¶þ¼×ÃÑÓëË®ÕôÆøÖØÕûÖÆÇâÆø×÷ΪȼÁϵç³ØµÄÇâÔ´£¬±ÈÆäËûÖÆÇâ¼¼Êõ¸üÓÐÓÅÊÆ£®Ö÷Òª·´Ó¦Îª£º

¢ÙCH3OCH3(g)£«H2O(g)2CH3OH(g)¡¡¦¤H£½£«37 kJ¡¤mol£­1

¢ÚCH3OH(g)£«H2O(g)3H2(g)£«CO2(g)¡¡¦¤H£½£«49 kJ¡¤mol£­1

¢ÛCO2(g)£«H2(g)CO(g)£«H2O(g)¡¡¦¤H£½£«41.3 kJ¡¤mol£­1

ÆäÖз´Ó¦¢ÛÊÇÖ÷ÒªµÄ¸±·´Ó¦£¬²úÉúµÄCOÄܶ¾º¦È¼Áϵç³ØPtµç¼«£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺

(1)

¶þ¼×ÃÑ¿ÉÒÔͨ¹ýÌìÈ»ÆøºÍCO2ºÏ³ÉÖƵ㬸÷´Ó¦µÄ»¯Ñ§·½³ÌʽΪ________£®

(2)

CH3OCH3(g)ÓëË®ÕôÆøÖØÕûÖÆÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽΪ________£®

(3)

ÏÂÁвÉÈ¡µÄ´ëÊ©ºÍ½âÊÍÕýÈ·µÄÊÇ

A£®

·´Ó¦¹ý³ÌÔÚµÍνøÐУ¬¿É¼õÉÙCOµÄ²úÉú

B£®

Ôö¼Ó½øË®Á¿£¬ÓÐÀûÓÚ¶þ¼×ÃѵÄת»¯£¬²¢¼õÉÙCOµÄ²úÉú

C£®

Ñ¡ÔñÔÚµÍξßÓнϸ߻îÐԵĴ߻¯¼Á£¬ÓÐÖúÓÚÌá¸ß·´Ó¦¢ÚCH3OHµÄת»¯ÂÊ

D£®

ÌåϵѹǿÉý¸ß£¬¶ÔÖÆÈ¡ÇâÆø²»Àû£¬ÇÒ¶Ô¼õÉÙCOµÄ²úÉú¼¸ºõÎÞÓ°Ïì

(4)

ÔÚζÈÏàͬµÄ3¸öÃܱÕÈÝÆ÷ÖУ¬°´²»Í¬·½Ê½Í¶Èë·´Ó¦Î±£³ÖºãΡ¢ºãѹ£¬·¢Éú·´Ó¦¢Ù£¬²âµÃ·´Ó¦´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏ£®

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ

A£®

a£«2c£½37

B£®

¦Á1£«¦Á2£½1

C£®

V1£¾V3

D£®

c1£½2c3

(5)

ÒÔ¶þ¼×ÃÑ¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪԭÁÏ£¬Ê¯Ä«Îªµç¼«Ò²¿ÉÖ±½Ó¹¹³ÉȼÁϵç³Ø£®¸Ãµç³ØÖиº¼«Éϵĵ缫·´Ó¦Ê½ÊÇ________£®

´ð°¸£º3£®ABD;4£®ABC;
½âÎö£º

(1)

3CH4£«CO2¡ú2CH3OCH3(3·Ö)

(2)

CH3OCH3(g)£«3H2O(g)6H2(g)£«2CO2(g)¡¡¦¤H£½135 kJ¡¤mol£­1(3·Ö)

(5)

CH3OCH3£«16OH£­£­12e£­2CO£«11H2O£®(3·Ö)


Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨2011?ÑïÖÝÈýÄ££©¶þ¼×ÃÑÓëË®ÕôÆøÖØÕûÖÆÇâÆø×÷ΪȼÁϵç³ØµÄÇâÔ´£¬±ÈÆäËûÖÆÇâ¼¼Êõ¸üÓÐÓÅÊÆ£®Ö÷Òª·´Ó¦Îª£º
¢ÙCH3OCH3£¨g£©+H2O£¨g£©?2CH3OH£¨g£©¡÷H=+37kJ?mol-1
¢ÚCH3OH£¨g£©+H2O£¨g£©?3H2£¨g£©+CO2£¨g£©¡÷H=+49kJ?mol-1
¢ÛCO2£¨g£©+H2£¨g£©?CO£¨g£©+H2O£¨g£©¡÷H=+41.3kJ?mol-1
ÆäÖз´Ó¦¢ÛÊÇÖ÷ÒªµÄ¸±·´Ó¦£¬²úÉúµÄCOÄܶ¾º¦È¼Áϵç³ØPtµç¼«£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©CH3OCH3£¨g£©ÓëË®ÕôÆøÖØÕûÖÆÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽΪ
CH3OCH3£¨g£©+3H2O£¨g£©?6H2£¨g£©+2CO2£¨g£©¡÷H=135kJ?mol-1
CH3OCH3£¨g£©+3H2O£¨g£©?6H2£¨g£©+2CO2£¨g£©¡÷H=135kJ?mol-1
£®
£¨2£©ÏÂÁвÉÈ¡µÄ´ëÊ©ºÍ½âÊÍÕýÈ·µÄÊÇ
ABD
ABD
£®£¨Ìî×ÖĸÐòºÅ£©
A£®·´Ó¦¹ý³ÌÔÚµÍνøÐУ¬¿É¼õÉÙCOµÄ²úÉú
B£®Ôö¼Ó½øË®Á¿£¬ÓÐÀûÓÚ¶þ¼×ÃѵÄת»¯£¬²¢¼õÉÙCOµÄ²úÉú
C£®Ñ¡ÔñÔÚµÍξßÓнϸ߻îÐԵĴ߻¯¼Á£¬ÓÐÖúÓÚÌá¸ß·´Ó¦¢ÚCH3OHµÄת»¯ÂÊ
D£®ÌåϵѹǿÉý¸ß£¬¶ÔÖÆÈ¡ÇâÆø²»Àû£¬ÇÒ¶Ô¼õÉÙCOµÄ²úÉú¼¸ºõÎÞÓ°Ïì
£¨3£©ÔÚζÈÏàͬµÄ3¸öÃܱÕÈÝÆ÷ÖУ¬°´²»Í¬·½Ê½Í¶Èë·´Ó¦Î±£³ÖºãΡ¢ºãѹ£¬·¢Éú·´Ó¦¢Ù£¬²âµÃ·´Ó¦´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏ£®
ÈÝÆ÷ ¼× ÒÒ ±û
·´Ó¦ÎïͶÈëÁ¿ 1mol CH3OCH3¡¢1mol H2O 2mol CH3OH 1mol CH3OH
CH3OHµÄŨ¶È£¨mol/L£© c1 c2 c3
·´Ó¦µÄÄÜÁ¿±ä»¯ ÎüÊÕa kJ ·Å³öb kJ ·Å³öc kJ
ƽºâʱÌå»ý£¨L£© V1 V2 V3
·´Ó¦Îïת»¯ÂÊ ¦Á 1 ¦Á 2 ¦Á 3
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
ABC
ABC
£®£¨Ìî×ÖĸÐòºÅ£©
A£®a+2c=37    B£®¦Á1+¦Á2=1    C£®V1£¾V3   D£®c1=2c3
£¨4£©ÒÔ¶þ¼×ÃÑ¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪԭÁÏ£¬Ê¯Ä«Îªµç¼«Ò²¿ÉÖ±½Ó¹¹³ÉȼÁϵç³Ø£®¸Ãµç³ØÖиº¼«Éϵĵ缫·´Ó¦Ê½ÊÇ
CH3OCH3+16OH--12e-=2CO32-+11H2O
CH3OCH3+16OH--12e-=2CO32-+11H2O
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

£¨15·Ö£©¶þ¼×ÃÑÓëË®ÕôÆøÖØÕûÖÆÇâÆø×÷ΪȼÁϵç³ØµÄÇâÔ´£¬±ÈÆäËûÖÆÇâ¼¼Êõ¸üÓÐÓÅÊÆ¡£Ö÷Òª·´Ó¦Îª£º

