ÌâÄ¿ÄÚÈÝ
£¨16·Ö£©ÏÂͼÊÇú»¯¹¤²úÒµÁ´µÄÒ»²¿·Ö£¬ÊÔÔËÓÃËùѧ֪ʶ£¬½â¾öÏÂÁÐÎÊÌ⣺
£¨1£©¸Ã²úÒµÁ´Öкϳɰ±µÄ·´Ó¦ÔڽϵÍÎÂÏÂÄÜ·ñ×Ô·¢½øÐУ¿ ¡£
£¨2£©ÒÑÖª¸Ã²úÒµÁ´ÖÐij·´Ó¦µÄƽºâ±í´ïʽΪ£º
ËüËù¶ÔÓ¦µÄ»¯Ñ§·´Ó¦Îª£º ¡£
£¨3£©ÒÑÖªÔÚÒ»¶¨Î¶ÈÏ£¬¸÷·´Ó¦µÄƽºâ³£ÊýÈçÏ£º
C£¨s£©+CO2£¨g£© 2CO£¨g£©£¬K1
CO£¨g£©+H2O£¨g H2£¨g£©+CO2£¨g£©£¬K2
C£¨s£©+H2O£¨g£© CO£¨g£©+H2£¨g£©£¬K3
ÔòK1¡¢K2¡¢K3Ö®¼äµÄ¹ØϵÊÇ£º ¡£
£¨4£©Ãº»¯¹¤Í¨³£Í¨¹ýÑо¿²»Í¬Î¶ÈÏÂƽºâ³£ÊýÒÔ½â¾ö¸÷ÖÖʵ¼ÊÎÊÌâ¡£ÒÑÖªµÈÌå»ýµÄÒ»Ñõ»¯Ì¼ºÍË®ÕôÆø½øÈë·´Ó¦Æ÷ʱ£¬»á·¢ÉúÈçÏ·´Ó¦£ºCO(g)+H2O(g) H2(g)+CO2(g)£¬¸Ã·´Ó¦Æ½ºâ³£ÊýËæζȵı仯ÈçÏ£º
¸Ã·´Ó¦µÄÕý·´Ó¦·½ÏòÊÇ ·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©£¬ÈôÔÚ500¡æʱ½øÐУ¬ÉèÆðʼʱCOºÍH2OµÄÆðʼŨ¶È¾ùΪ0.020mol/L£¬ÔÚ¸ÃÌõ¼þÏ£¬COµÄƽºâת»¯ÂÊΪ£º ¡£
£¨5£©´ÓͼÖп´³ö°±´ß»¯Ñõ»¯¿ÉÒÔÖÆÏõËᣬ´Ë¹ý³ÌÖÐÉæ¼°µªÑõ»¯ÎÈçNO¡¢NO2¡¢N2O4µÈ¡£ÒÑÖªNO2ºÍN2O4µÄ½á¹¹Ê½·Ö±ðÊǺ͡£ÒÑÖªN£N¼ü¼üÄÜΪ167kJ¡¤mol£1£¬NO2ÖеªÑõ¼üµÄ¼üÄÜΪ466kJ¡¤mol£1£¬N2O4ÖеªÑõ¼üµÄ¼üÄÜΪ438.5kJ¡¤mol£1¡£Çëд³öNO2ת»¯ÎªN2O4µÄÈÈ»¯Ñ§·½³ÌʽΪ ¡£
¶Ô·´Ó¦N2O4(g) 2NO2(g)£¬ÔÚζÈΪT1¡¢T2ʱ£¬Æ½ºâÌåϵÖÐNO2µÄÌå»ý·ÖÊýËæѹǿ±ä»¯ÇúÏßÈçͼËùʾ¡£
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ¡¡¡¡¡¡¡¡¡£
A£®A¡¢CÁ½µãµÄ·´Ó¦ËÙÂÊ£ºA£¾C
B£®A¡¢CÁ½µãÆøÌåµÄÑÕÉ«£ºAÉCdz
C£®B¡¢CÁ½µãµÄÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿£ºB£¼C
D£®ÓÉ״̬Bµ½×´Ì¬A£¬¿ÉÒÔÓüÓÈȵķ½·¨
£¨6£©ÒÔÉÏÊö²úÒµÁ´Öм״¼ÎªÈ¼ÁÏÖƳÉȼÁϵç³Ø£¬Çëд³öÔÚÇâÑõ»¯¼Ø½éÖÊÖиõç³ØµÄ¸º¼«·´Ó¦Ê½ ¡£
£¨1£©¸Ã²úÒµÁ´Öкϳɰ±µÄ·´Ó¦ÔڽϵÍÎÂÏÂÄÜ·ñ×Ô·¢½øÐУ¿ ¡£
£¨2£©ÒÑÖª¸Ã²úÒµÁ´ÖÐij·´Ó¦µÄƽºâ±í´ïʽΪ£º
