ÒÑÖª£ºH2O£¨g£©¨TH2O£¨l£©¡÷H1=Q 1 kJ/mol¢Ù
C2H5OH£¨g£©¨TC2H5OH£¨l£©¡÷H2=Q 2 kJ/mol¢Ú
C2H5OH£¨g£©+3O2£¨g£©¨T2CO2£¨g£©+3H2O£¨g£©¡÷H3=Q 3 kJ/mol¢Û
Èôʹ1molÒÒ´¼ÒºÌåÍêȫȼÉÕ£¬×îºó»Ö¸´µ½ÊÒΣ¬Ôò·Å³öµÄÈÈÁ¿Îª£¨¡¡¡¡£©
·ÖÎö£ºÀûÓÃÒÑÖªµÄ·´Ó¦µÃ³öC2H5OH£¨l£©+3O2£¨g£©¨T2CO2£¨g£©+3H2O£¨l£©µÄ·´Ó¦ÈÈ£¬ÀûÓÃÎïÖʵÄÁ¿Óë·´Ó¦·Å³öµÄÈÈÁ¿³ÉÕý±ÈÀ´½â´ð£®
½â´ð£º½â£º¢ÙH2O£¨g£©¨TH2O£¨l£©¡÷H1=-Q1kJ?mol-1£¨Q1£¾0£©£¬
¢ÚC2H5OH£¨g£©¨TC2H5OH£¨l£©¡÷H2=Q2kJ?mol-1£¨Q2£¼0£©£¬
¢ÛC2H5OH£¨g£©+3O2£¨g£©¨T2CO2£¨g£©+3H2O£¨g£©¡÷H3=-Q3kJ?mol-1£¨Q3£¾0£©£¬
¸ù¾Ý¸Ç˹¶¨ÂÉ¿ÉÖª£¬¢Ù¡Á3-¢Ú+¢ÛµÃC2H5OH£¨l£©+3O2£¨g£©¨T2CO2£¨g£©+3H2O£¨l£©¡÷H=£¨-3Q1+Q2-Q3£©kJ/mol£¬
¼´1molҺ̬ÒÒ´¼ÍêȫȼÉÕ²¢»Ö¸´ÖÁÊÒΣ¬Ôò·Å³öµÄÈÈÁ¿Îª£¨3Q1-Q2+Q3£©kJ£¬
¹ÊÑ¡B£®
µãÆÀ£º±¾Ì⿼²éѧÉúÀûÓøÇ˹¶¨ÂɼÆËã·´Ó¦ÈÈ£¬Ã÷È·ÒÑÖª·´Ó¦ºÍÄ¿±ê·´Ó¦µÄ¹ØÏµÊǽâ´ð±¾ÌâµÄ¹Ø¼ü£¬ÌâÄ¿ÄѶÈÖеȣ®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ºÏ³É°±Éú²ú¼¼ÊõµÄ´´Á¢¿ª±ÙÁËÈ˹¤¹ÌµªµÄ;¾¶£¬¶Ô»¯Ñ§¹¤Òµ¼¼ÊõÒ²²úÉúÁËÖØ´óÓ°Ï죮ºÏ³É°±·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºN2£¨g£©+3H2£¨g£©3NH3£¨g£©¡÷H=-92.2kJ/mol£®ºÏ³É°±¹¤ÒµÖÐÔ­ÁÏÆøN2¿É´Ó¿ÕÆøÖзÖÀëµÃµ½£¬H2¿ÉÓü×ÍéÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦ÖƵã®ÎÒ¹úºÏ³É°±¹¤ÒµÄ¿Ç°µÄÉú²úÌõ¼þΪ£º´ß»¯¼Á-Ìú´¥Ã½£¬Î¶È-400¡«500¡æ£¬Ñ¹Ç¿-30¡«50MPa£®
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ºÏ³É°±¹¤ÒµÖÐÔ­ÁÏÆøÑ¹Ëõµ½30¡«50MPaµÄÔ­ÒòÊÇ
¼ÓѹÓÐÀûÓÚÆ½ºâÕýÏòÒÆ¶¯£¬Ìá¸ßÔ­ÁÏÆøµÄת»¯ÂÊ
