1¡¢Ìî¿ÕÌâ £¨12·Ö£©ÔÚ450¡æ²¢Óд߻¯¼Á´æÔÚÏ£¬ÓÚÒ»ÈÝ»ýºã¶¨µÄÃܱÕÈÝÆ÷ÄÚ½øÐÐÏÂÁз´Ó¦£º
2SO2(g)£«O2(g)
2SO3(g)?¡÷H£½¨D190 kJ¡¤mol¡ª1
¢Å¸Ã·´Ó¦500¡æÊ±µÄƽºâ³£Êý________450¡æÊ±µÄƽºâ³£Êý£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°£½¡±£©¡£
¢ÆÅжϸ÷´Ó¦´ïµ½Æ½ºâ״̬µÄ±êÖ¾ÊÇ_______________¡££¨Ìî×Öĸ£©
a£®SO2ºÍSO3Ũ¶ÈÏàµÈ? b£®SO2°Ù·Öº¬Á¿±£³Ö²»±ä
c£®ÈÝÆ÷ÖÐÆøÌåµÄѹǿ²»±ä? d£®v(O2)Õý£½2v(SO3)Äæ
e£®ÈÝÆ÷ÖлìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
¢ÇÓûÌá¸ßSO2µÄת»¯ÂÊ£¬ÏÂÁдëÊ©¿ÉÐÐ
µÄÊÇ_______________¡££¨Ìî×Öĸ£©
a£®Ïò×°ÖÃÖÐÔÙ³äÈëN2? b£®Ïò×°ÖÃÖÐÔÙ³äÈëO2
c£®¸Ä±ä·´Ó¦µÄ´ß»¯¼Á? d£®Éý¸ßζÈ
¢ÈÔÚÒ»¸ö¹Ì¶¨ÈÝ»ýΪ5 LµÄÃܱÕÈÝÆ÷ÖгäÈë0.20 mol SO2ºÍ0.10 mol O2£¬°ë·ÖÖÓºó´ïµ½Æ½ºâ£¬²âµÃÈÝÆ÷Öк¬SO3 0.18 mol£»Èô¼ÌÐøÍ¨Èë0.20 mol SO2ºÍ0.10 mol O2£¬Ôòƽºâ_________________ÒÆ¶¯£¨Ìî¡°ÏòÕý·´Ó¦·½Ïò¡±¡¢¡°ÏòÄæ·´Ó¦·½Ïò¡±»ò¡°²»¡±£©£¬Ôٴδﵽƽºâºó£¬______ mol£¼n(SO3)£¼______mol¡£
²Î¿¼´ð°¸£º
±¾Ìâ½âÎö£ºÂÔ
±¾ÌâÄѶȣºÒ»°ã
2¡¢Ñ¡ÔñÌâ ¶ÔÓÚ¿ÉÄæ·´Ó¦? 2AB3(g)
?A2(g) £« 3B2(g) ; ¦¤H£¾0£¬ÏÂÁÐͼÏñÕýÈ·µÄÊÇ
²Î¿¼´ð°¸£ºD
±¾Ìâ½âÎö£ºÂÔ
±¾ÌâÄѶȣºÒ»°ã
3¡¢Ìî¿ÕÌâ
¢ñ. ¼ÓÈë3mol SO2ºÍ2mol O2·¢Éú·´Ó¦£¬´ïµ½Æ½ºâʱ·Å³öµÄÈÈÁ¿Îª ____________
¢ò. ±£³Öͬһ·´Ó¦Î¶ȣ¬ÔÚÏàͬÈÝÆ÷ÖУ¬½«ÆðʼÎïÖʵÄÁ¿¸ÄΪ? a mol SO2?¡¢ b mol O2 ¡¢c mol SO3 (g) £¬ÓûʹƽºâʱSO3µÄÌå»ý·ÖÊýΪ2/9¡£
¢Å´ïµ½Æ½ºâʱ£¬¢ñºÍ¢ò·Å³öµÄÈÈÁ¿ ______£¨ÌîÐòºÅ£©
A£®¡ª¶¨ÏàµÈ? B£®Ç°ÕßÒ»¶¨Ð¡? C. ǰÕß´óÓÚ»òµÈÓÚºóÕß
