ÌâÄ¿ÄÚÈÝ
T¡æÊ±£¬Óмס¢ÒÒÁ½¸öÃܱÕÈÝÆ÷£¬¼×ÈÝÆ÷µÄÌå»ýΪ1L£¬ÒÒÈÝÆ÷µÄÌå»ýΪ2L£¬·Ö±ðÏò¼×¡¢ÒÒÁ½ÈÝÆ÷ÖмÓÈë6mol AºÍ3mol B£¬·¢Éú·´Ó¦ÈçÏ£º3A£¨g£©+bB£¨g£©?3C£¨g£©+2D£¨g£©¡÷H£¼0£» 4minʱ¼×ÈÝÆ÷Äڵķ´Ó¦Ç¡ºÃ´ïµ½Æ½ºâ£¬AµÄŨ¶ÈΪ2.4mol/L£¬BµÄŨ¶ÈΪ1.8mol/L£» t minʱÒÒÈÝÆ÷Äڵķ´Ó¦´ïƽºâ£¬¸ù¾ÝÌâ¸øÐÅÏ¢»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¼×ÈÝÆ÷Öз´Ó¦µÄƽ¾ùËÙÂÊv£¨B£©= £¬»¯Ñ§·½³ÌʽÖмÆÁ¿Êýb= £®
£¨2£©ÒÒÈÝÆ÷Öз´Ó¦´ïµ½Æ½ºâʱËùÐèʱ¼ät 4min£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£®
£¨3£©ÈôҪʹ¼×¡¢ÒÒÈÝÆ÷ÖÐBµÄƽºâŨ¶ÈÏàµÈ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊÇ £®
A£®±£³ÖζȲ»±ä£¬Ôö´ó¼×ÈÝÆ÷µÄÌå»ýÖÁ2L
B£®±£³ÖÈÝÆ÷Ìå»ý²»±ä£¬Ê¹¼×ÈÝÆ÷Éý¸ßζÈ
C£®±£³ÖÈÝÆ÷ѹǿºÍζȶ¼²»±ä£¬Ïò¼×ÖмÓÈëÒ»¶¨Á¿µÄAÆøÌå
D£®±£³ÖÈÝÆ÷ѹǿºÍζȶ¼²»±ä£¬Ïò¼×ÖмÓÈëÒ»¶¨Á¿µÄBÆøÌå
£¨4£©Ð´³öƽºâ³£Êý±í´ïʽK= £¬²¢¼ÆËãÔÚT¡æÊ±µÄ»¯Ñ§Æ½ºâ³£ÊýK= £®
£¨1£©¼×ÈÝÆ÷Öз´Ó¦µÄƽ¾ùËÙÂÊv£¨B£©=
£¨2£©ÒÒÈÝÆ÷Öз´Ó¦´ïµ½Æ½ºâʱËùÐèʱ¼ät
£¨3£©ÈôҪʹ¼×¡¢ÒÒÈÝÆ÷ÖÐBµÄƽºâŨ¶ÈÏàµÈ£¬¿ÉÒÔ²ÉÈ¡µÄ´ëÊ©ÊÇ
A£®±£³ÖζȲ»±ä£¬Ôö´ó¼×ÈÝÆ÷µÄÌå»ýÖÁ2L
B£®±£³ÖÈÝÆ÷Ìå»ý²»±ä£¬Ê¹¼×ÈÝÆ÷Éý¸ßζÈ
C£®±£³ÖÈÝÆ÷ѹǿºÍζȶ¼²»±ä£¬Ïò¼×ÖмÓÈëÒ»¶¨Á¿µÄAÆøÌå
D£®±£³ÖÈÝÆ÷ѹǿºÍζȶ¼²»±ä£¬Ïò¼×ÖмÓÈëÒ»¶¨Á¿µÄBÆøÌå
£¨4£©Ð´³öƽºâ³£Êý±í´ïʽK=
¿¼µã£º·´Ó¦ËÙÂʵ͍Á¿±íʾ·½·¨,»¯Ñ§Æ½ºâµÄÓ°ÏìÒòËØ
