ÌâÄ¿ÄÚÈÝ

ÏÖÓз´Ó¦£ºmA(g)£«nB(g) pC(g)£¬´ïµ½Æ½ºâºó£¬µ±Éý¸ßζÈʱ£¬BµÄת»¯Âʱä´ó£»µ±¼õСѹǿʱ£¬»ìºÏÌåϵÖÐCµÄÖÊÁ¿·ÖÊý¼õС£¬Ôò£º
(1)¸Ã·´Ó¦µÄÄ淴ӦΪ________ÈÈ·´Ó¦£¬ÇÒm£«n______p (Ìî¡°£¾¡±¡¢¡°£½¡±»ò¡°£¼¡±) ¡£
(2)¼õѹʱ£¬AµÄÖÊÁ¿·ÖÊý__________¡£(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£¬ÏÂͬ)
(3)ÈôÈÝ»ý²»±ä¼ÓÈëB£¬ÔòAµÄת»¯ÂÊ__________£¬BµÄת»¯ÂÊ__________¡£
(4)ÈôÉý¸ßζȣ¬ÔòƽºâʱB¡¢CµÄŨ¶ÈÖ®±È½«__________¡£
(5)Èô¼ÓÈë´ß»¯¼Á£¬Æ½ºâʱÆøÌå»ìºÏÎïµÄ×ÜÎïÖʵÄÁ¿__________¡£
(6)ÈôBÊÇÓÐÉ«ÎïÖÊ£¬A¡¢C¾ùÎÞÉ«£¬Ôò¼ÓÈëC(Ìå»ý²»±ä)ʱ»ìºÏÎïÑÕÉ«__________£»¶øά³ÖÈÝÆ÷ÄÚѹǿ²»±ä£¬³äÈëÄÊÆøʱ£¬»ìºÏÎïÑÕÉ«         ¡£(Ìî¡°±äÉ¡¢¡°±ädz¡±»ò¡°²»±ä¡±)

(1)·Å £¾ (2)Ôö´ó (3)Ôö´ó ¼õС (4)¼õС (5)²»±ä (6)±äÉî ±ädz

½âÎöÊÔÌâ·ÖÎö£º£¨1£©´ïµ½Æ½ºâºó£¬µ±Éý¸ßζÈʱ£¬BµÄת»¯Âʱä´ó£¬ËµÃ÷ζÈÉý¸ßƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬ÔòÕý·´Ó¦ÎüÈÈ£¬Ä淴ӦΪ·ÅÈÈ·´Ó¦¡£µ±¼õСѹǿʱ£¬»ìºÏÌåϵÖÐCµÄÖÊÁ¿·ÖÊýÒ²¼õС£¬ËµÃ÷ѹǿ¼õСƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬Ôò·½³ÌʽÖз´Ó¦ÎïµÄÆøÌåµÄ¼ÆÁ¿ÊýÖ®ºÍ´óÓÚÉú³ÉÎïÆøÌåµÄ»¯Ñ§¼ÆÁ¿ÊýÖ®ºÍ£¬¼´m£«n£¾p¡£
£¨2£©¼õѹƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬ÔòAµÄÖÊÁ¿·ÖÊýÔö´ó¡£
£¨3£©ÔÚ·´Ó¦ÈÝÆ÷ÖмÓÈëÒ»¶¨Á¿µÄB£¬·´Ó¦ÎïBµÄŨ¶ÈÔö´ó£¬Æ½ºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬ÔòAµÄת»¯ÂÊÔö´ó£¬B¼ÓÈëµÄ¶à£¬¶øת»¯µÄÉÙ£¬ÔòBµÄת»¯ÂÊ·´¶ø¼õС¡£
£¨4£©Õý·´Ó¦ÎüÈÈ£¬ÔòÉý¸ßζÈƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯£¬BµÄÎïÖʵÄÁ¿¼õС£¬CµÄÎïÖʵÄÁ¿Ôö¶à£¬ËùÒÔ¶þÕߵıÈÖµ½«¼õС¡£
£¨5£©´ß»¯¼Á¶Ô»¯Ñ§Æ½ºâÒƶ¯Ã»ÓÐÓ°Ï죬ËùÒÔÈô¼ÓÈë´ß»¯¼Á£¬Æ½ºâʱÆøÌå»ìºÏÎïµÄ×ÜÎïÖʵÄÁ¿²»±ä¡£
£¨6£©ÈôBÊÇÓÐÉ«ÎïÖÊ£¬A¡¢C¾ùÎÞÉ«£¬Ôò¼ÓÈëCƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬BµÄŨ¶ÈÔö´ó£¬ÔòÑÕÉ«¼ÓÉî¡£¶øά³ÖÈÝÆ÷ÄÚѹǿ²»±ä£¬³äÈëÄÊÆøʱ£¬Ìå»ýÔö´ó£¬¶ÔÓÚ·´Ó¦ÌåϵÀ´Ëµ£¬Ï൱ÓÚ¼õСѹǿ£¬ÔòƽºâÏòÄæ·´Ó¦·½ÏòÒƶ¯£¬µ«Òƶ¯µÄÁ¿Ô¶Ð¡ÓÚÌå»ýÔö´óµÄÒòËØ£¬ÔòBµÄŨ¶È¼õС£¬ÑÕÉ«±ädz¡£
¿¼µã£º¿¼²é¿ÉÄæ·´Ó¦µÄÅжϡ¢Íâ½çÌõ¼þ¶Ôƽºâ״̬µÄÓ°Ïì

