ÌâÄ¿ÄÚÈÝ

ÈçͼËùʾ£¬ÒÑÖªAµÄ²úÁ¿ÊÇÒ»¸ö¹ú¼ÒʯÓÍ»¯¹¤Ë®Æ½µÄ±êÖ¾£¬D¾ßÓÐËáÐÔ£®EÊǾßÓÐÏãζµÄ²»ÈÜÓÚË®µÄÒºÌ壮
£¨1£©Ð´³öDµÄ¹ÙÄÜÍÅÃû³Æ£º
õ¥»ù
õ¥»ù
      DµÄ½á¹¹¼òʽ£º
CH3COOCH2CH3
CH3COOCH2CH3

£¨2£©A¡úBµÄ»¯Ñ§·´Ó¦·½³Ìʽ£º
CH2=CH2+H2O
Ò»¶¨Ìõ¼þÏÂ
CH3CH2OH
CH2=CH2+H2O
Ò»¶¨Ìõ¼þÏÂ
CH3CH2OH
 ·´Ó¦ÀàÐÍΪ£º
¼Ó³É·´Ó¦
¼Ó³É·´Ó¦

£¨3£©Ð´³öB+D¡úEµÄ»¯Ñ§·´Ó¦·½³Ìʽ£º
CH3COOH+CH3CH2OH
ŨH2SO4
¡÷
CH3COOCH2CH3+H2O
CH3COOH+CH3CH2OH
ŨH2SO4
¡÷
CH3COOCH2CH3+H2O
  ·´Ó¦ÀàÐÍΪ£º
õ¥»¯·´Ó¦£¨»òÈ¡´ú·´Ó¦£©
õ¥»¯·´Ó¦£¨»òÈ¡´ú·´Ó¦£©
£®
·ÖÎö£ºAµÄ²úÁ¿ÊÇÒ»¸ö¹ú¼ÒʯÓÍ»¯¹¤Ë®Æ½µÄ±êÖ¾£¬ÔòAÊÇCH2=CH2£¬ÒÒÏ©ºÍË®·¢Éú¼Ó³É·´Ó¦Éú³ÉB£¬ÔòBÊÇCH3CH2OH£¬ÔÚÍ­×÷´ß»¯¼Á¡¢¼ÓÈÈÌõ¼þÏ£¬ÒÒ´¼±»Ñõ»¯Éú³ÉC£¬CÊÇÒÒÈ©£¬Æä½á¹¹¼òʽΪ£ºCH3CHO£¬ÒÒÈ©±»Ñõ»¯Éú³ÉD£¬D¾ßÓÐËáÐÔ£¬ÔòDÊÇÒÒËᣬÆä½á¹¹¼òʽΪ£ºCH3COOH£¬ÔÚŨÁòËá×÷´ß»¯¼Á¡¢¼ÓÈÈÌõ¼þÏÂÒÒ´¼ºÍÒÒËá·¢Éúõ¥»¯·´Ó¦Éú³ÉE£¬EÊǾßÓÐÏãζµÄ²»ÈÜÓÚË®µÄÒºÌ壬EÊÇÒÒËáÒÒõ¥£¬Æä½á¹¹¼òʽΪ£ºCH3COOCH2CH3£®
½â´ð£º½â£ºAµÄ²úÁ¿ÊÇÒ»¸ö¹ú¼ÒʯÓÍ»¯¹¤Ë®Æ½µÄ±êÖ¾£¬ÔòAÊÇCH2=CH2£¬ÒÒÏ©ºÍË®·¢Éú¼Ó³É·´Ó¦Éú³ÉB£¬ÔòBÊÇCH3CH2OH£¬ÔÚÍ­×÷´ß»¯¼Á¡¢¼ÓÈÈÌõ¼þÏ£¬ÒÒ´¼±»Ñõ»¯Éú³ÉC£¬CÊÇÒÒÈ©£¬Æä½á¹¹¼òʽΪ£ºCH3CHO£¬ÒÒÈ©±»Ñõ»¯Éú³ÉD£¬D¾ßÓÐËáÐÔ£¬ÔòDÊÇÒÒËᣬÆä½á¹¹¼òʽΪ£ºCH3COOH£¬ÔÚŨÁòËá×÷´ß»¯¼Á¡¢¼ÓÈÈÌõ¼þÏÂÒÒ´¼ºÍÒÒËá·¢Éúõ¥»¯·´Ó¦Éú³ÉE£¬EÊǾßÓÐÏãζµÄ²»ÈÜÓÚË®µÄÒºÌ壬EÊÇÒÒËáÒÒõ¥£¬Æä½á¹¹¼òʽΪ£ºCH3COOCH2CH3£®