רÌ⣺»¯Ñ§Æ½ºâרÌâ
·ÖÎö£º£¨1£©¸ù¾Ýv=
¼ÆËã·´Ó¦ËÙÂÊ£¬¸ù¾ÝËÙÂÊÖ®±ÈµÈÓÚ¼ÆÁ¿ÊýÖ®±ÈÈ·¶¨bµÄÖµ£»
£¨2£©¸ù¾ÝÓ°Ï컯ѧ·´Ó¦ËÙÂʵÄÒòËØÅжϣ»
£¨3£©¸ù¾ÝµÈЧƽºâµÄÌõ¼þÅжϣ»
£¨4£©¸ù¾Ýƽºâ³£ÊýµÄ¶¨ÒåÊéд±í´ïʽ£¬²¢½øÐмÆË㣻
| ¡÷c |
| t |
£¨2£©¸ù¾ÝÓ°Ï컯ѧ·´Ó¦ËÙÂʵÄÒòËØÅжϣ»
£¨3£©¸ù¾ÝµÈЧƽºâµÄÌõ¼þÅжϣ»
£¨4£©¸ù¾Ýƽºâ³£ÊýµÄ¶¨ÒåÊéд±í´ïʽ£¬²¢½øÐмÆË㣻
½â´ð£º
½â£º£¨1£©ÆðʼʱA¡¢BµÄŨ¶È·Ö±ðΪ6mol/LºÍ3mol/L£¬´ïµ½Æ½ºâʱ£¬A¡¢BµÄŨ¶È·Ö±ðΪ2.4mol/LºÍ1.8mol/L£¬¸ù¾Ýv=
¿ÉÖª£¬AµÄ·´Ó¦ËÙÂÊΪ
=0.9mol/£¨L?min£©£¬BµÄ·´Ó¦ËÙÂÊΪ
=0.3mol/£¨L?min£©£¬¸ù¾ÝËÙÂÊÖ®±ÈµÈÓÚ¼ÆÁ¿ÊýÖ®±È¿ÉµÃ3£ºb=3£º1£¬ËùÒÔb=1£¬¹Ê´ð°¸Îª£º0.3mol/£¨L?min£©£»1£»
£¨2£©ÒÒÈÝÆ÷µÄÌå»ý´óÓÚ¼×ÈÝÆ÷£¬ËùÒÔ¼×ÈÝÆ÷ÖÐѹǿ´óÓÚÒÒ£¬ËùÒÔ·´Ó¦ËÙÂʿ죬ÒÒÈÝÆ÷Öз´Ó¦´ïµ½Æ½ºâʱËùÐèʱ¼ät£¾4min£¬¹Ê´ð°¸Îª£º´óÓÚ£»
£¨3£©ÒªÊ¹¼×¡¢ÒÒÈÝÆ÷ÖÐBµÄƽºâŨ¶ÈÏàµÈ£¬¼´Ê¹ËüÃdzÉΪµÈЧƽºâ£¬
A£®±£³ÖζȲ»±ä£¬Ôö´ó¼×ÈÝÆ÷µÄÌå»ýÖÁ2L£¬ÔòÆðʼ״̬ºÍÒÒÈÝÆ÷Ò»Ñù£¬ËùÒÔËüÃÇÆ½ºâʱ¸÷³É·ÖµÄŨ¶ÈÒ²Ò»Ñù£¬¹ÊAÕýÈ·£»
B£®ÔڸıäÌõ¼þǰ£¬¼×ÈÝÆ÷ÖÐBµÄŨ¶È´óÓÚÒÒÈÝÆ÷£¬ÓÉÓڸ÷´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬Ê¹¼×ÈÝÆ÷Éý¸ßζȣ¬Æ½ºâÄæÏòÒÆ¶¯£¬¼×ÈÝÆ÷ÖÐBµÄŨ¶È¸ü´ó£¬¹ÊB´íÎó£»
C£®ÔڸıäÌõ¼þǰ£¬¼×ÈÝÆ÷ÖÐBµÄŨ¶È´óÓÚÒÒÈÝÆ÷£¬±£³ÖÈÝÆ÷ѹǿºÍζȶ¼²»±ä£¬Ïò¼×ÖмÓÈëÒ»¶¨Á¿µÄAÆøÌ壬ƽºâÕýÏòÒÆ¶¯£¬BµÄŨ¶È¼õС£¬ÓпÉÄܺÍÒÒÈÝÆ÷ÖÐBµÄŨ¶ÈÏàµÈ£¬¹ÊCÕýÈ·£»
