ ___
(2)25¡æ¡¢101 kPaÌõ¼þϳä·ÖȼÉÕÒ»¶¨Á¿µÄ¶¡ÍéÆøÌå·Å³öÈÈÁ¿ÎªQ kJ£¬¾²â¶¨£¬½«Éú³ÉµÄCO2ͨÈë×ãÁ¿³ÎÇåʯ»ÒË®ÖвúÉú25 g°×É«³Áµí£¬Ð´³ö±íʾ¶¡ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ_________________________
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