ÌâÄ¿ÄÚÈÝ

11£®ºÏ³É°±¹¤Òµ¶Ô¹úÃñ¾­¼ÃºÍÉç»á·¢Õ¹¾ßÓÐÖØÒªµÄÒâÒ壮¸ù¾ÝÒÑѧ֪ʶ»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©¶ÔÓÚÃܱÕÈÝÆ÷Öеķ´Ó¦£ºN2£¨g£©+3H2£¨g£©¨T2NH3£¨g£©¡÷H£¼0£¬673K£¬30MPaÏÂn£¨NH3£©ºÍn£¨H2£©Ëæʱ¼ä±ä»¯µÄ¹ØϵÈçͼËùʾ£®
ÏÂÁÐÐðÊöÕýÈ·µÄÊÇAD£¨ÌîÑ¡Ï£®
A£®µãaµÄÕý·´Ó¦ËÙÂʱȵãbµÄ´ó
B£®µãc´¦·´Ó¦´ïµ½Æ½ºâ
C£®µãd£¨t1ʱ¿Ì£©ºÍµãe£¨t2ʱ¿Ì£©´¦n£¨N2£©²»Ò»Ñù
D£®ÆäËûÌõ¼þ²»±ä£¬773KÏ·´Ó¦ÖÁt1ʱ¿Ì£¬n£¨H2£©±ÈÉÏͼÖÐdµãµÄÖµ´ó
£¨2£©ÒÑÖªN2£¨g£©+3H2 £¨g£©¨T2NH3£¨g£©¡÷H=-92.4kJ•mol-1
¢ÙºÏ³É°±¹¤Òµ²ÉÈ¡µÄÏÂÁдëÊ©²»¿ÉÓÃƽºâÒƶ¯Ô­Àí½âÊ͵ÄÊÇBC£¨ÌîÑ¡Ï£®
A£®²ÉÓýϸßѹǿ£¨20MPa¡«50MPa£©    B£®²ÉÓÃ500¡æµÄ¸ßÎÂ
C£®ÓÃÌú´¥Ã½×÷´ß»¯¼Á                D£®½«Éú³ÉµÄ°±Òº»¯²¢¼°Ê±´ÓÌåϵÖзÖÀë³öÀ´
¢ÚÔÚÈÝ»ý¾ùΪ2L£¨ÈÝÆ÷Ìå»ý²»¿É±ä£©µÄ¼×¡¢ÒÒÁ½¸öÈÝÆ÷ÖУ¬·Ö±ð¼ÓÈë2molN2¡¢6molH2ºÍ1molN2¡¢3molH2£¬ÔÚÏàͬζȡ¢´ß»¯¼ÁÏÂʹÆä·´Ó¦£®×îÖմﵽƽºâºó£¬Á½ÈÝÆ÷N2ת»¯ÂÊ·Ö±ðΪ¦Á¼×¡¢¦ÁÒÒ£¬Ôò¼×ÈÝÆ÷ÖÐƽºâ³£Êý±í´ïʽΪ$\frac{4{{¦Á}_{¼×}}^{2}}{27£¨1-{¦Á}_{¼×}£©^{4}}$£¨Óú¬¦Á¼×µÄ´úÊýʽ±íʾ£¬»¯¼òΪ×î¼òʽ£©£¬´Ëʱ¦Á¼×£¾¦ÁÒÒ£¨Ìî¡°£¾¡±¡¢¡°£¼¡±¡°=¡±£©£®