D£®ÔڸıäÌõ¼þǰ£¬¼×ÈÝÆ÷ÖÐBµÄŨ¶È´óÓÚÒÒÈÝÆ÷£¬±£³ÖÈÝÆ÷ѹǿºÍζȶ¼²»±ä£¬Ïò¼×ÖмÓÈëÒ»¶¨Á¿µÄBÆøÌ壬Ôò¼×ÈÝÆ÷ÖÐBµÄŨ¶È¸ü´ó£¬ËùÒÔD´íÎó£»
¹ÊÑ¡AC£»
£¨4£©Æ½ºâ³£ÊýµÈÓÚÉú³ÉÎïŨ¶ÈµÄϵÊý´ÎÃÝÖ®»ý³ýÒÔ·´Ó¦ÎïŨ¶ÈµÄϵÊý´ÎÃÝÖ®»ý£¬ËùÒÔK=
£¬ÔÚ¼×ÈÝÆ÷ÖУ¬Æ½ºâʱ·´Ó¦ÓÃÈ¥µÄAµÄŨ¶ÈΪ3.6mol/L£¬¸ù¾Ý»¯Ñ§·½³Ìʽ¿ÉÖª£¬Æ½ºâʱCµÄŨ¶ÈΪ3.6mol/L£¬DµÄŨ¶ÈΪ2.4mol/L£¬ËùÒÔK=
=10.8mol/L£¬¹Ê´ð°¸Îª£º
£»10.8mol/L£»
| ¡÷c |
| t |
| 6mol/L-2.4mol/L |
| 4min |
| 3mol/L-1.8mol/L |
| 4min |
£¨2£©ÒÒÈÝÆ÷µÄÌå»ý´óÓÚ¼×ÈÝÆ÷£¬ËùÒÔ¼×ÈÝÆ÷ÖÐѹǿ´óÓÚÒÒ£¬ËùÒÔ·´Ó¦ËÙÂʿ죬ÒÒÈÝÆ÷Öз´Ó¦´ïµ½Æ½ºâʱËùÐèʱ¼ät£¾4min£¬¹Ê´ð°¸Îª£º´óÓÚ£»
£¨3£©ÒªÊ¹¼×¡¢ÒÒÈÝÆ÷ÖÐBµÄƽºâŨ¶ÈÏàµÈ£¬¼´Ê¹ËüÃdzÉΪµÈЧƽºâ£¬
A£®±£³ÖζȲ»±ä£¬Ôö´ó¼×ÈÝÆ÷µÄÌå»ýÖÁ2L£¬ÔòÆðʼ״̬ºÍÒÒÈÝÆ÷Ò»Ñù£¬ËùÒÔËüÃÇÆ½ºâʱ¸÷³É·ÖµÄŨ¶ÈÒ²Ò»Ñù£¬¹ÊAÕýÈ·£»
B£®ÔڸıäÌõ¼þǰ£¬¼×ÈÝÆ÷ÖÐBµÄŨ¶È´óÓÚÒÒÈÝÆ÷£¬ÓÉÓڸ÷´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬Ê¹¼×ÈÝÆ÷Éý¸ßζȣ¬Æ½ºâÄæÏòÒÆ¶¯£¬¼×ÈÝÆ÷ÖÐBµÄŨ¶È¸ü´ó£¬¹ÊB´íÎó£»
C£®ÔڸıäÌõ¼þǰ£¬¼×ÈÝÆ÷ÖÐBµÄŨ¶È´óÓÚÒÒÈÝÆ÷£¬±£³ÖÈÝÆ÷ѹǿºÍζȶ¼²»±ä£¬Ïò¼×ÖмÓÈëÒ»¶¨Á¿µÄAÆøÌ壬ƽºâÕýÏòÒÆ¶¯£¬BµÄŨ¶È¼õС£¬ÓпÉÄܺÍÒÒÈÝÆ÷ÖÐBµÄŨ¶ÈÏàµÈ£¬¹ÊCÕýÈ·£»
D£®ÔڸıäÌõ¼þǰ£¬¼×ÈÝÆ÷ÖÐBµÄŨ¶È´óÓÚÒÒÈÝÆ÷£¬±£³ÖÈÝÆ÷ѹǿºÍζȶ¼²»±ä£¬Ïò¼×ÖмÓÈëÒ»¶¨Á¿µÄBÆøÌ壬Ôò¼×ÈÝÆ÷ÖÐBµÄŨ¶È¸ü´ó£¬ËùÒÔD´íÎó£»
¹ÊÑ¡AC£»