¼ÓѹÓÐÀûÓÚÆ½ºâÕýÏòÒÆ¶¯£¬Ìá¸ßÔ­ÁÏÆøµÄת»¯ÂÊ
£®´ÓƽºâÒÆ¶¯Ô­Àí·ÖÎö£¬µÍÎÂÓÐÀûÓÚÔ­ÁÏÆøµÄת»¯£¬Êµ¼ÊÉú²úÖвÉÓÃ400¡«500¡æµÄ¸ßΣ¬Ô­ÒòÖ®Ò»ÊÇ¿¼Âǵ½´ß»¯¼ÁµÄ´ß»¯»îÐÔ£¬Ô­ÒòÖ®¶þÊÇ
Ôö¼Ó·´Ó¦ËÙÂÊ£¬Ëõ¶Ì´ïµ½Æ½ºâµÄʱ¼ä
Ôö¼Ó·´Ó¦ËÙÂÊ£¬Ëõ¶Ì´ïµ½Æ½ºâµÄʱ¼ä
£®
£¨2£©500¡æ¡¢50MPaʱ£¬ÔÚÈÝ»ýΪVLµÄÈÝÆ÷ÖмÓÈën mol N2¡¢3n mol H2£¬·´Ó¦´ïƽºâºó²âµÃƽºâ³£ÊýΪK£¬´ËʱN2µÄת»¯ÂÊΪx£®ÔòKºÍxµÄ¹ØÏµÂú×ãK=
4xV2
27n2(1-x)4
4xV2
27n2(1-x)4
£®
£¨3£©¼×ÍéÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦·´Ó¦·½³ÌʽΪ£ºCH4£¨g£©+H2O£¨g£©=CO£¨g£©+3H2£¨g£©£®²¿·ÖÎïÖʵÄȼÉÕÈÈÊý¾ÝÈçÏÂ±í£º
Îï  ÖÊ È¼ÉÕÈÈ£¨kJ?mol-1£©
H2£¨g£© -285.8
CO£¨g£© -283.0
CH4£¨g£© -890.3
ÒÑÖª1mol H2O£¨g£©×ª±äΪ1mol H2O£¨l£©Ê±·Å³ö44.0kJÈÈÁ¿£®Ð´³öCH4ºÍH2OÔÚ¸ßÎÂÏ·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
CH4£¨g£©+H2O£¨g£©=CO£¨g£©+3H2£¨g£©¡÷H=+206.1kJ?mol-1
CH4£¨g£©+H2O£¨g£©=CO£¨g£©+3H2£¨g£©¡÷H=+206.1kJ?mol-1
£®
£¨4£©ÓÐÈËÉèÏëѰÇóºÏÊʵĴ߻¯¼ÁºÍµç¼«²ÄÁÏ£¬ÒÔN2¡¢H2Ϊµç¼«·´Ó¦ÎÒÔHCl-NH4ClΪµç½âÖÊÈÜÒºÖÆÈ¡ÐÂÐÍȼÁÏµç³Ø£®Çëд³ö¸Ãµç³ØµÄÕý¼«·´Ó¦Ê½
N2+6e-+8H+=2NH4+
N2+6e-+8H+=2NH4+
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÑÖª£ºH2O£¨g£©¨TH2O£¨l£©¡÷H=Q1 kJ?mol-1
C2H5OH£¨g£©¨TC2H5OH£¨l£©¡÷H=Q2 kJ?mol-1
C2H5OH£¨g£©+3O2£¨g£©¨T2CO2£¨g£©+3H2O£¨g£©¡÷H=Q3 kJ?mol-1
Èôʹ46g¾Æ¾«ÒºÌåÍêȫȼÉÕ£¬×îºó»Ö¸´µ½ÊÒΣ¬Ôò·Å³öµÄÈÈÁ¿Îª£¨¡¡¡¡£©