¢Ù CH3OCH3(g) + H2O(g)2CH3OH(g)        ¦¤H£½+37 kJ¡¤mol£­1

¢Ú CH3OH(g)+ H2O(g) 3H2(g)+ CO2(g)   ¦¤H£½+49 kJ¡¤mol£­1

¢Û CO2(g)+ H2(g)CO(g) + H2O(g)         ¦¤H£½+41.3 kJ¡¤mol£­1

ÆäÖз´Ó¦¢ÛÊÇÖ÷ÒªµÄ¸±·´Ó¦£¬²úÉúµÄCOÄܶ¾º¦È¼Áϵç³ØPtµç¼«¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¶þ¼×ÃÑ¿ÉÒÔͨ¹ýÌìÈ»ÆøºÍCO2ºÏ³ÉÖƵ㬸÷´Ó¦µÄ»¯Ñ§·½³ÌʽΪ

                                                                ¡£

£¨2£©CH3OCH3(g)ÓëË®ÕôÆøÖØÕûÖÆÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽΪ

                                                                ¡£

£¨3£©ÏÂÁвÉÈ¡µÄ´ëÊ©ºÍ½âÊÍÕýÈ·µÄÊÇ            ¡££¨Ìî×ÖĸÐòºÅ£©

A£®·´Ó¦¹ý³ÌÔÚµÍνøÐУ¬¿É¼õÉÙCOµÄ²úÉú 

    B£®Ôö¼Ó½øË®Á¿£¬ÓÐÀûÓÚ¶þ¼×ÃѵÄת»¯£¬²¢¼õÉÙCOµÄ²úÉú

C£®Ñ¡ÔñÔÚµÍξßÓнϸ߻îÐԵĴ߻¯¼Á£¬ÓÐÖúÓÚÌá¸ß·´Ó¦¢ÚCH3OHµÄת»¯ÂÊ

D£®ÌåϵѹǿÉý¸ß£¬¶ÔÖÆÈ¡ÇâÆø²»Àû£¬ÇÒ¶Ô¼õÉÙCOµÄ²úÉú¼¸ºõÎÞÓ°Ïì

£¨4£©ÔÚζÈÏàͬµÄ3¸öÃܱÕÈÝÆ÷ÖУ¬°´²»Í¬·½Ê½Í¶Èë·´Ó¦Î±£³ÖºãΡ¢ºãѹ£¬·¢Éú·´Ó¦¢Ù£¬²âµÃ·´Ó¦´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏ¡£