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¼×´¼±»³ÆΪ2lÊÀ¼ÍµÄÐÂÐÍȼÁÏ£¬¹¤ÒµÉÏͨ¹ýÏÂÁз´Ó¦IºÍII£¬ÓÃCH4ºÍH2OΪԭÁÏÀ´ÖƱ¸¼×´¼£º
CH4(g)+H2O(g)  CO(g)+3H2 (g)¡­¡­I   CO(g)+2H2(g) CH3OH(g) ¡­¡­II¡£
£¨1£©½«1.0 mol CH4ºÍ2.0 mol H2O(g)ͨÈëÈÝ»ýΪ100L·´Ó¦ÊÒ£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦I£¬CH4µÄƽºâת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼ¡£

¢ÙÒÑÖª100¡æʱ´ïµ½Æ½ºâËùÐèµÄʱ¼äΪ5min£¬ÔòÓÃH2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ____________________¡£
¢ÚͼÖеÄP1_________P2£¨Ìî¡°<¡±¡¢¡°>¡±»ò¡°=¡±£©£¬100¡æʱƽºâ³£ÊýµÄֵΪ__________ ¡£
£¨2£©ÔÚѹǿΪ0.1 MPaÌõ¼þÏ£¬ ½«a mol COÓë 3a mol H2µÄ»ìºÏÆøÌåÔÚ´ß»¯¼Á×÷ÓÃÏ£¬×Ô·¢·´Ó¦¢ò£¬Éú³É¼×´¼¡£
¢Û¸Ã·´Ó¦µÄ¡÷H ____ 0£»ÈôÈÝÆ÷ÈÝ»ý²»±ä£¬ÏÂÁдëÊ©¿ÉÔö¼Ó¼×´¼²úÂʵÄÊÇ________¡£

A£®Éý¸ßÎÂ¶È B£®½«CH3OH(g)´ÓÌåϵÖзÖÀë
C£®³äÈëHe£¬Ê¹Ìåϵ×ÜѹǿÔö´ó D£®ÔÙ³äÈë1mol COºÍ3mol H2
¢ÜΪÁËÑ°ÕҺϳɼ״¼µÄζȺÍѹǿµÄÊÊÒËÌõ¼þ£¬Ä³Í¬Ñ§Éè¼ÆÈý×éʵÑ飬²¿·ÖʵÑéÌõ¼þÒѾ­ÌîÔÚÏÂÃæʵÑéÉè¼Æ±íÖС£
ʵÑé±àºÅ
T(¡æ)
N(CO)/n(H2)
P£¨Mpa£©
i
150
1/3
0.1
ii
 
