ÌâÄ¿ÄÚÈÝ

£¨1£©¹Ì¶¨ºÍÀûÓÃCO2ÄÜÓÐЧµØÀûÓÃ×ÊÔ´£¬²¢¼õÉÙ¿ÕÆøÖеÄÎÂÊÒÆøÌå¡£¹¤ÒµÉÏÓÐÒ»ÖÖÓÃCO2À´Éú²ú¼×´¼È¼Áϵķ½·¨£ºCO2(g)£«3H2(g)CH3OH(g)£«H2O(g)¡÷H£½£­49£®0kJ¡¤mol£­1¡£Ä³¿ÆѧʵÑ齫6molCO2ºÍ8molH2³äÈë2LÃܱÕÈÝÆ÷ÖУ¬²âµÃH2µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçÓÒͼËùʾ£¨ÊµÏߣ©¡£Í¼ÖÐÊý¾Ýa£¨1£¬6£©´ú±íµÄÒâ˼ÊÇ£ºÔÚl minʱH2µÄÎïÖʵÄÁ¿ÊÇ6mol¡£

¢ÙÏÂÁÐʱ¼ä¶Îƽ¾ù·´Ó¦ËÙÂÊ×î´óµÄÊÇ__________£¬×îСµÄÊÇ______________¡£
A£®0¡«1minB£®1¡«3minC£®3¡«8minD£®8¡«11min
¢Ú½ö¸Ä±äijһʵÑéÌõ¼þÔÙ½øÐÐÁ½´ÎʵÑé²âµÃH2µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçͼÖÐÐéÏßËùʾ£¬ÇúÏߢñ¶ÔÓ¦µÄʵÑé¸Ä±äµÄÌõ¼þÊÇ________£¬ÇúÏߢò¶ÔÓ¦µÄʵÑé¸Ä±äµÄÌõ¼þÊÇ_________¡£
£¨2£©ÀûÓùâÄܺ͹â´ß»¯¼Á£¬¿É½«CO2ºÍH2O(g)ת»¯ÎªCH4ºÍO2¡£×ÏÍâ¹âÕÕÉäʱ£¬µÈÁ¿µÄCO2ºÍH2O(g)ÔÚ²»Í¬´ß»¯¼Á(¢ñ¡¢¢ò¡¢¢ó)×÷ÓÃÏ£¬CH4²úÁ¿Ëæ¹âÕÕʱ¼äµÄ±ä»¯ÈçͼËùʾ¡£ÔÚ0~30 hÄÚ£¬CH4µÄƽ¾ùÉú³ÉËÙÂÊv(¢ñ)¡¢v(¢ò)ºÍv(¢ó)´Ó´óµ½Ð¡µÄ˳ÐòΪ                  ¡£·´Ó¦¿ªÊ¼ºóµÄ12СʱÄÚ£¬ÔÚµÚ___________ÖÖ´ß»¯¼ÁµÄ×÷ÓÃÏ£¬ÊÕ¼¯µÄCH4×î¶à¡£
£¨12·Ö£©£¨1£©¢ÙA¡¢D  ¢ÚÉý¸ßζȣ¬Ôö´óѹǿ»òÕßÊÇÔö´óCO2Ũ¶È£¨¸÷2·Ö£©
£¨2£©v(¢ó)>v(¢ò)>v(¢ñ)£¨2·Ö£©;¢ò  £¨2·Ö£©

