ÌâÄ¿ÄÚÈÝ


ÔÚºãÈÝÃܱÕÈÝÆ÷ÖÐͨÈëX²¢·¢Éú·´Ó¦£º2X(g)Y(g)£¬Î¶ÈT1¡¢T2ÏÂXµÄÎïÖʵÄÁ¿Å¨¶Èc(X)ËæÊ±¼ät±ä»¯µÄÇúÏßÈçͼËùʾ£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ(¡¡¡¡)

A£®¸Ã·´Ó¦½øÐе½Mµã·Å³öµÄÈÈÁ¿´óÓÚ½øÐе½Wµã·Å³öµÄÈÈÁ¿

B£®T2Ï£¬ÔÚ0¡«t1ʱ¼äÄÚ£¬v(Y)£½mol¡¤L£­1¡¤min£­1

C£®MµãµÄÕý·´Ó¦ËÙÂÊvÕý´óÓÚNµãµÄÄæ·´Ó¦ËÙÂÊvÄæ

D£®MµãʱÔÙ¼ÓÈëÒ»¶¨Á¿X£¬Æ½ºâºóXµÄת»¯ÂʼõС


C¡¡[½âÎö] ¸ù¾ÝͼÏñ¿ÉÖª£¬T1£¼T2£¬¦¤H<0£¬Òò´Ë¸Ã·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬¹Êc(X)±ä»¯Ô½´ó£¬·Å³öÈÈÁ¿Ô½¶à£¬¹ÊMµã·Å³öµÄÈÈÁ¿Ð¡ÓÚWµã·Å³öµÄÈÈÁ¿£¬AÏî´íÎó£»T2ζÈÏ£¬ÔÚ0¡«t1ʱ¼äÄÚ£¬v(Y)£½v(X)£½ mol¡¤L£­1¡¤min£­1£¬BÏî´íÎó£»ÒòT1>T2£¬vMÕý£½vMÄæ>vWÄæ£¬ÓÖÒòvNÄæ£¼vWÄæ£¬ËùÒÔvNÄæ<vWÕý£¬CÕýÈ·£»ÓÉÓÚ·´Ó¦Ç°ºó¾ùÖ»ÓÐÒ»ÖÖÎïÖÊ£¬Òò´ËMµãʱÔÙÔö¼ÓÒ»¶¨Á¿µÄX£¬ÔòÏ൱ÓÚÔö´óѹǿ£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒÆ¶¯£¬XµÄת»¯ÂÊÉý¸ß£¬DÏî´íÎó¡£


Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÒÒ´¼ÊÇÖØÒªµÄÓлú»¯¹¤Ô­ÁÏ£¬¿ÉÓÉÒÒÏ©ÆøÏàÖ±½ÓË®ºÏ·¨Éú²ú»ò¼ä½ÓË®ºÏ·¨Éú²ú¡£»Ø´ðÏÂÁÐÎÊÌ⣺

(1)¼ä½ÓË®ºÏ·¨ÊÇÖ¸ÏȽ«ÒÒÏ©ÓëŨÁòËá·´Ó¦Éú³ÉÁòËáÇâÒÒõ¥(C2H5OSO3H)£¬ÔÙË®½âÉú³ÉÒÒ´¼¡£Ð´³öÏàÓ¦·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º________________________________________¡£

(2)ÒÑÖª£º

¼×´¼µÄÍÑË®·´Ó¦

2CH3OH(g)===CH3OCH3(g)£«H2O(g)

¦¤H1£½£­23.9 kJ¡¤mol£­1

¼×´¼ÖÆÏ©ÌþµÄ·´Ó¦

2CH3OH(g)===C2H4(g)£«2H2O(g)

¦¤H2£½£­29.1 kJ¡¤mol£­1

ÒÒ´¼µÄÒì¹¹»¯·´Ó¦¡¡C2H5OH(g)===CH3OCH3(g)

¦¤H3£½£«50.7 kJ¡¤mol£­1

ÔòÒÒÏ©ÆøÏàÖ±½ÓË®ºÏ·´Ó¦C2H4(g)£«H2O(g)===C2H5OH(g)µÄ¦¤H________kJ¡¤mol£­1¡£Óë¼ä½ÓË®ºÏ·¨Ïà±È£¬ÆøÏàÖ±½ÓË®ºÏ·¨µÄÓŵãÊÇ____________________________________¡£

(3)ÈçͼËùÊ¾ÎªÆøÏàÖ±½ÓË®ºÏ·¨ÖÐÒÒÏ©µÄƽºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØÏµ[ÆäÖÐn(H2O)¡Ãn(C2H4)£½1¡Ã1]¡£

¢ÙÁÐʽ¼ÆËãÒÒϩˮºÏÖÆÒÒ´¼·´Ó¦ÔÚͼÖÐAµãµÄƽºâ³£ÊýKp£½____________________(ÓÃÆ½ºâ·Öѹ´úÌæÆ½ºâŨ¶È¼ÆË㣬·Öѹ£½×Üѹ¡ÁÎïÖʵÄÁ¿·ÖÊý)¡£