£¨1£©Í¨¹ýÒÔÉÏ·ÖÎöÖª£¬DµÄ½á¹¹¼òʽΪ£ºCH3COOCH2CH3£¬DÖйÙÄÜÍÅÃû³ÆÊÇõ¥»ù£¬
¹Ê´ð°¸Îª£ºõ¥»ù£¬CH3COOCH2CH3£»
£¨2£©·´Ó¦¢ÙÊÇÒÒÏ©ÓëË®·¢Éú¼Ó³É·´Ó¦Éú³ÉÒÒ´¼£¬·´Ó¦·½³ÌʽΪ£ºCH2=CH2+H2O
Ò»¶¨Ìõ¼þÏÂ
CH3CH2OH£¬¸Ã·´Ó¦ÊôÓڼӳɷ´Ó¦£¬
¹Ê´ð°¸Îª£ºCH2=CH2+H2O
Ò»¶¨Ìõ¼þÏÂ
CH3CH2OH£¬¼Ó³É·´Ó¦£»
£¨3£©·´Ó¦¢ÜÊÇÒÒËáÓëÒÒ´¼·¢Éúõ¥»¯·´Ó¦Éú³ÉÒÒËáÒÒõ¥£¬·´Ó¦·½³ÌʽΪ£ºCH3COOH+CH3CH2OH
ŨH2SO4
¡÷
CH3COOCH2CH3+H2O£¬¸Ã·´Ó¦ÊôÓÚõ¥»¯·´Ó¦£¨»òÈ¡´ú·´Ó¦£©£¬
¹Ê´ð°¸Îª£ºCH3COOH+CH3CH2OH
ŨH2SO4
¡÷
CH3COOCH2CH3+H2O£¬õ¥»¯·´Ó¦£¨»òÈ¡´ú·´Ó¦£©£®
µãÆÀ£º¿¼²éÓлúÎïµÄÍƶϣ¬±È½Ï»ù´¡£¬Éæ¼°Ï©Ìþ¡¢´¼¡¢È©¡¢ôÈËá¡¢õ¥µÄÐÔÖÊÓëת»¯£¬×¢Òâ»ù´¡ÖªÊ¶µÄÕÆÎÕ£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
»¯Ñ§ÊµÑéÊÒͨ³£ÓôÖпºÍÏ¡ÁòËá·´Ó¦ÖÆÇâÆø£¬Òò´ËÔÚÖÆÇâ·ÏÒºÖк¬ÓдóÁ¿µÄÁòËáп£®Í¬Ê±£¬ÓÉÓÚ´ÖпÖл¹º¬ÓÐÌúµÈÔÓÖÊ£¬Ê¹µÃÈÜÒºÖлìÓÐÒ»¶¨Á¿µÄÁòËáÑÇÌú£¬ÎªÁ˳ä·ÖÀûÓÃÖÆÇâ·ÏÒº£¬³£ÓÃÆäÖƱ¸ð©·¯£¨ZnSO4?7H2O£©£®Ä³Ð£»¯Ñ§ÐËȤС×éµÄͬѧÒÔÖÆÇâÆøµÄ·ÏҺΪԭÁÏÀ´ÖÆÈ¡ð©·¯£®ÖƱ¸ð©·¯µÄʵÑéÁ÷³ÌÈçͼËùʾ£®