 
5
iii
350
 
5
 
a£®ÇëÔÚÉϱí¿Õ¸ñÖÐÌîÈëÊ£ÓàµÄʵÑéÌõ¼þÊý¾Ý¡£
b£®¸ù¾Ý·´Ó¦IIµÄÌص㣬ÔÚ¸ø³öµÄ×ø±êͼÖУ¬»­³öÔÚ5MPaÌõ¼þÏÂCOµÄת»¯ÂÊËæζȱ仯µÄÇ÷ÊÆÇúÏßʾÒâͼ£¬²¢±êÃ÷ѹǿ¡£

(12·Ö)ÏòÌå»ýΪ2LµÄ¹Ì¶¨ÃܱÕÈÝÆ÷ÖÐͨÈë3molXÆøÌå,ÔÚÒ»¶¨Î¶ÈÏ·¢ÉúÈçÏ·´Ó¦£º
2X(g)Y(g)+3Z(g)
£¨1£©¾­5minºó·´Ó¦´ïµ½Æ½ºâ,´Ëʱ²âµÃÈÝÆ÷ÄÚµÄѹǿΪÆðʼʱµÄ1.2±¶,ÔòÓÃYµÄÎïÖʵÄÁ¿Å¨¶È±ä»¯±íʾµÄËÙÂÊΪ             ¡£
£¨2£©ÈôÉÏÊö·´Ó¦Ôڼס¢ÒÒ¡¢±û¡¢¶¡ËĸöͬÑùµÄÃܱÕÈÝÆ÷ÖнøÐÐ,ÔÚͬһ¶Îʱ¼äÄÚ²âµÃÈÝÆ÷Äڵķ´Ó¦ËÙÂÊ·Ö±ðΪ£º¼×v(X)£½3.5mol/(L?min)£»ÒÒv(Y)£½2mol/(L?min)£»±ûv(Z)=4.5mol/(L?min)£»¶¡v(X)£½0.075mol/(L?s)¡£ÈôÆäËüÌõ¼þÏàͬ,ζȲ»Í¬£¬ÔòζÈÓɸߵ½µÍµÄ˳ÐòÊÇ(ÌîÐòºÅ)            ¡£
£¨3£©ÈôÏò´ïµ½(1)µÄƽºâÌåϵÖгäÈëë²Æø,ÔòƽºâÏò             (Ìî"×ó"»ò"ÓÒ"»ò"²»")Òƶ¯£»ÈôÏò
´ïµ½(1)µÄƽºâÌåϵÖÐÒÆ×ß²¿·Ö»ìºÏÆøÌå,ÔòƽºâÏò          (Ìî" ×ó " »ò " ÓÒ " »ò " ²»")Òƶ¯¡£
£¨4£©ÈôÔÚÏàͬÌõ¼þÏÂÏò´ïµ½(1)ËùÊöµÄƽºâÌåϵÖÐÔÙ³äÈë0.5molXÆøÌå,ÔòƽºâºóXµÄת»¯ÂÊ¢ÈÓë¢ÅµÄ
ƽºâÖеÄXµÄת»¯ÂÊÏà±È½Ï              ¡£

A£®ÎÞ·¨È·¶¨B£®¢ÈÒ»¶¨´óÓÚ¢ÅC£®¢ÈÒ»¶¨µÈÓÚ¢ÅD£®¢ÈÒ»¶¨Ð¡ÓÚ¢Å
£¨5£©Èô±£³ÖζȺÍѹǿ²»±ä,Æðʼʱ¼ÓÈëX¡¢Y¡¢ZÎïÖʵÄÁ¿·Ö±ðΪamol¡¢bmol¡¢cmol,´ïµ½Æ½ºâʱÈÔ
Óë(1)µÄƽºâµÈЧ,Ôòa¡¢b¡¢cÓ¦¸ÃÂú×ãµÄ¹ØϵΪ                        ¡£
£¨6£©Èô±£³ÖζȺÍÌå»ý²»±ä£¬Æðʼʱ¼ÓÈëX¡¢Y¡¢ZÎïÖʵÄÁ¿·Ö±ðΪamol¡¢bmol¡¢cmol,´ïµ½Æ½ºâʱÈÔ
Óë(1)µÄƽºâµÈЧ,ÇÒÆðʼʱά³Ö»¯Ñ§·´Ó¦ÏòÄæ·´Ó¦·½Ïò½øÐÐ,ÔòcµÄÈ¡Öµ·¶Î§Ó¦¸ÃΪ              ¡£

