ÌâÄ¿ÄÚÈÝ
¡¾ÌâÄ¿¡¿ÒÔº¬1¸ö̼Ô×ÓµÄÎïÖÊ(ÈçCO¡¢CO2¡¢CH4¡¢CH3OHµÈ)ΪÔÁϵÄ̼һ»¯Ñ§´¦ÓÚδÀ´»¯Ñ§²úÒµµÄºËÐÄ£¬³ÉΪ¿Æѧ¼ÒÑо¿µÄÖØÒª¿ÎÌâ¡£
(1))ÒÑÖªCO¡¢H2¡¢CH3OH(g)µÄȼÉÕÈÈ·Ö±ðΪ-283.0 kJ¡¤mol£1¡¢-285.8 kJ¡¤mol£1¡¢-764.5 kJ¡¤mol£1¡£Ôò·´Ó¦¢ñ£ºCO(g)£«2H2(g)CH3OH(g)¡¡¦¤H£½_____£»¡£
(2)ÔÚT1ʱ£¬ÏòÌå»ýΪ2 LµÄºãÈÝÈÝÆ÷ÖгäÈëÎïÖʵÄÁ¿Ö®ºÍΪ3 molµÄCOºÍH2£¬·¢Éú·´Ó¦CO(g)£«2H2(g)CH3OH(g)£¬·´Ó¦´ïµ½Æ½ºâʱCH3OH(g)µÄÌå»ý·ÖÊý(¦Õ)Óën(H2)/n(CO)µÄ¹ØϵÈçͼËùʾ¡£
¢Ùµ±Æðʼn(H2)/n(CO)£½2ʱ£¬¾¹ý5 min´ïµ½Æ½ºâ£¬COµÄת»¯ÂÊΪ0.6£¬Ôò0¡«5 minÄÚƽ¾ù·´Ó¦ËÙÂÊv(H2)£½______¡£Èô´Ë¿ÌÔÙÏòÈÝÆ÷ÖмÓÈëCO(g)ºÍCH3OH(g)¸÷0.4 mol£¬´ïµ½ÐÂƽºâʱH2µÄת»¯Âʽ«____(Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±)¡£
¢Úµ±n(H2)/n(CO)£½3.5ʱ£¬´ïµ½Æ½ºâºó£¬CH3OHµÄÌå»ý·ÖÊý¿ÉÄÜÊÇͼÏñÖеÄ________(Ìî¡°D¡±¡°E¡±»ò¡°F¡±)µã¡£
(3)ÔÚÒ»ÈÝ»ý¿É±äµÄÃܱÕÈÝÆ÷ÖгäÓÐ10 mol COºÍ20 mol H2¡£COµÄƽºâת»¯ÂÊ(¦Á)ÓëζÈ(T)¡¢Ñ¹Ç¿(p)µÄ¹ØϵÈçͼËùʾ¡£
¢ÙA¡¢B¡¢CÈýµãµÄƽºâ³£ÊýKA¡¢KB¡¢KCµÄ´óС¹ØϵΪ________¡£
¢ÚÈô´ïµ½Æ½ºâ״̬Aʱ£¬ÈÝÆ÷µÄÌå»ýΪ10 L£¬ÔòÔÚƽºâ״̬BʱÈÝÆ÷µÄÌå»ýΪ_____L¡£
(4)ÒÔ¼×´¼ÎªÖ÷ÒªÔÁÏ£¬µç»¯Ñ§ºÏ³É̼Ëá¶þ¼×õ¥µÄ¹¤×÷ÔÀíÈçͼËùʾ¡£ÔòµçÔ´µÄ¸º¼«Îª__(Ìî¡°A¡±»ò¡°B¡±)£¬Ð´³öÑô¼«µÄµç¼«·´Ó¦Ê½____¡£
¡¾´ð°¸¡¿£90.1 kJ¡¤mol£1 0.12 mol¡¤L£1¡¤min£1 Ôö´ó F KA£½KB>KC 2 B 2CH3OH£«CO£2e£===(CH3O)2CO£«2H£«
¡¾½âÎö¡¿
(1)¸ù¾ÝCO ¡¢H2ºÍCH3OHµÄȼÉÕÈÈÏÈÊéдÈÈ·½³Ìʽ£¬ÔÙÀûÓøÇ˹¶¨ÂɼÆËãCO(g)£«2H2(g)CH3OH(g)¡¡µÄ¦¤H£»
(2) ¢Ù¸ù¾Ý ¼ÆËãËÙÂÊ£»¸ù¾ÝQºÍKµÄ¹ØϵÅжϷ´Ó¦·½Ïò£»¢Ú¸ù¾ÝºãÈÝÈÝÆ÷ÖУ¬Í¶ÁϱȵÈÓÚϵÊý±È£¬´ïµ½Æ½ºâ״̬ʱ²úÎïµÄ°Ù·Öº¬Á¿×î´ó£»
(3) ¢ÙÏàͬζÈÏÂƽºâ³£ÊýÏàµÈ£»¸ù¾ÝͼÏñ£¬COµÄƽºâת»¯ÂÊ(¦Á)ËæζÈÉý¸ß¶ø¼õС£¬¿ÉÖªÉý¸ßζÈƽºâÄæÏòÒƶ¯£»
¢Ú¸ù¾ÝA¡¢BÁ½µãµÄƽºâ³£ÊýÏàµÈ¼ÆËãÔÚƽºâ״̬BʱÈÝÆ÷µÄÌå»ý£»
£¨4£©ÓɽṹʾÒâͼ¿ÉÖª£¬µç½â³Ø×ó²à·¢ÉúÑõ»¯·´Ó¦¡¢ÓҲ෢Éú»¹Ô·´Ó¦£¬Ôòµç½â³Ø×ó²àΪÑô¼«£¬ÓÒ²àΪÒõ¼«¡£