ÈÝÆ÷

¼×

ÒÒ

±û

·´Ó¦ÎïͶÈëÁ¿

1mol CH3OCH3 ¡¢1mol H2O

2mol CH3OH

1mol CH3OH

CH3OHµÄŨ¶È£¨mol/L£©

c1

c2

c3

·´Ó¦µÄÄÜÁ¿±ä»¯

ÎüÊÕa kJ

·Å³öb kJ

·Å³öc kJ

ƽºâʱÌå»ý£¨L£©

V1

V2

V3

·´Ó¦Îïת»¯ÂÊ

¦Á 1

¦Á 2

¦Á 3

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ          ¡££¨Ìî×ÖĸÐòºÅ£©

A. a+2c=37       B. ¦Á1 + ¦Á2=1       C. V1 > V3       D.c1 = 2c3

£¨5£©ÒÔ¶þ¼×ÃÑ¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪԭÁÏ£¬Ê¯Ä«Îªµç¼«Ò²¿ÉÖ±½Ó¹¹³ÉȼÁϵç³Ø¡£¸Ãµç³ØÖиº¼«Éϵĵ缫·´Ó¦Ê½ÊÇ                                             ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìºÓ±±Ê¡Õý¶¨ÖÐѧ¸ßÈýµÚÈý´Î¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨15·Ö£©¶þ¼×ÃÑÓëË®ÕôÆøÖØÕûÖÆÇâÆø×÷ΪȼÁϵç³ØµÄÇâÔ´£¬±ÈÆäËûÖÆÇâ¼¼Êõ¸üÓÐÓÅÊÆ¡£Ö÷Òª·´Ó¦Îª£º
¢Ù CH3OCH3(g) + H2O(g)2CH3OH(g)         ¦¤H£½+37 kJ¡¤mol£­1
¢Ú CH3OH(g) + H2O(g) 3H2(g) + CO2(g)   ¦¤H£½+49 kJ¡¤mol£­1
¢Û CO2(g) + H2(g)CO(g) + H2O(g)         ¦¤H£½+41.3 kJ¡¤mol£­1
ÆäÖз´Ó¦¢ÛÊÇÖ÷ÒªµÄ¸±·´Ó¦£¬²úÉúµÄCOÄܶ¾º¦È¼Áϵç³ØPtµç¼«¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¶þ¼×ÃÑ¿ÉÒÔͨ¹ýÌìÈ»ÆøºÍCO2ºÏ³ÉÖƵ㬸÷´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
                                                                ¡£
£¨2£©CH3OCH3(g)ÓëË®ÕôÆøÖØÕûÖÆÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽΪ
                                                                ¡£
£¨3£©ÏÂÁвÉÈ¡µÄ´ëÊ©ºÍ½âÊÍÕýÈ·µÄÊÇ           ¡££¨Ìî×ÖĸÐòºÅ£©
A£®·´Ó¦¹ý³ÌÔÚµÍνøÐУ¬¿É¼õÉÙCOµÄ²úÉú 
B£®Ôö¼Ó½øË®Á¿£¬ÓÐÀûÓÚ¶þ¼×ÃѵÄת»¯£¬²¢¼õÉÙCOµÄ²úÉú
C£®Ñ¡ÔñÔÚµÍξßÓнϸ߻îÐԵĴ߻¯¼Á£¬ÓÐÖúÓÚÌá¸ß·´Ó¦¢ÚCH3OHµÄת»¯ÂÊ
D£®ÌåϵѹǿÉý¸ß£¬¶ÔÖÆÈ¡ÇâÆø²»Àû£¬ÇÒ¶Ô¼õÉÙCOµÄ²úÉú¼¸ºõÎÞÓ°Ïì
£¨4£©ÔÚζÈÏàͬµÄ3¸öÃܱÕÈÝÆ÷ÖУ¬°´²»Í¬·½Ê½Í¶Èë·´Ó¦Î±£³ÖºãΡ¢ºãѹ£¬·¢Éú·´Ó¦¢Ù£¬²âµÃ·´Ó¦´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏ¡£

ÈÝÆ÷
¼×
ÒÒ
±û
·´Ó¦ÎïͶÈëÁ¿
1mol CH3OCH3¡¢1mol H2O
2mol CH3OH
1mol CH3OH
CH3OHµÄŨ¶È£¨mol/L£©
c1
c2
c3
·´Ó¦µÄÄÜÁ¿±ä»¯
ÎüÊÕa kJ
·Å³öb kJ
·Å³öc kJ
ƽºâʱÌå»ý£¨L£©
V1
V2
V3
·´Ó¦Îïת»¯ÂÊ
¦Á 1
¦Á 2
¦Á 3
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ          ¡££¨Ìî×ÖĸÐòºÅ£©
A. a+2c="37 "      B. ¦Á1 + ¦Á2="1"       C. V1 > V3      D. c1=2c3
£¨5£©ÒÔ¶þ¼×ÃÑ¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪԭÁÏ£¬Ê¯Ä«Îªµç¼«Ò²¿ÉÖ±½Ó¹¹³ÉȼÁϵç³Ø¡£¸Ãµç³ØÖиº¼«Éϵĵ缫·´Ó¦Ê½ÊÇ                                             ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2011-2012ѧÄêºÓ±±Ê¡¸ßÈýµÚÈý´Î¿¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

