ÌâÄ¿ÄÚÈÝ

ÓÐÁ½¸öÃܱÕÈÝÆ÷AºÍB£¬AÈÝÆ÷ÖÐÓÐÒ»¸öÒƶ¯µÄ»îÈûÄÜʹÈÝÆ÷ÄÚ±£³Öºãѹ£¬BÈÝÆ÷Äܱ£³ÖºãÈÝ¡£ÆðʼʱÏòÕâÁ½¸öÈÝÆ÷Öзֱð³äÈëµÈÁ¿µÄÌå»ý±ÈΪ2:lµÄSO2ºÍO2µÄ»ìºÏÆøÌ壬²¢Ê¹AºÍBÈÝ»ýÏàµÈ(ÈçͼËùʾ)£¬ÔÚ±£³Ö400¡æµÄζÈÏÂʹ֮·¢ÉúÈçÏ·´Ó¦£º2SO2+O22SO3¡£ÌîдÏÂÁпհףº

(1)SO2µÄת»¯ÂʦÁ(A)     ¦Á(B)¡£(Ìî¡°£¼¡±¡¢¡°£¾¡±»ò¡°=¡±)
(2)´ïµ½(1)ËùÊöƽºâʱ£¬ÈôÏòÁ½ÈÝÆ÷ÖÐͨÈëÉÙÁ¿µÄë²Æø£¬AÈÝÆ÷»¯Ñ§Æ½ºâ        Òƶ¯¡£(Ìî¡°ÕýÏò¡±¡¢¡°ÄæÏò¡±»ò¡°²»¡±)
(3)´ïµ½(1)ƽºâºó£¬ÏòÈÝÆ÷ÖгäÈëµÈÁ¿µÄÔ­·´Ó¦ÆøÌ壬Ôٴδﵽƽºâºó£¬BÈÝÆ÷ÖÐSO3ÔÚ»ìºÏÆøÌåÖеÄÌå»ý·ÖÊý      ¡£(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±)

(1)  >   (2) ÄæÏò   (3) Ôö´ó

½âÎöÊÔÌâ·ÖÎö£º(1) ¿ÉÒÔÏÈÀí½âΪAÈÝÆ÷µÄÌå»ý±£³Ö²»±ä£¬¶þÕ߶¼´ïµ½µÈЧƽºâ£¬´Ëʱ£¬¶ÔÓÚAÖÐÎïÖʵÄÁ¿À´½²£¬±ÈÔ­À´¼õС£¬ÔÙÈÃÈÝÆ÷Ìå»ý·¢Éú±ä»¯£¬¿ÉÒÔÀí½âΪ¶ÔÓÚAÀ´ËµÊÇÔö´óѹǿƽºâÓÒÒÆ£¬×ª»¯ÂÊÔö´ó£¬ËùÒÔÓЦÁ(A) > ¦Á(B)¡£ (2) ³äÈëÉÙÁ¿ë²Æøºó£¬AµÄÌå»ýÔö´ó£¬Æ½ºâÄæÏòÒƶ¯¡£(3) ´ïµ½(1)ƽºâºó£¬ÏòÈÝÆ÷ÖгäÈëµÈÁ¿µÄÔ­·´Ó¦ÆøÌ壬Èôƽºâ²»Òƶ¯£¬ÈýÑõ»¯ÁòµÄÌå»ý·ÖÊý¶¼²»±ä£¬µ«¶ÔÓÚBÈÝÆ÷À´½²£¬Ï൱ÓÚÔö´óѹǿ£¬Æ½ºâÓÒÒÆ£¬ÖÂʹBÈÝÆ÷ÖÐSO3ÔÚ»ìºÏÆøÌåÖеÄÌå»ý·ÖÊýÔö´ó¡£
¿¼µã£ºµç½âÖʵÄÓйØ֪ʶ¡££¬