(2) a¡¢b¡¢c±ØÐëÂú×ãµÄ¹ØÏµÊÇ(Ò»¸öÓÃa¡¢c±íʾ£¬ÁíÒ»¸öÓÃb¡¢c±íʾ)________________
²Î¿¼´ð°¸£º¢Ç ÓûʹÆðʼ·´Ó¦±íÏÖΪÏòÕý·´Ó¦·½Ïò½øÐУ¬a µÄȡֵ·¶Î§ÊÇ____________________¢ñ? 98.3KJ?¢ò¢Å C?¢Æa+c="3?" 2b+c="4?" ¢Ç 2<a¡Ü3
±¾Ìâ½âÎö£º±¾ÌâΪ»¯Ñ§Æ½ºâºÍ·´Ó¦ÈÈ×ÛºÏÊÔÌ⣬Ϊ½ÏÄÑÊÔÌâ¡£
¢ñ¡¢ÓÉÌâÒâÆøÌåѹǿΪÆðʹʱµÄ90%£¬Ôò·´Ó¦ºó×ܵÄÎïÖʵÄÁ¿ÎªÆðʹʱµÄ90%£¬Éè·´Ó¦¹ý³ÌÖÐO2ÏûºÄµÄÎïÖʵÄÁ¿Îªx£¬Ôòƽºâʱ×ÜÎïÖʵÄÁ¿Îª£¨5-x£©£¬£¨5-x£©=4.5molÖªx=0.5mol¼´·´Ó¦Á˵ÄO2Ϊ0.5mol£¬´Ëʱ·Å³öµÄÈÈÁ¿Îª196.6KJ¨M2= 98.3KJ¡£Í¬Ê±¿ÉÇó³öSO3µÄÌå»ý·ÖÊýΪ2/9¡£
¢ò¡¢Æ½ºâʱSO3µÄÌå»ý·ÖÊýҲΪ2/9£¬ÓÉ¢ñÖªÁ½ÕßΪµÈЧƽºâ¡£¢ÅÈôcΪÁ㣬Ôò·Å³öµÄÈÈÁ¿Á½ÕßÏàµÈ£¬Èôc²»ÎªÁ㣬Ôò·Å³öµÄÈÈÁ¿Ç°Õß´óÓÚºóÕß¡£(2)ÓɵÈЧƽºâÖеÈÁ¿µÈЧ¹ØÏµ²»ÄÑÕÒ³öa+c="3?" 2b+c=4¹ØÏµ£»(3)Óɼ«Öµ·¨¿ÉÇó2<a¡Ü3¡£
±¾ÌâÄѶȣºÒ»°ã
4¡¢Ñ¡ÔñÌâ ÔÚÒ»¸ö²»µ¼ÈȵÄÃܱշ´Ó¦Æ÷ÖÐ,Ö»·¢ÉúÁ½¸ö·´Ó¦:
a(g)+b(g)
2c(g)¡¡¦¤H1<0
x(g)+3y(g)
2z(g)¡¡¦¤H2>0
½øÐÐÏà¹Ø²Ù×÷ÇҴﵽƽºâºó(ºöÂÔÌå»ý¸Ä±äËù×öµÄ¹¦),ÏÂÁÐÐðÊö´íÎóµÄÊÇ(¡¡¡¡)
A£®µÈѹʱ,ͨÈë¶èÐÔÆøÌå,cµÄÎïÖʵÄÁ¿²»±ä
B£®µÈѹʱ,ͨÈëzÆøÌå,·´Ó¦Æ÷ÖÐζÈÉý¸ß
C£®µÈÈÝʱ,ͨÈë¶èÐÔÆøÌå,¸÷·´Ó¦ËÙÂʲ»±ä
D£®µÈÈÝʱ,ͨÈëzÆøÌå,yµÄÎïÖʵÄÁ¿Å¨¶ÈÔö´ó
²Î¿¼´ð°¸£ºA
±¾Ìâ½âÎö£º±¾ÌâÒªÌØ±ð×¢ÒâÌâ¸ÉÖеÄÐÅÏ¢¡°²»µ¼ÈȵÄÃܱշ´Ó¦Æ÷¡±¡£AÏî,µÈѹʱ,ͨÈë¶èÐÔÆøÌå,ÆøÌåµÄÌå»ýÔö´ó,ƽºâx(g)+3y(g)
2z(g)(¦¤H2>0)Ïò×óÒÆ¶¯,·´Ó¦·ÅÈÈ,·´Ó¦ÌåϵµÄζÈÉý¸ß,ÓÉÓڸ÷´Ó¦ÈÝÆ÷ÊÇÒ»¸ö²»µ¼ÈȵÄÈÝÆ÷,ËùÒÔÆ½ºâa(g)+b(g)
2c(g)Ò²Ïò×ó(ÎüÈÈ·½Ïò)ÒÆ¶¯,cµÄÎïÖʵÄÁ¿¼õС,¹ÊA²»ÕýÈ·;BÏî,µÈѹʱ,ͨÈëzÆøÌå,Ôö´óÁËÉú³ÉÎïµÄŨ¶È,ƽºâx(g)+3y(g)