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÔËÓÃËùѧµÄ»¯Ñ§·´Ó¦Ô­ÀíÑо¿»¯Ñ§·´Ó¦ºÜÓÐÒâÒ壮Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ºÏ³É°±Ô­ÁÏÖеÄH2¿ÉÓü×ÍéÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦ÖƵ㮲¿·Ö1molÎïÖÊÍêȫȼÉÕÉú³É³£ÎÂÏÂÎȶ¨Ñõ»¯ÎïµÄ¡÷HÊý¾ÝÈçÏÂ±í£º
ÎïÖÊ ¡÷H£¨kJ/mol£©
H2£¨g£© -285.8
CO£¨g£© -283.0
CH4£¨g£© -890.3
ÒÑÖª1mol H2O£¨g£©×ª±äΪ1molH2O£¨l£©Ê±·ÅÈÈ44.0kJ£®Ð´³öCH4ºÍË®ÕôÆøÔÚ¸ßÎÂÏ·´Ó¦µÃµ½COºÍH2µÄÈÈ»¯Ñ§·½³Ìʽ
CH4£¨g£©+H2O£¨g£©¨TCO£¨g£©+3H2£¨g£©¡÷H=+206.1kJ/mol
CH4£¨g£©+H2O£¨g£©¨TCO£¨g£©+3H2£¨g£©¡÷H=+206.1kJ/mol
£®
£¨2£©500¡ãC¡¢50Mpaʱ£¬ÔÚÈÝ»ýΪVLµÄ¶¨ÈÝÈÝÆ÷ÖмÓÈën mol N2¡¢3n mol H2£¬·´Ó¦´ïµ½Æ½ºâºóN2µÄת»¯ÂÊΪa£®ÔòÈÝÆ÷ÄÚÆøÌåµÄѹǿ·´Ó¦Ç°ÓëÆ½ºâʱµÄ±ÈֵΪ
4n
4n-2na
4n
4n-2na
£®
£¨3£©Èô°Ñ±ê×¼×´¿öÏÂ4.48L°±ÆøÍ¨Èëµ½100g19.6%µÄÁòËáÈÜÒºÖУ¬ËùµÃÈÜÒºÖеÄÈÜÖÊÊÇ
NH4HSO4
NH4HSO4
£¬ÈÜÒºÖÐÀë×ÓŨ¶È´Ó´óµ½Ð¡µÄ˳ÐòÊÇ
c£¨H+£©£¾c£¨SO42-£©£¾c£¨NH+4£©£¾c£¨OH-£©
c£¨H+£©£¾c£¨SO42-£©£¾c£¨NH+4£©£¾c£¨OH-£©
£®
£¨4£©Ä³Î¶Èʱ£¬BaSO4ÔÚË®ÖеijÁµíÈÜ½âÆ½ºâÇúÏßÈçͼËùʾ£®ÓгÁµíÎö³öµÄÊÇ
b
b
µã£¨Ìî×Öĸ£©£®aµã¶ÔÓ¦µÄKsp
µÈÓÚ
µÈÓÚ
cµã¶ÔÓ¦µÄKsp £¨Ìî´óÓÚ¡¢µÈÓÚ»òСÓÚ£©
£¨5£©ÒÔÁ×ËáÑÇÌúﮣ¨LiFePO4£©Îªµç¼«²ÄÁϵÄÐÂÐÍï®Àë×Óµç³Ø¾ßÓÐÎȶ¨ÐԸߡ¢°²È«¡¢¶Ô»·¾³ÓѺõÈÓŵ㣬¿ÉÓÃÓڵ綯Æû³µ£®¸Ãï®Àë×Óµç³ØÔÚ³äµç¹ý³ÌÖУ¬Ñô¼«µÄÁ×ËáÑÇÌúï®Éú³ÉÁ×ËáÌú£¬Ôò¸Ãµç³Ø·ÅµçʱÕý¼«µÄµç¼«·´Ó¦Ê½Îª
FePO4+Li++e-¨TLiFePO4
FePO4+Li++e-¨TLiFePO4
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÔĶÁÀí½â