£¨4£©Æ½ºâ³£ÊýµÈÓÚÉú³ÉÎïŨ¶ÈµÄϵÊý´ÎÃÝÖ®»ý³ýÒÔ·´Ó¦ÎïŨ¶ÈµÄϵÊý´ÎÃÝÖ®»ý£¬ËùÒÔK=
| c3(C)?c2(D) |
| c3(A)?c(B) |
| 3£®63¡Á2£®42 |
| 2£®43¡Á1.8 |
| c3(C)?c2(D) |
| c3(A)?c(B) |
µãÆÀ£º±¾ÌâÖ÷Òª¿¼²éÁË»¯Ñ§·´Ó¦ËÙÂʵļÆËã¡¢»¯Ñ§·´Ó¦ËÙÂÊÖ®±ÈµÈÓÚ¼ÆÁ¿ÊýÖ®±È¡¢Ó°Ï컯ѧ·´Ó¦ËÙÂʵÄÒòËØ¡¢µÈЧƽºâµÈ֪ʶµã£¬ÖеÈÄѶȣ¬½âÌâʱעÒâ»ù±¾¸ÅÄîºÍ»ù±¾¹«Ê½µÄÁé»îÔËÓã®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ÒÑÖª298K¡¢101kPaʱÏÂÁз´Ó¦£º¢Ù2H2£¨g£©+O2£¨g£©=2H2O£¨l£©¡÷H=-571.6kJ?mol-1¢ÚC2H4£¨g£©+3O2£¨g£©=2CO2£¨g£©+2H2O£¨l£©¡÷H=-1 411.0kJ?mol-1¢ÛC2H5OH£¨l£©+3O2£¨g£©=2CO2£¨g£©+3H2O£¨l£©¡÷H=-1 366.8kJ?mol-1ÔòC2H4£¨g£©+H2O£¨l£©=C2H5OH£¨l£©µÄ¡÷H Ϊ£¨¡¡¡¡£©
| A¡¢-44.2 kJ?mol-1 |
| B¡¢+44.2 kJ?mol-1 |
| C¡¢-330 kJ?mol-1 |
| D¡¢+330 kJ?mol |
Ò»¶¨Î¶ÈÏÂ2SO2+O2?2SO3£¬´ïµ½Æ½ºâʱ£¬n£¨SO2£©£ºn£¨O2£©£ºn£¨SO3£©=3£º2£º4£®ËõСÌå»ý£¬·´Ó¦Ôٴδﵽƽºâʱ£¬n£¨O2£©=0.4mol£¬n £¨SO3£©=1.6mol£¬´ËʱSO2µÄÎïÖʵÄÁ¿Ó¦ÊÇ£¨¡¡¡¡£©
| A¡¢0.5mol |
| B¡¢0.6moL |
| C¡¢0.7mol |
| D¡¢1.2mol |
ÓÃNA±íʾ°¢·ü¼ÓµÂÂÞ³£ÊýµÄÖµ£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©
| A¡¢22.4LÇâÆøÖк¬ÓÐÇâ·Ö×ÓÖÂĿΪNA |
| B¡¢0.5mol Na2CO3Öк¬ÓеÄNa+ÊýĿΪ0.5 NA |
| C¡¢³£Î³£Ñ¹Ï£¬14gµªÆøº¬ÓеÄÔ×ÓÊýĿΪNA |
| D¡¢0.5 mol/L Fe2£¨SO4£©3ÈÜÒºÖУ¬SµÄÊýĿΪ1.5NA |