2z(g)Ïò×óÒÆ¶¯,ÓÉÓڸ÷´Ó¦µÄÄæ·´Ó¦ÊÇ·ÅÈÈ·´Ó¦,ËùÒÔ·´Ó¦Æ÷µÄζÈÉý¸ß,BÏîÕýÈ·;µÈÈÝʱ,ͨÈë¶èÐÔÆøÌå,¸÷·´Ó¦ÎïºÍÉú³ÉÎïµÄÎïÖʵÄÁ¿Ã»Óб仯,¼´¸÷×é·ÖµÄŨ¶ÈûÓз¢Éú±ä»¯,ËùÒÔ¸÷×é·ÖµÄ·´Ó¦ËÙÂʲ»·¢Éú±ä»¯,CÏîÕýÈ·;µÈÈÝʱ,ͨÈëzÆøÌå,Ôö´óÁËÉú³ÉÎïzµÄŨ¶È,ƽºâÄæÏòÒÆ¶¯,ËùÒÔyµÄÎïÖʵÄÁ¿Å¨¶ÈÔö´ó,DÏîÕýÈ·¡£
±¾ÌâÄѶȣºÒ»°ã
5¡¢Ìî¿ÕÌâ ÁòµâÑ»··Ö½âË®ÖÆÇâÖ÷񻃾¼°ÏÂÁз´Ó¦£º
I SO2+2H2O+I2=H2SO4+2HI
II 2HI
H2+I2
¢ó 2H2SO4=2SO2+O2+2H2O
£¨1£©·ÖÎöÉÏÊö·´Ó¦£¬ÏÂÁÐÅжÏÕýÈ·µÄÊÇ____¡£
a£®·´Ó¦¢óÒ×ÔÚ³£ÎÂϽøÐÐ
b£®·´Ó¦IÖÐSO2Ñõ»¯ÐÔ±ÈHIÇ¿
c£®Ñ»·¹ý³ÌÖÐÐè²¹³äH2O
d£®Ñ»·¹ý³Ì²úÉú1molO2µÄͬʱ²úÉú1mol H2
£¨2£©Ò»¶¨Î¶ÈÏ£¬Ïò1LÃܱÕÈÝÆ÷ÖмÓÈë1molHI(g)£¬·¢Éú ·´Ó¦II£¬H2ÎïÖʵÄÁ¿ËæÊ±¼äµÄ±ä»¯ÈçÓÒͼËùʾ¡£0~2 minÄ򵀮½¾ù·´Ó¦ËÙÂÊv(HI)=____¡£¸ÃζÈÏÂµÄÆ½ºâ³£ÊýK=____ ÏàͬζÈÏ£¬Èô¿ªÊ¼¼ÓÈëHI(g)µÄÎïÖʵÄÁ¿ÊÇÔÀ´µÄ2±¶£¬Ôò___ÊÇÔÀ´µÄ2±¶¡£
a£®Æ½ºâ³£Êý
b£®HIµÄƽºâŨ¶È
c£®´ïµ½Æ½ºâµÄʱ¼ä
d£®Æ½ºâʱH2µÄÌå»ý·ÖÊý

£¨3£©ÊµÑéÊÒÓÃZnºÍÏ¡ÁòËáÖÆÈ¡H2£¬·´Ó¦Ê±ÈÜÒºÖÐË®µÄµçÀëÆ½ºâ___ÒÆ¶¯£¨Ìî ¡°Ïò×󡱡°ÏòÓÒ¡±»ò¡°²»¡±£©£»Èô¼ÓÈëÉÙÁ¿ÏÂÁÐÊÔ¼ÁÖеÄ____£¬²úÉúH2µÄËÙÂʽ«Ôö´ó¡£
a£®NaNO3
b£®CuSO4
c£®Na2SO4
d£®NaHSO3
£¨4£©ÒÔH2ΪȼÁÏ¿ÉÖÆ×÷ÇâÑõȼÁÏµç³Ø¡£ÒÑÖª2H2(g)+O2(g) = 2H2O(1) ¡÷H=-572kJ¡¤mol-1 ijÇâÑõȼÁÏµç³ØÊÍ·Å228.8 kJµçÄÜʱ£¬Éú³É1molҺ̬ˮ£¬¸Ãµç³ØµÄÄÜÁ¿×ª»¯ÂÊΪ________¡£
²Î¿¼´ð°¸£º£¨1£©c
£¨2£©0.1 mol.L-1¡¤min-1 £»64£»b
£¨3£©ÏòÓÒ£»b ¡¢f
£¨4£©80%
±¾Ìâ½âÎö£º
±¾ÌâÄѶȣºÒ»°ã