ÌâÄ¿ÄÚÈÝ

¡°C1»¯Ñ§¡±ÊÇÖ¸ÒÔ·Ö×ÓÖÐÖ»º¬Ò»¸ö̼ԭ×ÓµÄÎïÖÊΪԭÁϽøÐÐÎïÖʺϳɵĻ¯Ñ§¡£¡°C1»¯Ñ§¡±¶ÔÓÚ»º½âÈÕÒæ ÑÏÖصÄÄÜԴΣ»ú¡¢ºÏÀíÀûÓÃúºÍÌìÈ»ÆøµÈ»¯Ê¯È¼ÁÏ¡¢±£»¤»·¾³µÈ¶¼Óзdz£ÖØÒªµÄÒâÒå¡£ºÏ³ÉÆø(CO+H2)ÊÇ ¡°C1»¯Ñ§¡±Öеij£ÓÃÔ­ÁÏ¡£
(1)úÆø»¯¿ÉÉú³ÉºÏ³ÉÆø£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________________£¬Óø÷½·¨Éú²úºÏ³ÉÆøµÄÒ»¸öµäÐÍȱÏÝÊÇ___________¡£
(2)½«¼×ÍéÑõ»¯¿ÉÖƺϳÉÆø£ºCH4(g)+1/2O2(g)CO(g)+2H2(g) ¡÷H=-35.6 kJ/mol¡£¸Ã·´Ó¦ÊÇ_____£¨Ìî¡°×Ô·¢¡±»ò¡°·Ç×Ô·¢¡±£©·´Ó¦¡£
(3)ͨ¹ýÒÒ´¼ÖÆÈ¡ºÏ³ÉÆø¾ßÓÐÁ¼ºÃµÄÓ¦ÓÃÇ°¾°¡£ÓÉÒÒ´¼ÖÆÈ¡ºÏ³ÉÆøÓÐÈçÏÂÁ½Ìõ·Ïߣº
a£®Ë®ÕôÆø´ß»¯ÖØÕû£ºCH3CH2OH(g)+H2O(g)¡ú4H2(g)+2CO(g) ¡÷H=+255.58 kJ/mol
b£®²¿·Ö´ß»¯Ñõ»¯£ºCH3CH2OH(g)+1/2O2(g)¡ú 3H2(g)+2CO(g) ¡÷H=+13.76 kJ/mol
ÏÂÁÐ˵·¨´íÎóµÄÊÇ____¡£
A£®´ÓÔ­ÁÏÏûºÄµÄ½Ç¶ÈÀ´¿´£¬a·ÏßÖÆÇâ¸üÓмÛÖµ
B£®´ÓÄÜÁ¿ÏûºÄµÄ½Ç¶ÈÀ´¿´£¬b·ÏßÖÆÇâ¸ü¼ÓÓÐÀû
C£®a·ÏßÖÆÇâÓÉÓÚÒªÏûºÄºÜ¶àÄÜÁ¿£¬ËùÒÔÔÚʵ¼ÊÉú²úÖÐÒâÒå²»´ó
D£®ÔÚÒÔÉÏÁ½¸ö·´Ó¦ÖУ¬Ô­×ÓÀûÓÃÂʽϸߵÄÊÇb·´Ó¦
(4)¹¤ÒµÓúϳÉÆøÖƱ¸¶þ¼×ÃѵÄÉú²úÁ÷³ÌÈçÏ£º

´ß»¯·´Ó¦ÊÒÖÐ(ѹÁ¦2.0~ 10.0 MPa£¬Î¶È230~ 280¡æ)½øÐÐÏÂÁз´Ó¦£º
CO(g)+2H2(g)CH3OH(g) ¡÷H=-90.7 kJ/mol ¢Ù
2CH3OH(g)CH3OCH3(g)+H2O(g) ¡÷H = -23.5 kJ/mol ¢Ú
CO(g) +H2O(g)CO2(g)+H2(g) ¡÷H=-41.2 kJ/mol ¢Û
¢Ù´ß»¯·´Ó¦ÊÒÖÐ×Ü·´Ó¦3CO(g)+3H2(g)CH3OCH3(g)+CO2(g)µÄ¡÷H=_____¡£830¡æ ʱ·´Ó¦¢ÛµÄK=1.0£¬ÔòÔÚ´ß»¯·´Ó¦ÊÒÖз´Ó¦¢ÛµÄK_______£¨Ìî¡°>¡±¡¢¡°<¡±»ò¡°=¡±£©1.0¡£
¢ÚÉÏÊöÁ÷³ÌÖУ¬¿ÉÒÔÑ­»·Ê¹ÓõÄÎïÖÊÓÐ____________¡£
(1)C+H2O(g)CO+H2£»ÄܺĴó
(2)×Ô·¢
(3)D
(4)¢Ù-246.1 kJ/mol£»>£»¢ÚCO¡¢H2¡¢¼×´¼ºÍË®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÊÔÍê³ÉÏÂÁÐÁ½Ð¡Ì⣺