ÒÑÖª£º¿ªÊ¼Éú³ÉÇâÑõ»¯Îï³Áµíµ½³ÁµíÍêÈ«µÄpH·¶Î§·Ö±ðΪ£ºFe£¨OH£©3£º2.7-3.7Fe£¨OH£©2£º7.6-9.6  Zn£¨OH£©2£º5.7-8.0ÊԻشðÏÂÁÐÎÊÌ⣺?
£¨1£©¼ÓÈëµÄÊÔ¼Á¢Ù£¬¹©Ñ¡ÔñʹÓõÄÓУº°±Ë®¡¢NaClOÈÜÒº¡¢20%µÄH2O2¡¢Å¨ÁòËᡢŨÏõËáµÈ£¬×îºÃÑ¡ÓÃ
20%µÄH2O2
20%µÄH2O2
£¬ÆäÀíÓÉÊÇ
½«ÖÆÇâ·ÏÒºÖеÄFe2+Ñõ»¯³ÉFe3+£¬Í¬Ê±±ÜÃâÒýÈëеÄÔÓÖÊ
½«ÖÆÇâ·ÏÒºÖеÄFe2+Ñõ»¯³ÉFe3+£¬Í¬Ê±±ÜÃâÒýÈëеÄÔÓÖÊ
£»
£¨2£©¼ÓÈëµÄÊÔ¼Á¢Ú£¬¹©Ñ¡ÔñʹÓõÄÓУºa¡¢Zn·Û£¬b¡¢ZnO£¬c¡¢Zn£¨OH£©2£¬d¡¢ZnCO3£¬e¡¢ZnSO4µÈ£¬¿ÉÑ¡ÓÃ
bcd
bcd
£»
£¨3£©´Ó¾§Ìå1¡ú¾§Ìå2£¬¸Ã¹ý³ÌµÄÃû³ÆÊÇ
Öؽᾧ
Öؽᾧ
£»
£¨4£©Ôڵõ½ð©·¯Ê±£¬Ïò¾§ÌåÖмÓÈëÉÙÁ¿¾Æ¾«Ï´µÓ¶ø²»ÓÃË®µÄÔ­ÒòÊÇ
ΪÁ˳åÏ´µô¾§Ìå±íÃæµÄÔÓÖÊÀë×Ó£»·ÀÖ¹¾§ÌåÈܽ⣬ӰÏì²úÂÊ
ΪÁ˳åÏ´µô¾§Ìå±íÃæµÄÔÓÖÊÀë×Ó£»·ÀÖ¹¾§ÌåÈܽ⣬ӰÏì²úÂÊ
£®

(8·Ö)¼×´¼ÊÇÒ»ÖֺܺõÄȼÁÏ£¬¹¤ÒµÉÏÓÃCH4ºÍH2O(g)ΪԭÁÏ£¬Í¨¹ý·´Ó¦IºÍIIÀ´ÖƱ¸¼×´¼¡£
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
(1)½«1.0molCH4ºÍ2.0molH2O(g)ͨÈë·´Ó¦ÊÒ(ÈÝ»ýΪl00L)£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º
cCH4(g)+H2O(g) CO(g)+3H2(g)   I¡£
CH4µÄת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼËùʾ¡£

¢ÙÒÑÖª100¡æʱ´ïµ½Æ½ºâËùÐèµÄʱ¼äΪ5min£¬ÔòÓÃH2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ        
¢ÚͼÖеÄP1          P2(Ìî¡°<¡±¡¢¡°>¡±»ò¡°=¡±)£¬100¡æʱƽºâ³£ÊýΪ          ¡£
¢Û¸Ã·´Ó¦µÄH       0¡£(Ìî¡°<¡±¡¢¡°>¡±»ò¡°=¡±)
(2)ÔÚѹǿΪ0.1MPaÌõ¼þÏ£¬a molCOÓë3a mol H2µÄ»ìºÏÆøÌåÔÚ´ß»¯¼Á×÷ÓÃÏÂÄÜ×Ô·¢·´Ó¦Éú³É¼×´¼£ºCO(g)+2H2(g) CH30H(g) H<0  ¢ò¡£
¢ÙÈôÈÝÆ÷ÈÝ»ý²»±ä£¬ÏÂÁдëÊ©¿ÉÔö¼Ó¼×´¼²úÂʵÄÊÇ        (ÌîÐòºÅ)¡£