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

ÒÑÖª2NO2(g)N2O4(g) ¦¤H= -52.7kJ¡¤mol-1£¬Ä³¿ÎÍâ»î¶¯Ð¡×éΪÁË̽¾¿Î¶ȺÍѹǿ¶Ô»¯Ñ§Æ½ºâµÄÓ°Ï죬×öÁËÈçÏÂÁ½×éʵÑ飺 
¢ñ.¸ÃС×éµÄͬѧȡÁËÁ½¸öÉÕÆ¿AºÍB£¬·Ö±ð¼ÓÈëÏàͬŨ¶ÈµÄNO2ÓëN2O4µÄ»ìºÏÆøÌ壬ÖмäÓüÐ×Ӽнô£¬ ²¢½«AºÍB½þÈëµ½ÒÑÊ¢ÓÐË®µÄÁ½¸öÉÕ±­ÖУ¨ÈçͼËùʾ£©£¬È»ºó·Ö±ðÏòÁ½¸öÉÕ±­ÖмÓÈëŨÁòËáºÍNH4NO3¹ÌÌå¡£Çë»Ø´ðÏÂÁÐÓйØÎÊÌâ

£¨1£©¹Û²ìµ½µÄʵÑéÏÖÏóΪ£º_____________________________ 
£¨2£©ÓÉʵÑéÏÖÏó¿ÉÖª£¬Éý¸ßζȣ¬¸Ã»¯Ñ§Æ½ºâÏò___________£¨Ìî¡°Õý¡±»ò¡°Ä桱£©·´Ó¦·½ÏòÒƶ¯£¬·´Ó¦ÖÐNO2µÄת»¯Âʽ«_______________£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©¡£ 
¢ò.ÔÚÈýÖ§ÈÝ»ý¾ùΪ30mLÕëͲÖзֱð³éÈë10mLNO2ÆøÌ壬½«ÕëͲǰ¶Ë·â±Õ¡£ÊµÑé¹ý³ÌÖеÚÒ»Ö§ÕëͲ²»×öÈκβÙ×÷£¬½ö×÷ΪʵÑéÏÖÏó¹Û²ìʱµÄ²ÎÕÕ¶ÔÏó¡£
£¨3£©Ä³Í¬Ñ§½«µÚ¶þÖ§ÕëͲ»îÈûѸËÙÍÆÖÁ5mL´¦£¬´ËʱÆøÌåµÄÑÕÉ«±äÉһ¶Îʱ¼äºóÆøÌåÑÕÉ«ÓÖ±ädzÁË¡£ÊÔ½âÊÍÒ»¶Îʱ¼äºóÆøÌåÑÕÉ«ÓÖ±ädzµÄÔ­Òò£º________________________¡£
£¨4£©Ä³Í¬Ñ§½«µÚÈýÖ§ÕëͲ»îÈûѸËÙÀ­ÖÁ20mL´¦¡£
¢Ù¸Ãͬѧ¹Û²ìµ½µÄÏÖÏóÊÇ£º__________________________
¢ÚÔڴ˹ý³ÌÖУ¬¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£Êý½«______________(Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±»ò¡°²»±ä¡±£©¡£

ÄÜÔ´¶ÌȱÊÇÈËÀàÉç»áÃæÁÙµÄÖØ´óÎÊÌâ¡£¼×´¼ÊÇÒ»ÖÖ¿ÉÔÙÉúÄÜÔ´£¬¾ßÓй㷺µÄ¿ª·¢ºÍÓ¦ÓÃÇ°¾°¡£¹¤ÒµÉϺϳɼ״¼µÄ·´Ó¦Îª£ºCO(g)£«2H2(g)CH3OH(g)£»¦¤H
£¨1£©ÒÑÖª£¬¸Ã·´Ó¦ÔÚ300¡æ£¬5MPaÌõ¼þÏÂÄÜ×Ô·¢½øÐУ¬Ôò¦¤H     0£¬¡÷S    0£¨Ìî¡°£¼£¬£¾»ò£½¡±£©¡£
£¨2£©ÔÚ300¡æ£¬5MPaÌõ¼þÏ£¬½«0.20molµÄ COÓë0.58mol H2µÄ»ìºÏÆøÌå³äÈë2LÃܱÕÈÝÆ÷·¢Éú·´Ó¦£¬·´Ó¦¹ý³ÌÖм״¼µÄÎïÖʵÄÁ¿Å¨¶ÈËæʱ¼äµÄ±ä»¯ÈçͼËùʾ¡£