£¨1£©C1»¯Ñ§ÊÇÖ¸´ÓÒ»¸ö̼ԭ×ӵĻ¯ºÏÎÈçCH4£¬CO£¬CO2£¬CH3OH£¬HCHOµÈ£©³ö·¢ºÏ³É¸÷ÖÖ»¯Ñ§Æ·µÄ¼¼Êõ¡£´Óú¡¢ÌìÈ»ÆøÖƺϳÉÆøÔÙ½øÒ»²½ÖƱ¸¸÷ÖÖ»¯¹¤²úÆ·ºÍ½à¾»È¼ÁÏ£¬ÒѳÉΪµ±½ñ»¯Ñ§¹¤Òµ·¢Õ¹µÄ±ØÈ»Ç÷ÊÆ¡£ÆäÖм״¼ÊÇC1»¯Ñ§µÄ»ù´¡¡£

¢ÙCOÓëH2°´Ò»¶¨±ÈÀý¿ÉÉú³ÉÒÒ¶þ´¼£¬Ôòn(CO)/n(H2)=_____________£¨ÌîÊý×Ö£©¡£

¢ÚÈôÆûÓÍƽ¾ù×é³ÉÓÃCmHn±íʾ£¬ÔòºÏ³ÉÆûÓÍ¿ØÖÆn(CO)/n(H2)=(ÓÃm¡¢n±íʾ)¡£

¢Û¼×´¼ÔÚÒ»¶¨Ìõ¼þÏÂÓëCO¡¢H2×÷ÓÃÉú³ÉÓлúÎïA£¬A·¢Éú¼Ó¾Û¿ÉÉú³É¸ß·Ö×Óд³öÉú³ÉAµÄ»¯Ñ§·½³Ìʽ£º_________________________________¡£

£¨2£©ÒÑÖª´¼È©ÔÚÒ»¶¨Ìõ¼þÏÂÄÜ·¢ÉúËõºÏ·´Ó¦£¬Ê¾ÀýÈçÏ£º

ÒÑÖª£º

¢Ù1827ÄêÈËÃǾͷ¢ÏÖÓлúÎïA£¬ËüµÄ·Ö×ÓʽΪC13H18O7£¬ÓëÒ»·Ö×ÓË®×÷Óã¬Ë®½âÉú³ÉBºÍC¡£

¢ÚBÄÜ·¢ÉúÒø¾µ·´Ó¦£¬BÒ²¿ÉÓɵí·ÛË®½âµÃµ½£¬BµÄ·Ö×ÓʽΪC6H12O6¡£

¢ÛCÓöÂÈ»¯ÌúÈÜÒºÄÜ·¢ÉúÏÔÉ«·´Ó¦£¬1 mol CÓë×ãÁ¿ÄÆ·´Ó¦¿É²úÉú1 mol H2¡£

¢ÜCÔÚÊʵ±µÄÌõ¼þÏÂÓÃÊʵ±Ñõ»¯¼ÁÑõ»¯£¬¿ÉµÃD£¬DµÄ·Ö×ÓʽΪC7H6O3£¬Ïà¶Ô·Ö×ÓÖÊÁ¿D±ÈC´ó14¡£

¢ÝDÓÐÁ½¸öÈ¡´ú»ù£¬µ«²»ÊǼä룬ËüÓëBr2ÔÚ´ß»¯¼Á×÷ÓÃÏ·¢ÉúÒ»äåÈ¡´ú£¬²úÎïÓÐËÄÖÖ£¬DÄÜÓë̼ËáÇâÄÆÈÜÒº·´Ó¦¡£

¢ÞDÓëÒÒËáôû¡²(CH3CO)2O¡³·´Ó¦£¬¿ÉµÃ³£¼ûÒ©ÎïEºÍÒÒËᣬEÄÜÓë̼ËáÇâÄÆ·´Ó¦·Å³ö¶þÑõ»¯Ì¼¡£

ÊԻشðÏÂÁÐÎÊÌ⣺

¢Ùд³ö½á¹¹¼òʽ£ºC_________________£¬E_________________¡£

¢Úд³öÓëD»¥ÎªÍ¬·ÖÒì¹¹Ì壬º¬Óб½»·ÇÒº¬ÓÐõ¥½á¹¹µÄ½á¹¹¼òʽ£º_________________£¨Ö»ÐèдһÖÖ£©¡£

¢ÛBͨ³£ÒÔÁùÔª»·×´½á¹¹´æÔÚ£¬Ð´³öBµÄ»·×´½á¹¹¼òʽ£º_________________________¡£