ºÏ³É°±Éú²ú¼¼ÊõµÄ´´Á¢¿ª±ÙÁËÈ˹¤¹ÌµªµÄ;¾¶£¬¶Ô»¯Ñ§¹¤Òµ¼¼ÊõÒ²²úÉúÁËÖØ´óÓ°Ï죮ºÏ³É°±·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºN2£¨g£©+3H2£¨g£©2£¨£¨£¨NH3£¨g£©¡÷H=-92.2kJ/mol£®ºÏ³É°±¹¤ÒµÖÐÔ­ÁÏÆøN2¿É´Ó¿ÕÆøÖзÖÀëµÃµ½£¬H2¿ÉÓü×ÍéÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦ÖƵã®

£¨1£©ÔÚÒ»ÈÝ»ý¹Ì¶¨µÄÃܱÕÈÝÆ÷ÖÐ×¢ÈëN2ºÍH2Á½ÖÖÆøÌ壬·¢ÉúÉÏÊö·´Ó¦£®ÔÚijζÈÏ´ﵽƽºâʱ¸÷ÎïÖʵÄŨ¶È·Ö±ðΪc£¨H2£©=9.00mol/L£¬c£¨N2£©=3.00mol/L£¬c£¨NH3£©=4.00mol/L£¬´ËζÈϸ÷´Ó¦µÄƽºâ³£ÊýΪ
7.32¡Á10-3
7.32¡Á10-3
£®
£¨2£©Èý¸öÏàͬÈÝÆ÷Öи÷³äÈë1molN2ºÍ3molH2£¬ÔÚ²»Í¬Ìõ¼þÏ·´Ó¦²¢´ïµ½Æ½ºâ£¬°±µÄÌå»ý·ÖÊýËæÊ±¼äµÄ±ä»¯ÈçͼËùʾ£®ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ
D
D
£¨ÌîÐòºÅ£©£®
A£®Í¼I¿ÉÄÜÊDz»Í¬Ñ¹Ç¿¶Ô·´Ó¦µÄÓ°Ï죬ÇÒP2£¾P1
B£®Í¼¢ò¿ÉÄÜÊDz»Í¬Ñ¹Ç¿¶Ô·´Ó¦µÄÓ°Ï죬ÇÒP1£¾P2
C£®Í¼¢ó¿ÉÄÜÊDz»Í¬Î¶ȶԷ´Ó¦µÄÓ°Ï죬ÇÒT1£¾T2
D£®Í¼¢òÒ»¶¨ÊÇÔÚͬÎÂͬѹϲ»Í¬´ß»¯¼Á¶Ô·´Ó¦µÄÓ°Ï죬ÇÒ´ß»¯¼ÁЧ¹û1£¾2
£¨3£©½«Ë®ÕôÆøÍ¨¹ýºìÈȵÄÌ¿¼´²úÉúË®ÕôÆø£¬»¯Ñ§·½³ÌʽΪ£ºC£¨s£©+H2O£¨g£© H2£¨g£©+CO£¨g£©¡÷H=+131.3kJ£¬¡÷S=+133.7J/K¸Ã·´Ó¦ÔÚµÍÎÂÏÂÄÜ·ñ×Ô·¢
·ñ
·ñ
£¨ÌÄÜ»ò·ñ£©£®
£¨4£©¼×ÍéÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦·´Ó¦·½³ÌʽΪ£ºCH4£¨g£©+H2O£¨g£©=CO£¨g£©+3H2£¨g£©£®²¿·ÖÎïÖʵÄȼÉÕÈÈÊý¾ÝÈçÏÂ±í£º
Îï  ÖÊ È¼ÉÕÈÈ£¨kJ?mol-1£©
H2£¨g£© -285.8
CO£¨g£© -283.0
CH4£¨g£© -890.3
ÒÑÖª1mol H2O£¨g£©×ª±äΪ1mol H2O£¨l£©Ê±·Å³ö44.0kJÈÈÁ¿£®Ð´³öCH4ºÍH2OÔÚ¸ßÎÂÏ·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
CH4£¨g£©+H2O£¨g£©=CO£¨g£©+3H2£¨g£©¡÷H=+206.1kJ?mol-1
