ÌâÄ¿ÄÚÈÝ

¡¾ÌâÄ¿¡¿ÔÚ2 LµÄÃܱÕÈÝÆ÷ÖÐÓÐ3ÖÖÎïÖʽøÐз´Ó¦£¬X¡¢Y¡¢ZµÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯µÄÇúÏßÈçͼËùʾ£¬·´Ó¦ÔÚt1minʱ´ïµ½»¯Ñ§Æ½ºâ״̬¡£

(1) 0¡«t1minÄÚ£¬XµÄŨ¶È±ä»¯Á¿Îª___________YµÄŨ¶È±ä»¯Á¿______________

(2) 0¡«t1minÄÚ£¬YµÄƽ¾ù·´Ó¦ËÙÂÊΪ_________________________________¡£

(3) X¡¢Y¡¢ZÈýÕßµÄËÙÂÊÖ®±ÈΪ_____________________¡£

(4) ¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ_______________________________________¡£

(5) ÏÂÁйØÓڸ÷´Ó¦µÄ˵·¨ÕýÈ·µÄÊÇ__________________________________¡£

A£®t1minʱ£¬¸Ã·´Ó¦ÒÑÍ£Ö¹

B£®t1min֮ǰ£¬XµÄÏûºÄËÙÂÊ´óÓÚËüµÄÉú³ÉËÙÂÊ

C£®t1minʱ£¬Õý·´Ó¦ËÙÂʵÈÓÚÄæ·´Ó¦ËÙÂÊ

¡¾´ð°¸¡¿ 0.4mol/L 0.6mol/L 0.6/t1mol¡¤L£­1¡¤min£­1 V(X):V(Y):V(Z)=2:3:1 2X3Y£«Z BC

¡¾½âÎö¡¿(1) ¸ù¾ÝͼÏñ£¬0¡«t1minÄÚ£¬XµÄŨ¶È±ä»¯Á¿Îª==0.4 mol/L£¬YµÄŨ¶È±ä»¯Á¿==0.6 mol/L£¬¹Ê´ð°¸Îª£º0.4mol/L£»0.6mol/L£»

(2) 0¡«t1minÄÚ£¬YµÄƽ¾ù·´Ó¦ËÙÂÊΪ== mol/(Lmin)£¬¹Ê´ð°¸Îª£º mol/(Lmin)£»

(3) X¡¢Y¡¢ZÈýÕßµÄËÙÂÊÖ®±ÈµÈÓÚÎïÖʵÄÁ¿µÄ±ä»¯Á¿Ö®±È=0.8mol£º1.2mol£º0.4mol=2:3:1£¬¹Ê´ð°¸Îª£º2:3:1£»

(4)ͼÏó¿ÉÖª´Ó0¡«tʱ¿Ì£¬n(X)¼õС£¬n(Y)Ôö´ó£¬n(Z)Ôö´ó£¬¿ÉÒÔÈ·¶¨XΪ·´Ó¦ÎYZΪÉú³ÉÎ¡÷n(X)£º¡÷n(Y)£º¡÷n(Z)=0.8£º1.2£º0.4=2£º3£º1£¬ËùÒÔ·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£º2X3Y+Z£»¹Ê´ð°¸Îª£º2X3Y+Z£»

(5)A£®»¯Ñ§Æ½ºâÊÇÒ»ÖÖ¶¯Ì¬Æ½ºâ£¬´ïµ½Æ½ºâ״̬ʱ£¬ÕýÄæ·´Ó¦ËÙÂÊÏàµÈ£¬µ«²»Îª0£¬¹ÊA´íÎó£»B£®ÔÚt1min֮ǰ£¬¿ÉÄæ·´Ó¦ÉÐδ´ïµ½»¯Ñ§Æ½ºâ£¬Æ½ºâÕýÏòÒƶ¯£¬Ôò´ËʱvÕý(X)£¾vÄæ(X)£¬¹ÊBÕýÈ·£»C£®ÔÚt1min£¬¿ÉÄæ·´Ó¦ÒÑ´ïµ½»¯Ñ§Æ½ºâ£¬´ËʱÕý¡¢Äæ·´Ó¦ËÙÂÊÏàµÈ£¬¹ÊCÕýÈ·£»¹Ê´ð°¸Îª£ºBC¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

¡¾ÌâÄ¿¡¿Ì¼¼°Æ仯ºÏÎïÔÚ»¯¹¤Éú²úÖÐÓÐ׏㷺µÄÓ¦Óá£

I.Ϊ½â¾ö´óÆøÖÐCO2µÄº¬Á¿Ôö´óµÄÎÊÌ⣬ij¿Æѧ¼ÒÌá³öÈçϹ¹Ï룺

°Ñ¹¤³§ÅųöµÄ¸»º¬CO2µÄ·ÏÆø¾­¾»»¯´µÈë̼Ëá¼ØÈÜÒºÎüÊÕ£¬È»ºóÔÙ°ÑCO2´ÓÈÜÒºÖÐÌáÈ¡³öÀ´£¬ÔںϳÉËþÖо­»¯Ñ§·´Ó¦Ê¹·ÏÆøÖеÄCO2ת±äΪȼÁϼ״¼¡£