£¨15·Ö£©¶þ¼×ÃÑÓëË®ÕôÆøÖØÕûÖÆÇâÆø×÷ΪȼÁϵç³ØµÄÇâÔ´£¬±ÈÆäËûÖÆÇâ¼¼Êõ¸üÓÐÓÅÊÆ¡£Ö÷Òª·´Ó¦Îª£º

¢Ù CH3OCH3(g) + H2O(g)2CH3OH(g)         ¦¤H£½+37 kJ¡¤mol£­1

¢Ú CH3OH(g) + H2O(g) 3H2(g) + CO2(g)   ¦¤H£½+49 kJ¡¤mol£­1

¢Û CO2(g) + H2(g)CO(g) + H2O(g)          ¦¤H£½+41.3 kJ¡¤mol£­1

ÆäÖз´Ó¦¢ÛÊÇÖ÷ÒªµÄ¸±·´Ó¦£¬²úÉúµÄCOÄܶ¾º¦È¼Áϵç³ØPtµç¼«¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¶þ¼×ÃÑ¿ÉÒÔͨ¹ýÌìÈ»ÆøºÍCO2ºÏ³ÉÖƵ㬸÷´Ó¦µÄ»¯Ñ§·½³ÌʽΪ

                                                                 ¡£

£¨2£©CH3OCH3(g)ÓëË®ÕôÆøÖØÕûÖÆÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽΪ

                                                                 ¡£

£¨3£©ÏÂÁвÉÈ¡µÄ´ëÊ©ºÍ½âÊÍÕýÈ·µÄÊÇ            ¡££¨Ìî×ÖĸÐòºÅ£©

A£®·´Ó¦¹ý³ÌÔÚµÍνøÐУ¬¿É¼õÉÙCOµÄ²úÉú 

     B£®Ôö¼Ó½øË®Á¿£¬ÓÐÀûÓÚ¶þ¼×ÃѵÄת»¯£¬²¢¼õÉÙCOµÄ²úÉú

C£®Ñ¡ÔñÔÚµÍξßÓнϸ߻îÐԵĴ߻¯¼Á£¬ÓÐÖúÓÚÌá¸ß·´Ó¦¢ÚCH3OHµÄת»¯ÂÊ

D£®ÌåϵѹǿÉý¸ß£¬¶ÔÖÆÈ¡ÇâÆø²»Àû£¬ÇÒ¶Ô¼õÉÙCOµÄ²úÉú¼¸ºõÎÞÓ°Ïì

£¨4£©ÔÚζÈÏàͬµÄ3¸öÃܱÕÈÝÆ÷ÖУ¬°´²»Í¬·½Ê½Í¶Èë·´Ó¦Î±£³ÖºãΡ¢ºãѹ£¬·¢Éú·´Ó¦¢Ù£¬²âµÃ·´Ó¦´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏ¡£

ÈÝÆ÷

¼×

ÒÒ

±û

·´Ó¦ÎïͶÈëÁ¿

1mol CH3OCH3 ¡¢1mol H2O

2mol CH3OH

1mol CH3OH

CH3OHµÄŨ¶È£¨mol/L£©

c1

c2

c3

·´Ó¦µÄÄÜÁ¿±ä»¯

ÎüÊÕa kJ

·Å³öb kJ

·Å³öc kJ

ƽºâʱÌå»ý£¨L£©

V1

V2

V3

·´Ó¦Îïת»¯ÂÊ

¦Á 1

¦Á 2

¦Á 3

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ           ¡££¨Ìî×ÖĸÐòºÅ£©

A. a+2c=37       B. ¦Á1 + ¦Á2=1       C. V1 > V3       D. c1 = 2c3

£¨5£©ÒÔ¶þ¼×ÃÑ¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪԭÁÏ£¬Ê¯Ä«Îªµç¼«Ò²¿ÉÖ±½Ó¹¹³ÉȼÁϵç³Ø¡£¸Ãµç³ØÖиº¼«Éϵĵ缫·´Ó¦Ê½ÊÇ                                              ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