ÊÔÌâ·ÖÎö£º£¨1£©¢ÙÓÉͼ1¿ÉÖª£¬0¡«1minÄÚÇâÆøµÄ±ä»¯Á¿Îª8mol-6mol=2mol£»
B£®1¡«3minÄÚÇâÆøµÄ±ä»¯Á¿Îª6mol-3mol=3mol£¬Æ½¾ù1min±ä»¯Á¿Îª1.5mol£»
C£®3¡«8minÄÚÇâÆøµÄ±ä»¯Á¿Îª3mol-2mol=1mol£¬Æ½¾ù1min±ä»¯Á¿Îª0.2mol£»
D£®8¡«11min´ïƽºâ״̬£¬ÇâÆøµÄÎïÖʵÄÁ¿²»Ôٱ仯£®
¹Ê1¡«3minËÙÂÊÔö´ó£¬8¡«11minËÙÂÊ×îС£®
¹Ê´ð°¸Îª£ºA£»D£®
¢Ú¶ÔÓÚ¿ÉÄæ·´Ó¦CO2£¨g£©+3H2£¨g£©CH3OH£¨g£©+H2O£¨g£©¡÷H=-49.0kJ¡¤mol£­1£¬Õý·´Ó¦ÊÇÌå»ý¼õСµÄ·ÅÈÈ·´Ó¦£®
ÓÉͼ1¿ÉÖª£¬ÇúÏߢñ×îÏȵ½´ïƽºâ£¬Æ½ºâʱÇâÆøµÄÎïÖʵÄÁ¿Ôö´ó£¬¹Ê¸Ä±äÌõ¼þÓ¦Ôö´ó·´Ó¦ËÙÂÊÇÒƽºâÏòÄæ·´Ó¦Òƶ¯£¬¿ÉÒÔ²ÉÈ¡µÄ´ëʩΪ£ºÉý¸ßζȣ®
ÇúÏߢòµ½´ïƽºâµÄʱ¼ä±Èԭƽºâ¶Ì£¬Æ½ºâʱÇâÆøµÄÎïÖʵÄÁ¿¼õС£¬¹Ê¸Ä±äÌõ¼þÓ¦Ôö´ó·´Ó¦ËÙÂÊÇÒƽºâÏòÕý·´Ó¦Òƶ¯£¬¿ÉÒÔ²ÉÈ¡µÄ´ëʩΪ£ºÔö´óѹǿ»òÕßÊÇÔö´óCO2Ũ¶È£®
¹Ê´ð°¸Îª£ºÉý¸ßζȣ»Ôö´óѹǿ»òÕßÊÇÔö´óCO2Ũ¶È£®
£¨2£©ÓÉͼ2¿ÉÖª£¬ÔÚ0¡«30hÄÚ£¬¼×ÍéµÄÎïÖʵÄÁ¿±ä»¯Á¿Îª¡÷n£¨¢ñ£©£¼¡÷n£¨¢ò£©£¼¡÷n£¨¢ó£©£¬¹ÊÔÚ0¡«30hÄÚ£¬CH4µÄƽ¾ùÉú³ÉËÙÂÊv£¨¢ó£©£¾v£¨¢ò£©£¾v£¨¢ñ£©£»
ÓÉͼ2¿ÉÖª·´Ó¦¿ªÊ¼ºóµÄ12СʱÄÚ£¬ÔÚµÚ¢òÖÖ´ß»¯¼ÁµÄ×÷ÓÃÏ£¬ÊÕ¼¯µÄCH4×î¶à£®
¹Ê´ð°¸Îª£ºv£¨¢ó£©£¾v£¨¢ò£©£¾v£¨¢ñ£©£»¢ò£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
°±ÆøÊÇÒ»ÖÖÖØÒªµÄ»¯¹¤²úÆ·£¬ÊÇÉú²úï§ÑΡ¢ÄòËصȵÄÔ­ÁÏ¡£¹¤ÒµºÏ³É°±µÄ·´Ó¦ÈçÏÂ:N2(g) +3H2(g) 2NH3(g)  ¡÷H=Ò»92. 4 KJ¡¤mol-1
£¨1£©2NH3(g) N2(g) +3H2(g)ÔÚºãÈÝÃܱÕÈÝÆ÷ÖдﵽƽºâµÄ±êÖ¾ÓÐ
¢Ùµ¥Î»Ê±¼äÄÚÉú³É3n mol H2:ͬʱÉú³É2n mol NH3¢ÚÓÃNH3¡¢N2¡¢H2±íʾ·´Ó¦ËÙÂʱÈΪ2¡Ã1¡Ã3 ¢Û»ìºÏÆøÌåµÄÃܶȲ»ÔÙ¸Ä±ä ¢Ü»ìºÏÆøÌåѹǿ²»ÔÙ¸Ä±ä ¢Ý»ìºÏÆøÌåƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿²»Ôٸıä
A£®¢Ù¢Û¢ÜB£®¢Ù¢Ú¢Ü¢ÝC£®¢Ù¢Ü¢ÝD£®¢Ú¢Û¢Ü
£¨2£©¹¤ÒµÉϳ£ÓÃCO2ºÍNH3ͨ¹ýÈçÏ·´Ó¦ºÏ³ÉÄòËØ[CO(NH2)2]¡£

t¡æʱ£¬ÏòÈÝ»ýºã¶¨Îª2LµÄÃܱÕÈÝÆ÷ÖмÓÈë0.10 molCO:ºÍ0. 40 molNH3 ,70 min¿ªÊ¼´ïµ½Æ½ºâ¡£·´Ó¦ÖÐCO2 ( g)µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçϱíËùʾ:
ʱ¼ä£¯min
¡¡0
30
70
80
100
n(CO2) £¯mol
0.10
0.060
0.040
0.040
0.040
 
¢Ù20 minʱ£¬¦ÔÕý(CO2 )_¡¡¡¡¡¡¡¡¡¡80 minʱ¡£¦ÔÄæ(H2O)(Ìî¡°>¡±¡¢¡°=¡±»ò¡°<¡±)¡£
¢ÚÔÚ100 minʱ£¬±£³ÖÆäËüÌõ¼þ²»±ä£¬ÔÙÏòÈÝÆ÷ÖгäÈë0. 050 mo1CO2ºÍ0. 20 molNH3£¬ÖØн¨Á¢Æ½ºâºóCO2µÄת»¯ÂÊÓëԭƽºâÏà±È½«_    (Ìî¡°Ôö´ó¡±¡¢¡°²»±ä¡±»ò¡°¼õС¡±)¡£
¢ÛÉÏÊö¿ÉÄæ·´Ó¦µÄƽºâ³£ÊýΪ_      (±£Áô¶þλСÊý)¡£
¢Ü¸ù¾Ý±íÖÐÊý¾ÝÔÚͼ¼×ÖлæÖƳöÔÚt¡æÏÂNH3µÄת»¯ÂÊËæʱ¼ä±ä»¯µÄͼÏñ;±£³ÖÆäËüÌõ¼þ²»±ä;Ôò(t+10)¡æÏÂÕýÈ·µÄͼÏñ¿ÉÄÜÊÇ         (Ìîͼ¼×Öеġ°A¡±»ò¡°B¡±)¡£

¢ÝͼÒÒËùʾװÖÃ(Òõ¡¢Ñô¼«¾ùΪ¶èÐԵ缫)¿ÉÓÃÓÚµç½âÄòËØ¡²CO(NH2)2¡³µÄ¼îÐÔÈÜÒºÖÆÈ¡ÇâÆø¡£¸Ã×°ÖÃÖÐÑô¼«µÄµç¼«·´Ó¦Ê½Îª                £¬ÈôÁ½¼«¹²ÊÕ¼¯µ½ÆøÌå22. 4L(±ê
¿ö)£¬ÔòÏûºÄµÄÄòËØΪ            g(ºöÂÔÆøÌåµÄÈܽâ)¡£

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

¾«Ó¢¼Ò½ÌÍø