ËüËù¶ÔÓ¦µÄ»¯Ñ§·´Ó¦Îª£º ¡£
£¨3£©ÒÑÖªÔÚÒ»¶¨Î¶ÈÏ£¬¸÷·´Ó¦µÄƽºâ³£ÊýÈçÏ£º
C£¨s£©+CO2£¨g£© 2CO£¨g£©£¬K1
CO£¨g£©+H2O£¨g H2£¨g£©+CO2£¨g£©£¬K2
C£¨s£©+H2O£¨g£© CO£¨g£©+H2£¨g£©£¬K3
ÔòK1¡¢K2¡¢K3Ö®¼äµÄ¹ØϵÊÇ£º ¡£
£¨4£©Ãº»¯¹¤Í¨³£Í¨¹ýÑо¿²»Í¬Î¶ÈÏÂƽºâ³£ÊýÒÔ½â¾ö¸÷ÖÖʵ¼ÊÎÊÌâ¡£ÒÑÖªµÈÌå»ýµÄÒ»Ñõ»¯Ì¼ºÍË®ÕôÆø½øÈë·´Ó¦Æ÷ʱ£¬»á·¢ÉúÈçÏ·´Ó¦£ºCO(g)+H2O(g) H2(g)+CO2(g)£¬¸Ã·´Ó¦Æ½ºâ³£ÊýËæζȵı仯ÈçÏ£º
ζÈ/¡æ | 400 | 500 | 800 |
ƽºâ³£ÊýK | 9.94 | 9 | 1 |
£¨5£©´ÓͼÖп´³ö°±´ß»¯Ñõ»¯¿ÉÒÔÖÆÏõËᣬ´Ë¹ý³ÌÖÐÉæ¼°µªÑõ»¯ÎÈçNO¡¢NO2¡¢N2O4µÈ¡£ÒÑÖªNO2ºÍN2O4µÄ½á¹¹Ê½·Ö±ðÊǺ͡£ÒÑÖªN£N¼ü¼üÄÜΪ167kJ¡¤mol£1£¬NO2ÖеªÑõ¼üµÄ¼üÄÜΪ466kJ¡¤mol£1£¬N2O4ÖеªÑõ¼üµÄ¼üÄÜΪ438.5kJ¡¤mol£1¡£Çëд³öNO2ת»¯ÎªN2O4µÄÈÈ»¯Ñ§·½³ÌʽΪ ¡£
¶Ô·´Ó¦N2O4(g) 2NO2(g)£¬ÔÚζÈΪT1¡¢T2ʱ£¬Æ½ºâÌåϵÖÐNO2µÄÌå»ý·ÖÊýËæѹǿ±ä»¯ÇúÏßÈçͼËùʾ¡£
ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ¡¡¡¡¡¡¡¡¡£
A£®A¡¢CÁ½µãµÄ·´Ó¦ËÙÂÊ£ºA£¾C
B£®A¡¢CÁ½µãÆøÌåµÄÑÕÉ«£ºAÉCdz
C£®B¡¢CÁ½µãµÄÆøÌåµÄƽ¾ùÏà¶Ô·Ö×ÓÖÊÁ¿£ºB£¼C
D£®ÓÉ״̬Bµ½×´Ì¬A£¬¿ÉÒÔÓüÓÈȵķ½·¨
£¨6£©ÒÔÉÏÊö²úÒµÁ´Öм״¼ÎªÈ¼ÁÏÖƳÉȼÁϵç³Ø£¬Çëд³öÔÚÇâÑõ»¯¼Ø½éÖÊÖиõç³ØµÄ¸º¼«·´Ó¦Ê½ ¡£
£¨16·Ö£©
£¨1£©¿ÉÒÔ
£¨2£©C£¨s£©+H2O£¨g£© CO£¨g£©+H2£¨g£©
£¨3£©K3="K1¡ÁK2"
£¨4£©·ÅÈÈ£¬75%
£¨5£©2NO2(g) N2O4(g)¡¡¦¤H£½£57 kJ¡¤mol£1 £¬ D
£¨6£©CH3OH+8OH¡ª¡ª6e¡ª==CO32¡ª+6H2O
£¨1£©¿ÉÒÔ
£¨2£©C£¨s£©+H2O£¨g£© CO£¨g£©+H2£¨g£©
£¨3£©K3="K1¡ÁK2"
£¨4£©·ÅÈÈ£¬75%
£¨5£©2NO2(g) N2O4(g)¡¡¦¤H£½£57 kJ¡¤mol£1 £¬ D
£¨6£©CH3OH+8OH¡ª¡ª6e¡ª==CO32¡ª+6H2O
ÂÔ
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