£¨1£©ÓÉCO£¨g£©¡¢H2£¨g£©ºÍCH3OH£¨g£©µÄȼÉÕÈÈ¡÷H·Ö±ðΪ-283.0 kJ¡¤mol£1¡¢-285.8 kJ¡¤mol£1ºÍ-764.5 kJ¡¤mol£1£¬Ôò
¢ÙCO£¨g£©+ O2£¨g£©=CO2£¨g£©¡÷H=-283.0 kJ¡¤mol£1
¢ÚCH3OH£¨g£©+O2£¨g£©=CO2£¨g£©+2 H2O£¨l£©¡÷H=-764.5 kJ¡¤mol£1
¢ÛH2£¨g£©+O2£¨g£©=H2O£¨l£©¡÷H=-285.8 kJ¡¤mol£1
ÓɸÇ˹¶¨ÂÉ¿ÉÖªÓâÙ+¢Û¡Á2-¢ÚµÃ·´Ó¦CO£¨g£©+2H2£¨g£©=CH3OH£¨l£©£¬
¸Ã·´Ó¦µÄ·´Ó¦ÈÈ¡÷H=-283.0 kJ¡¤mol£1+£¨-285.8 kJ¡¤mol£1£©¡Á2-£¨-764.5 kJ¡¤mol£1£©=£90.1 kJ¡¤mol£1£»
(2) ¢Ùµ±Æðʼn(H2)/n(CO)£½2ʱ£¬ÔòÆðʼn(H2) £½2mol£¬n(CO)£½1mol£¬¾¹ý5 min´ïµ½Æ½ºâ£¬COµÄת»¯ÂÊΪ0.6£¬
CO(g)£«2H2(g)CH3OH(g)
Æðʼ 0.5 1 0
ת»¯ 0.3 0.6 0.3
ƽºâ 0.2 0.4 0.3
0.12 mol¡¤L£1¡¤min£1
K9.375
Èô´Ë¿ÌÔÙÏòÈÝÆ÷ÖмÓÈëCO(g)ºÍCH3OH(g)¸÷0.4 mol£¬Ôò7.8125£¼K£¬ËùÒÔ·´Ó¦ÕýÏò½øÐУ¬´ïµ½ÐÂƽºâʱH2µÄת»¯Âʽ«Ôö´ó£»
¢Ú¸ù¾ÝºãÈÝÈÝÆ÷ÖУ¬Í¶ÁϱȵÈÓÚϵÊý±È£¬´ïµ½Æ½ºâ״̬ʱ²úÎïµÄ°Ù·Öº¬Á¿×î´ó£¬ËùÒÔµ±n(H2)/n(CO)£½3.5ʱ£¬´ïµ½Æ½ºâºó£¬CH3OHµÄÌå»ý·ÖÊý¿ÉÄÜÊÇͼÏñÖеÄFµã£»
(3) ¢ÙÏàͬζÈÏÂƽºâ³£ÊýÏàµÈ£¬ËùÒÔKA=KB£»¸ù¾ÝͼÏñ£¬COµÄƽºâת»¯ÂÊ(¦Á)ËæζÈÉý¸ß¶ø¼õС£¬¿ÉÖªÉý¸ßζÈƽºâÄæÏòÒƶ¯£¬Æ½ºâ³£Êý¼õС£¬ËùÒÔKB£¾KC£¬¹ÊKA£½KB>KC£»
¢ÚA¡¢BÁ½µãµÄƽºâ³£ÊýÏàµÈ£¬ÉèBµãÈÝÆ÷µÄÌå»ýΪVL
CO(g)£«2H2(g)CH3OH(g)
Æðʼ 10 20 0
ת»¯ 5 10 5
ƽºâ 5 10 5
£»
CO(g)£«2H2(g)CH3OH(g)
Æðʼ 10 20 0
ת»¯ 8 16 8
ƽºâ 2 4 8
£»V=2L£»
£¨4£©ÓɽṹʾÒâͼ¿ÉÖª£¬µç½â³Ø×ó²à·¢ÉúÑõ»¯·´Ó¦¡¢ÓҲ෢Éú»¹Ô·´Ó¦£¬Ôòµç½â³Ø×ó²àΪÑô¼«£¬Á¬½ÓµçÔ´µÄÕý¼«£¬ÓÒ²àΪÒõ¼«£¬Á¬½ÓµçÔ´µÄ¸º¼«£¬BΪµçÔ´µÄ¸º¼«£¬Ñô¼«ÊǼ״¼¡¢COʧȥµç×ÓÉú³É£¨CH3
¡¾ÌâÄ¿¡¿³ôÑõÊÇÀíÏëµÄÑÌÆøÍÑÏõ¼Á£¬ÆäÍÑÏõ·´Ó¦Îª£º2NO2(g)+O3(g)N2O5(g)+O2(g)£¬·´Ó¦ÔÚºãÈÝÃܱÕÈÝÆ÷ÖнøÐУ¬ÏÂÁÐÓɸ÷´Ó¦Ïà¹ØͼÏñ×÷³öµÄÅжÏÕýÈ·µÄÊÇ£¨ £©
A | B | C | D |
Éý¸ßζȣ¬ | 0¡«3sÄÚ£¬·´Ó¦ËÙÂÊΪ£º | t1ʱ½ö¼ÓÈë´ß»¯¼Á£¬ | ´ïƽºâʱ£¬½ö¸Ä±äx£¬ÔòxΪc(O2) |