A£®Éý¸ßζÈB£®½«CH3OH(g)´ÓÌåϵÖзÖÀë
C£®³äÈëHe£¬Ê¹Ìåϵ×ÜѹǿÔö´óD£®ÔÙ³äÈËlmolCOºÍ3 mol H2
¢ÚΪÁËÑ°ÕҺϳɼ״¼µÄÊÊÒËζȺÍѹǿ£¬Ä³Í¬Ñ§Éè¼ÆÁËÈý×éʵÑ飬²¿·ÖʵÑéÌõ¼þÒѾ­ÌîÔÚÁËÏÂÃæµÄʵÑéÉè¼Æ±íÖС£
ʵÑé±àºÅ
T(¡æ)
n(CO)£¯n(H2)
p(MPa)
l
150
1£¯3
0.1
2
n
1£¯3
5
3
350
m
5
a£®ÉϱíÖÐÊ£ÓàµÄʵÑéÌõ¼þÊý¾Ý£ºn=         £¬m=          ¡£
b£®¸ù¾Ý·´Ó¦¢òµÄÌص㣬ÈçͼÊÇÔÚѹǿ·Ö±ðΪ0.1MPaºÍ5MPaÏÂCOµÄת»¯ÂÊËæζȱ仯µÄÇúÏßͼ£¬ÇëÖ¸Ã÷ͼÖеÄѹǿ=       MPa¡£

(8·Ö)¼×´¼ÊÇÒ»ÖֺܺõÄȼÁÏ£¬¹¤ÒµÉÏÓÃCH4ºÍH2O(g)ΪԭÁÏ£¬Í¨¹ý·´Ó¦IºÍIIÀ´ÖƱ¸¼×´¼¡£

Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)½«1.0molCH4ºÍ2.0molH2O(g)ͨÈë·´Ó¦ÊÒ(ÈÝ»ýΪl00L)£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º

cCH4(g)+H2O(g)  CO(g)+3H2(g)   I¡£

CH4µÄת»¯ÂÊÓëζȡ¢Ñ¹Ç¿µÄ¹ØϵÈçͼËùʾ¡£

¢ÙÒÑÖª100¡æʱ´ïµ½Æ½ºâËùÐèµÄʱ¼äΪ5min£¬ÔòÓÃH2±íʾµÄƽ¾ù·´Ó¦ËÙÂÊΪ         

¢ÚͼÖеÄP1           P2(Ìî¡°<¡±¡¢¡°>¡±»ò¡°=¡±)£¬100¡æʱƽºâ³£ÊýΪ           ¡£

¢Û¸Ã·´Ó¦µÄH        0¡£(Ìî¡°<¡±¡¢¡°>¡±»ò¡°=¡±)

(2)ÔÚѹǿΪ0.1MPaÌõ¼þÏ£¬a molCOÓë3a mol H2µÄ»ìºÏÆøÌåÔÚ´ß»¯¼Á×÷ÓÃÏÂÄÜ×Ô·¢·´Ó¦Éú³É¼×´¼£ºCO(g)+2H2(g) CH30H(g)  H<0  ¢ò¡£

¢ÙÈôÈÝÆ÷ÈÝ»ý²»±ä£¬ÏÂÁдëÊ©¿ÉÔö¼Ó¼×´¼²úÂʵÄÊÇ         (ÌîÐòºÅ)¡£

A£®Éý¸ßζȠ                                                         B£®½«CH3OH(g)´ÓÌåϵÖзÖÀë

C£®³äÈëHe£¬Ê¹Ìåϵ×ÜѹǿÔö´ó                   D£®ÔÙ³äÈËlmolCOºÍ3 mol H2

¢ÚΪÁËÑ°ÕҺϳɼ״¼µÄÊÊÒËζȺÍѹǿ£¬Ä³Í¬Ñ§Éè¼ÆÁËÈý×éʵÑ飬²¿·ÖʵÑéÌõ¼þÒѾ­ÌîÔÚÁËÏÂÃæµÄʵÑéÉè¼Æ±íÖС£

ʵÑé±àºÅ

T(¡æ)

n(CO)£¯n(H2)

p(MPa)

l

150

1£¯3

0.1

2

n

1£¯3

5

3

350

m

5

a£®ÉϱíÖÐÊ£ÓàµÄʵÑéÌõ¼þÊý¾Ý£ºn=          £¬m=           ¡£

b£®¸ù¾Ý·´Ó¦¢òµÄÌص㣬ÈçͼÊÇÔÚѹǿ·Ö±ðΪ0.1MPaºÍ5MPaÏÂCOµÄת»¯ÂÊËæζȱ仯µÄÇúÏßͼ£¬ÇëÖ¸Ã÷ͼÖеÄѹǿ=        MPa¡£

 

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

¾«Ó¢¼Ò½ÌÍø