½ðÊôÎÙÓÃ;¹ã·º£¬Ö÷ÒªÓÃÓÚÖÆÔìÓ²ÖÊ»òÄ͸ßεĺϽð£¬ÒÔ¼°µÆÅݵĵÆË¿¡£¸ßÎÂÏ£¬ÔÚÃܱÕÈÝÆ÷ÖÐÓÃH2»¹Ô­WO3¿ÉµÃµ½½ðÊôÎÙ£¬Æä×Ü·´Ó¦Îª£º
WO3 (s) + 3H2 (g)  W (s) + 3H2O (g)       Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÔÚÒ»¶¨Î¶ÈϽøÐÐÉÏÊö·´Ó¦£¬Èô·´Ó¦ÈÝÆ÷µÄÈÝ»ýΪ0£®5 L£¬2minºó´ïµ½Æ½ºâ£¬²âµÃ¹ÌÌåµÄÖÊÁ¿¼õÉÙÁË4£®80 g£¬ÔòH2µÄƽ¾ù·´Ó¦ËÙÂÊ_________£»¸Ã·´Ó¦µÄƽºâ³£Êý±í´ïʽK=___________
£¨2£©Ä³Î¶ÈÏ·´Ó¦´ïƽºâʱ£¬H2ÓëË®ÕôÆøµÄÌå»ý±ÈΪ2:3£¬ÔòH2µÄƽºâת»¯ÂÊΪ__________£»ËæζȵÄÉý¸ß£¬H2ÓëË®ÕôÆøµÄÌå»ý±È¼õС£¬Ôò¸Ã·´Ó¦Îª  ____£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£© ·´Ó¦¡£
£¨3£©Ò»¶¨Î¶ÈÏ£¬ÔÚÃܱպãÈݵÄÈÝÆ÷ÖУ¬ÄܱíʾÉÏÊö·´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÊÇ________¡£
A£®»ìºÏÆøÌåµÄ×Üѹǿ±£³Ö²»±ä      B£®vÕý(H20)= vÕý(H2)
C£®»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä        D£®»ìºÏÆøÌåµÄƽ¾ùʽÁ¿±£³Ö²»±ä
£¨4£©ÎÙË¿µÆ¹ÜÖеÄWÔÚʹÓùý³ÌÖлºÂý»Ó·¢£¬Ê¹µÆË¿±äϸ£¬¼ÓÈëI2¿ÉÑÓ³¤µÆ¹ÜµÄʹÓÃÊÙÃü£¬Æ乤×÷Ô­ÀíΪ£ºW (s) +2I2 (g)  WI4 (g)¡£ÏÂÁÐ˵·¨ÕýÈ·µÄÓÐ__________¡£
A£®Î¶ÈÉý¸ßʱ£¬WI4µÄ·Ö½âËÙÂʼӿ죬WºÍI2µÄ»¯ºÏËÙÂʼõÂý 
B£®WI4ÔÚµÆË¿ÉϷֽ⣬²úÉúµÄWÓÖ³Á»ýÔÚµÆË¿ÉÏ
C£®WI4ÔڵƹܱÚÉϷֽ⣬ʹµÆ¹ÜµÄÊÙÃüÑÓ³¤     
D£®µÆ¹ÜÄÚµÄI2¿ÉÑ­»·Ê¹ÓÃ

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

¾«Ó¢¼Ò½ÌÍø