¢ÙÔÚ0¡«2minÄÚ£¬ÒÔH2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ         mol¡¤L-1¡¤s-1 £¬COµÄת»¯ÂÊΪ        ¡£
¢ÚÁÐʽ¼ÆËã300¡æʱ¸Ã·´Ó¦µÄƽºâ³£ÊýK£½                                  ¡£
¢Û300¡æʱ£¬½«0.50mol CO¡¢1.00mol H2ºÍ1.00 mol CH3OH³äÈëÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖУ¬´Ëʱ·´Ó¦½«          ¡£
A£®ÏòÕý·½ÏòÒƶ¯   B£®ÏòÄæ·½ÏòÒƶ¯   C£®´¦ÓÚƽºâ״̬   D£®ÎÞ·¨ÅжÏ
¢ÜÏÂÁдëÊ©¿ÉÔö¼Ó¼×´¼²úÂʵÄÊÇ           ¡£
A£®Ñ¹ËõÈÝÆ÷Ìå»ý                B£®½«CH3OH(g)´ÓÌåϵÖзÖÀë
C£®³äÈëHe£¬Ê¹Ìåϵ×ÜѹǿÔö´ó   D£®ÔÙ³äÈë0.20mol COºÍ0.58mol H2
£¨3£©ÈôÆäËüÌõ¼þ²»±ä£¬Ê¹·´Ó¦ÔÚ500¡æϽøÐУ¬ÔÚͼÖÐ×÷³ö¼×´¼µÄÎïÖʵÄÁ¿Å¨¶ÈËæʱ¼äµÄ±ä»¯µÄʾÒâͼ¡£

(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              ¡£

£¨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¡£

¹¤Òµ·ÏË®Öг£º¬ÓÐÒ»¶¨Á¿µÄCr2O72-ºÍCrO42-£¬ËüÃÇ»á¶ÔÈËÀ༰Éú̬ϵͳ²úÉúºÜ´óË𺦣¬±ØÐë½øÐд¦Àí¡£³£ÓõĴ¦Àí·½·¨ÓÐÁ½ÖÖ¡£
·½·¨1£º»¹Ô­³Áµí·¨¡£¸Ã·¨µÄ¹¤ÒÕÁ÷³ÌΪCrO42-Cr2O72-Cr3£«Cr(OH)3¡ý
ÆäÖеڢٲ½´æÔÚƽºâ£º2CrO42- (»ÆÉ«)£«2H£«Cr2O72- (³ÈÉ«)£«H2O
£¨1£©ÈôƽºâÌåϵµÄpH=2£¬¸ÃÈÜÒºÏÔ________É«¡£
£¨2£©ÄÜ˵Ã÷µÚ¢Ù²½·´Ó¦´ïƽºâ״̬µÄÊÇ__________¡£
a£®Cr2O72-ºÍCrO42-µÄŨ¶ÈÏàͬ 
b£®2v(Cr2O72-)=v(CrO42-
c£®ÈÜÒºµÄÑÕÉ«²»±ä
£¨3£©µÚ¢Ú²½ÖУ¬»¹Ô­1 mol Cr2O72-Àë×Ó£¬ÐèÒª________molµÄFeSO4¡¤7H2O¡£
£¨4£©µÚ¢Û²½Éú³ÉµÄCr(OH)3ÔÚÈÜÒºÖдæÔÚÒÔϳÁµíÈܽâƽºâ£ºCr(OH)3(s) Cr3£«(aq)£«3OH£­(aq)£¬³£ÎÂÏ£¬Cr(OH)3µÄÈܶȻýKsp=c(Cr3£«)¡¤c3(OH£­)=10£­32£¬ÒªÊ¹c(Cr3£«)½µÖÁ10£­5mol/L£¬ÈÜÒºµÄpHÓ¦µ÷ÖÁ_______¡£
·½·¨2£ºµç½â·¨¡£¸Ã·¨ÓÃFe×öµç¼«µç½âº¬Cr2O72-µÄËáÐÔ·ÏË®£¬Ëæ×ŵç½â½øÐУ¬ÔÚÒõ¼«¸½½üÈÜÒºpHÉý¸ß£¬²úÉúCr(OH)3³Áµí¡£
£¨5£©ÓÃFe×öµç¼«µÄÔ­ÒòΪ__________________¡£
£¨6£©ÔÚÒõ¼«¸½½üÈÜÒºpHÉý¸ßµÄÔ­ÒòÊÇ(Óõ缫·´Ó¦½âÊÍ)_________________________________¡£ÈÜÒºÖÐͬʱÉú³ÉµÄ³Áµí»¹ÓÐ__________¡£

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

¾«Ó¢¼Ò½ÌÍø