¶þ¼×ÃÑÓëË®ÕôÆøÖØÕûÖÆÇâÆø×÷ΪȼÁϵç³ØµÄÇâÔ´£¬±ÈÆäËûÖÆÇâ¼¼Êõ¸üÓÐÓÅÊÆ¡£Ö÷Òª·´Ó¦Îª£º

¢Ù CH3OCH3(g) + H2O(g)2CH3OH(g)         ¦¤H£½+37 kJ¡¤mol£­1

¢Ú CH3OH(g) + H2O(g) 3H2(g) + CO2(g)       ¦¤H£½+49 kJ¡¤mol£­1

¢Û CO2(g) + H2(g)CO(g) + H2O(g)                ¦¤H£½+41.3 kJ¡¤mol£­1

ÆäÖз´Ó¦¢ÛÊÇÖ÷ÒªµÄ¸±·´Ó¦£¬²úÉúµÄCOÄܶ¾º¦È¼Áϵç³ØPtµç¼«¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©¶þ¼×ÃÑ¿ÉÒÔͨ¹ýÌìÈ»ÆøºÍCO2ºÏ³ÉÖƵ㬸÷´Ó¦µÄ»¯Ñ§·½³ÌʽΪ

                                                                 ¡£

£¨2£©CH3OCH3(g)ÓëË®ÕôÆøÖØÕûÖÆÇâÆøµÄÈÈ»¯Ñ§·½³ÌʽΪ

                                                                 ¡£

£¨3£©ÏÂÁвÉÈ¡µÄ´ëÊ©ºÍ½âÊÍÕýÈ·µÄÊÇ            ¡££¨Ìî×ÖĸÐòºÅ£©

A£®·´Ó¦¹ý³ÌÔÚµÍνøÐУ¬¿É¼õÉÙCOµÄ²úÉú 

     B£®Ôö¼Ó½øË®Á¿£¬ÓÐÀûÓÚ¶þ¼×ÃѵÄת»¯£¬²¢¼õÉÙCOµÄ²úÉú

C£®Ñ¡ÔñÔÚµÍξßÓнϸ߻îÐԵĴ߻¯¼Á£¬ÓÐÖúÓÚÌá¸ß·´Ó¦¢ÚCH3OHµÄת»¯ÂÊ

D£®ÌåϵѹǿÉý¸ß£¬¶ÔÖÆÈ¡ÇâÆø²»Àû£¬ÇÒ¶Ô¼õÉÙCOµÄ²úÉú¼¸ºõÎÞÓ°Ïì

£¨4£©ÔÚζÈÏàͬµÄ3¸öÃܱÕÈÝÆ÷ÖУ¬°´²»Í¬·½Ê½Í¶Èë·´Ó¦Î±£³ÖºãΡ¢ºãѹ£¬·¢Éú·´Ó¦¢Ù£¬²âµÃ·´Ó¦´ïµ½Æ½ºâʱµÄÓйØÊý¾ÝÈçÏ¡£

ÈÝÆ÷

¼×

ÒÒ

±û

·´Ó¦ÎïͶÈëÁ¿

1mol CH3OCH3 ¡¢1mol H2O

2mol CH3OH

1mol CH3OH

CH3OHµÄŨ¶È£¨mol/L£©

c1

c2

c3

·´Ó¦µÄÄÜÁ¿±ä»¯

ÎüÊÕa kJ

·Å³öb kJ

·Å³öc kJ

ƽºâʱÌå»ý£¨L£©

V1

V2

V3

·´Ó¦Îïת»¯ÂÊ

¦Á 1

¦Á 2

¦Á 3

ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ           ¡££¨Ìî×ÖĸÐòºÅ£©

A. a+2c=37       B. ¦Á1 + ¦Á2=1       C. V1 > V3       D. c1 = 2c3

£¨5£©ÒÔ¶þ¼×ÃÑ¡¢¿ÕÆø¡¢ÇâÑõ»¯¼ØÈÜҺΪԭÁÏ£¬Ê¯Ä«Îªµç¼«Ò²¿ÉÖ±½Ó¹¹³ÉȼÁϵç³Ø¡£¸Ãµç³ØÖиº¼«Éϵĵ缫·´Ó¦Ê½ÊÇ                                              ¡£

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