²¿·Ö¼¼ÊõÁ÷³ÌÈçÏÂ:

¢ÅºÏ³ÉËþÖз´Ó¦µÄ»¯Ñ§·½³ÌʽΪ____________£»¡÷H<0¡£¸Ã·´Ó¦Îª¿ÉÄæ·´Ó¦£¬´ÓƽºâÒƶ¯Ô­Àí·ÖÎö£¬µÍÎÂÓÐÀûÓÚÌá¸ßÔ­ÁÏÆøµÄƽºâת»¯ÂÊ¡£¶øʵ¼ÊÉú²úÖвÉÓÃ300¡æµÄζȣ¬³ý¿¼ÂÇζȶԷ´Ó¦ËÙÂʵÄÓ°ÏìÍ⣬»¹Ö÷Òª¿¼ÂÇÁË___________________________________________________________________¡£

(2)´ÓºÏ³ÉËþ·ÖÀë³ö¼×´¼µÄÔ­ÀíÓëÏÂÁÐ_______²Ù×÷µÄÔ­Àí±È½ÏÏà·û(Ìî×Öĸ£©

A.¹ýÂË B.·ÖÒº C.ÕôÀ¡ D.½á¾§

(3)È罫CO2ÓëH2ÒÔ1:4µÄÌå»ý±È»ìºÏ£¬ÔÚÊʵ±µÄÌõ¼þÏ¿ÉÖƵÃCH4¡£Ð´³öCO2(g)ÓëH2(g)·´Ó¦ÉúCH4(g)ÓëҺ̬ˮµÄÈÈ»¯Ñ§·½³Ìʽ:

ÒÑÖª£ºCH4(g)+2O2(g)=CO2(g)+2H2O(l) ¡÷H1=-890.3kJ/mol

2H2(g)+O2(g)=2H2O(l) ¡÷H2=-571.6kJ/mol

_______________________________________________________________________¡£

II.¼×ÍéȼÉÕ»á·Å³ö´óÁ¿µÄÈÈ£¬¿É×÷ΪÄÜÔ´Ó¦ÓÃÓÚÈËÀàµÄÉú²úºÍÉú»î¡£

¼ºÖª£º¢Ù2CH4(g)+3O2(g)=2CO(g)+4H2O(l)£» ¡÷H1=-1214.6kJ/mol

¢ÚCO2(g)=CO(g)+1/2O2(g)£» ¡÷H2=+283.0kJ/mol

Ôò±íʾ¼×ÍéȼÉÕÈȵÄÈÈ»¯Ñ§·½³Ìʽ______________________________¡£

III.ijÐËȤС×éÄ£Ä⹤ҵºÏ³É¼×´¼µÄ·´Ó¦:CO(g)+2H2(g)CH3OH(g)£¬ÔÚÈÝ»ý¹Ì¶¨Îª2LµÄÃܱÕÈÝÆ÷ÖгäÈë1mol CO 2mol H2£¬¼ÓÈëºÏÊʵĴ߻¯¼Á£¨´ß»¯¼ÁÌå»ýºöÂÔ²»¼Æ£©ºó¿ªÊ¼·´Ó¦¡£²âµÃÈÝÆ÷ÄÚµÄѹǿËæʱ¼ä±ä»¯ÈçÏÂ:

ʱ¼ä/min

0

5

10

15

20

25

ѹǿ/Mpa

12.6

10.8

9.5

8.7

8.4

8.4

(1)´Ó·´Ó¦¿ªÊ¼µ½20minʱ£¬ÒÔCO±íʾ·´Ó¦ËÙÂÊΪ_____________________¡£

(2)ÏÂÁÐÃèÊöÄÜ˵Ã÷·´Ó¦´ïµ½Æ½ºâµÄÊÇ_______________________

A.×°ÖÃÄÚÆøÌåÑÕÉ«²»Ôٸıä B.ÈÝÆ÷ÄÚÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿±£³Ö²»±ä

C.ÈÝÆ÷ÄÚÆøÌåµÄѹǿ±£³Ö²»±ä D.ÈÝÆ÷ÄÚÆøÌåÃܶȱ£³Ö²»±ä

(3)¸ÃζÈÏÂƽºâ³£ÊýK=____£¬Èô´ïµ½Æ½ºâºó¼ÓÈëÉÙÁ¿CH3OH(g)£¬´Ëʱƽºâ³£ÊýKÖµ½«____ (Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)

(4)¸Ã·´Ó¦´ïµ½Æ½ºâºó£¬ÔÙÏòÈÝÆ÷ÖгäÈë1mol CO 2mol H2£¬´ËʱCOµÄת»¯Âʽ«__________(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)

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

¾«Ó¢¼Ò½ÌÍø