¢ÚͼÖÐѹǿ(p1¡¢p2¡¢p3¡¢p4)µÄ´óС˳ÐòΪ____£¬ÀíÓÉÊÇ___________________¡£

¢ÛÆøÏàÖ±½ÓË®ºÏ·¨³£²ÉÓõŤÒÕÌõ¼þΪ£ºÁ×Ëá/¹èÔåÍÁΪ´ß»¯¼Á£¬·´Ó¦Î¶È290 ¡æ¡¢Ñ¹Ç¿6.9 MPa£¬n(H2O)¡Ãn(C2H4)£½0.6¡Ã1¡£ÒÒÏ©µÄת»¯ÂÊΪ5%£¬ÈôÒª½øÒ»²½Ìá¸ßÒÒϩת»¯ÂÊ£¬³ýÁË¿ÉÒÔÊʵ±¸Ä±ä·´Ó¦Î¶ȺÍѹǿÍ⣬»¹¿É²ÉÈ¡µÄ´ëÊ©ÓÐ________________________________________________________________________¡¢

________________________________________________________________________¡£


 ÃºÌ¿È¼ÉÕ¹ý³ÌÖлáÊͷųö´óÁ¿µÄSO2£¬ÑÏÖØÆÆ»µÉú̬»·¾³¡£²ÉÓÃÒ»¶¨µÄÍÑÁò¼¼Êõ¿ÉÒÔ°ÑÁòÔªËØÒÔCaSO4µÄÐÎʽ¹Ì¶¨£¬´Ó¶ø½µµÍSO2µÄÅÅ·Å¡£µ«ÊÇú̿ȼÉÕ¹ý³ÌÖвúÉúµÄCOÓÖ»áÓëCaSO4·¢Éú»¯Ñ§·´Ó¦£¬½µµÍÍÑÁòЧÂÊ¡£Ïà¹Ø·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÈçÏ£º

CaSO4(s)£«CO(g)CaO(s) £« SO2(g) £« CO2(g)¡¡¦¤H1£½218.4 kJ¡¤mol£­1(·´Ó¦¢ñ)

CaSO4(s)£«4CO(g)CaS(s) £« 4CO2(g)¡¡¦¤H2£½£­175.6 kJ¡¤mol£­1(·´Ó¦¢ò)

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)·´Ó¦¢ñÄܹ»×Ô·¢½øÐеķ´Ó¦Ìõ¼þÊÇ________¡£

(2)¶ÔÓÚÆøÌå²ÎÓëµÄ·´Ó¦£¬±íʾƽºâ³£ÊýKpʱÓÃÆøÌå×é·Ö(B)µÄƽºâѹǿp(B)´úÌæ¸ÃÆøÌåÎïÖʵÄÁ¿µÄŨ¶Èc(B)£¬Ôò·´Ó¦¢òµÄKp£½________(Óñí´ïʽ±íʾ)¡£

(3)¼ÙÉèijζÈÏ£¬·´Ó¦¢ñµÄËÙÂÊ(v1 )´óÓÚ·´Ó¦¢òµÄËÙÂÊ(v2 )£¬ÔòÏÂÁз´Ó¦¹ý³ÌÄÜÁ¿±ä»¯Ê¾ÒâͼÕýÈ·µÄÊÇ________¡£

(4)ͨ¹ý¼à²â·´Ó¦ÌåϵÖÐÆøÌåŨ¶ÈµÄ±ä»¯¿ÉÅжϷ´Ó¦¢ñºÍ¢òÊÇ·ñͬʱ·¢Éú£¬ÀíÓÉÊÇ____________________________________________________________________¡£

¡¡¡¡¡¡¡¡¡¡¡¡¡¡              ¡¡A¡¡¡¡¡¡¡¡¡¡¡¡B

¡¡¡¡¡¡¡¡¡¡¡¡¡¡               ¡¡C¡¡¡¡¡¡¡¡¡¡¡¡D

(5)ͼ(a)ΪʵÑé²âµÃ²»Í¬Î¶ÈÏ·´Ó¦ÌåϵÖÐCO³õʼÌå»ý°Ù·ÖÊýÓëÆ½ºâʱ¹ÌÌå²úÎïÖÐCaSÖÊÁ¿°Ù·ÖÊýµÄ¹ØÏµÇúÏß¡£Ôò½µµÍ¸Ã·´Ó¦ÌåϵÖÐSO2Éú³ÉÁ¿µÄ´ëÊ©ÓÐ________¡£

A£®Ïò¸Ã·´Ó¦ÌåϵÖÐͶÈëʯ»Òʯ

B£®ÔÚºÏÊʵÄζÈÇø¼ä¿ØÖƽϵ͵ķ´Ó¦Î¶È

C£®Ìá¸ßCOµÄ³õʼÌå»ý°Ù·ÖÊý

D£®Ìá¸ß·´Ó¦ÌåϵµÄζÈ

(6)ºãκãÈÝÌõ¼þÏ£¬¼ÙÉè·´Ó¦¢ñºÍ¢òͬʱ·¢Éú£¬ÇÒv1>v2£¬ÇëÔÚͼ(b)»­³ö·´Ó¦ÌåϵÖÐc(SO2)ËæÊ±¼ät±ä»¯µÄ×ÜÇ÷ÊÆÍ¼¡£

¡¡

¡¡¡¡¡¡¡¡¡¡(a)¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡¡(b)

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

¾«Ó¢¼Ò½ÌÍø