CH4£¨g£©+H2O£¨g£©=CO£¨g£©+3H2£¨g£©¡÷H=+206.1kJ?mol-1
£®
£¨5£©ÓÐÈËÉèÏëѰÇóºÏÊʵĴ߻¯¼ÁºÍµç¼«²ÄÁÏ£¬ÒÔN2¡¢H2Ϊµç¼«·´Ó¦ÎÒÔHCl-NH4ClΪµç½âÖÊÈÜÒºÖÆÈ¡ÐÂÐÍȼÁÏµç³Ø£®Çëд³ö¸Ãµç³ØµÄÕý¼«·´Ó¦Ê½
N2+6e-+8H+=2NH4+
N2+6e-+8H+=2NH4+
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÒÀ¾ÝÊÂʵ£¬Ð´³öÏÂÁз´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ£®
£¨1£©1mol N2 £¨g£©ÓëÊÊÁ¿O2£¨g£©Æð·´Ó¦£¬Éú³ÉNO2£¨g£©£¬ÎüÊÕ68kJÈÈÁ¿
N2£¨g£©+2O2£¨g£©=2NO2£¨g£©¡÷H=+68kJ?mol-1
N2£¨g£©+2O2£¨g£©=2NO2£¨g£©¡÷H=+68kJ?mol-1

£¨2£©1mol Cu£¨s£©ÓëÊÊÁ¿O2£¨g£©Æð·´Ó¦£¬Éú³ÉCuO£¨s£©£¬·Å³ö157kJÈÈÁ¿
Cu£¨s£©+
1
2
O2£¨g£©=CuO£¨s£©£»¡÷H=-157kJ?mol-1
Cu£¨s£©+
1
2
O2£¨g£©=CuO£¨s£©£»¡÷H=-157kJ?mol-1

£¨3£©ÎÀÐÇ·¢Éäʱ¿ÉÓÃ루N2H4£©×÷ȼÁÏ£¬1mol N2H4£¨l£©ÔÚO2£¨g£©ÖÐȼÉÕ£¬Éú³ÉN2£¨g£©ºÍH2O£¨l£©£¬·Å³ö622kJÈÈÁ¿£®
N2H4£¨g£©+O2£¨g£©=N2£¨g£©+2H2O£¨l£©¡÷H=-622kJ?mol-l
N2H4£¨g£©+O2£¨g£©=N2£¨g£©+2H2O£¨l£©¡÷H=-622kJ?mol-l

£¨4£©½«0.3molµÄÆøÌ¬¸ßÄÜȼÁÏÒÒÅðÍ飨B2H6£©ÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5kJÈÈÁ¿£¬¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽΪ
B2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol
B2H6£¨g£©+3O2£¨g£©=B2O3£¨s£©+3H2O£¨l£©¡÷H=-2165kJ/mol
£®ÓÖÒÑÖª£ºH2O£¨g£©=H2O£¨l£©£»¡÷H2=-44.0kJ/mol£¬Ôò11.2L£¨±ê×¼×´¿ö£©ÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆøÌ¬Ë®Ê±·Å³öµÄÈÈÁ¿ÊÇ
1016.5
1016.5
kJ£®

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