·ÖÎö £¨1£©A£®a¡¢bÁ½µãÏòÕý·´Ó¦½øÐУ¬Å¨¶ÈÔ½´ó·´Ó¦ËÙÂÊÔ½¿ì£»
B£®cµã·´Ó¦ÎïºÍÉú³ÉÎïÎïÖʵÄÁ¿ÈÔÔڱ仯£»
C£®dµãºÍeµã¶¼´¦ÓÚƽºâ״̬£¬n£¨N2£©²»±ä£»
D£®¸Ã·´Ó¦Õý·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬Î¶ÈÉý¸ßƽºâÏòÄæ·´Ó¦Òƶ¯£»
£¨2£©¢ÙÀÕɳÌØÁÐÔ­ÀíÊÇÈç¹û¸Ä±äÓ°ÏìƽºâµÄÒ»¸öÌõ¼þ£¨ÈçŨ¶È¡¢Ñ¹Ç¿»òζȵȣ©£¬Æ½ºâ¾ÍÏòÄܹ»¼õÈõÕâÖָıäµÄ·½ÏòÒƶ¯£¬ÀÕɳÌØÁÐÔ­ÀíÊÊÓõĶÔÏóÓ¦´æÔÚ¿ÉÄæ¹ý³Ì£¬ÈçÓë¿ÉÄæ¹ý³ÌÎ޹أ¬ÓëƽºâÒƶ¯Î޹أ¬Ôò²»ÄÜÓÃÀÕɳÌØÁÐÔ­Àí½âÊÍ£»
¢Ú¼×ÖÐN2µÄÆðʼŨ¶ÈΪ$\frac{2mol}{2L}$=1mol/L£¬H2µÄÆðʼŨ¶ÈΪ$\frac{6mol}{2L}$=3mol/L£¬Æ½ºâʱµªÆøŨ¶È±ä»¯Á¿Îª¦Á¼×mol/L£¬Ôò£º
               N2£¨g£©+3H2£¨g£©¨T2NH3£¨g£©
ÆðʼŨ¶È£¨mol/L£©£º1      3          0
±ä»¯Å¨¶È£¨mol/L£©£º¦Á¼×     3¦Á¼×       2¦Á¼×     
ƽºâŨ¶È£¨mol/L£©£º1-¦Á¼×   3£¨1-¦Á¼×£©  2¦Á¼×         
ÔÙ¸ù¾ÝK=$\frac{{c}^{2}£¨N{H}_{3}£©}{c£¨{N}_{2}£©¡Á{c}^{3}£¨{H}_{2}£©}$¼ÆËãƽºâ³£Êý£»
¼×µÈЧΪÒÒƽºâµÄ»ù´¡ÉÏѹǿÔö´óÒ»±¶£¬Ôö´óѹǿƽºâÕýÏòÒƶ¯£®

½â´ð ½â£º£¨1£©A£®a¡¢bÁ½µãÏòÕý·´Ó¦½øÐУ¬Ëæ×Å·´Ó¦µÄ½øÐУ¬·´Ó¦ÎïµÄŨ¶ÈÖð½¥¼õС£¬·´Ó¦ËÙÂÊÖð½¥¼õС£¬ÔòaµãµÄÕý·´Ó¦ËÙÂʱÈbµã´ó£¬¹ÊAÕýÈ·£»
B£®cµã·´Ó¦ÎïºÍÉú³ÉÎïÎïÖʵÄÁ¿ÈÔÔڱ仯£¬Ã»Óдﵽƽºâ״̬£¬¹ÊB´íÎó£»
C£®dµãºÍeµã¶¼´¦ÓÚƽºâ״̬£¬n£¨N2£©²»±ä£¬dµãºÍeµãn£¨N2£©ÏàµÈ£¬¹ÊC´íÎó£»
D£®¸Ã·´Ó¦Õý·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¬Î¶ÈÉý¸ßƽºâÏòÄæ·´Ó¦Òƶ¯£¬ÇâÆøµÄÎïÖʵÄÁ¿Ôö´ó£¬¹ÊDÕýÈ·£»
¹Ê´ð°¸Îª£ºAD£»
£¨2£©¢ÙA£®¸ßÓÚ³£Ñ¹£¬Ôö´óѹǿ£¬»¯Ñ§Æ½ºâ»áÕýÏòÒƶ¯£¬ÓÐÀûÓÚ°±ÆøµÄºÏ³É£¬ÄÜÓÃÀÕɳÌØÁÐÔ­Àí½âÊÍ£¬¹ÊA²»Ñ¡£»
B£®500¡æµÄ¸ßΣ¬²»ÓÐÀûÓÚ°±ÆøµÄºÏ³É£¬µ«ÊÇ¿ÉÒÔÌá¸ß´ß»¯¼ÁµÄ´ß»¯»îÐÔ£¬²»ÄÜÓÃÀÕɳÌØÁÐÔ­Àí½âÊÍ£¬¹ÊBÑ¡£»
C£®Ìú´¥Ã½×÷´ß»¯¼Á£¬²»»áÒýÆð»¯Ñ§Æ½ºâµÄÒƶ¯£¬²»ÄÜÓÃÀÕɳÌØÁÐÔ­Àí½âÊÍ£¬¹ÊCÑ¡£»
D£®½«Éú³ÉµÄ°±Òº»¯²¢¼°Ê±´ÓÌåϵÖзÖÀë³öÀ´£¬Î´·´Ó¦µÄN2¡¢H2Ñ­»·µ½ºÏ³ÉËþÖУ¬¶¼»áʹµÃ»¯Ñ§Æ½ºâÕýÏòÒƶ¯£¬ÓÐÀûÓÚ°±µÄºÏ³É£¬ÄÜÓÃÀÕɳÌØÁÐÔ­Àí½âÊÍ£¬¹ÊD²»Ñ¡£¬
¹ÊÑ¡£ºBC£»
¢Ú¼×ÖÐN2µÄÆðʼŨ¶ÈΪ$\frac{2mol}{2L}$=1mol/L£¬H2µÄÆðʼŨ¶ÈΪ$\frac{6mol}{2L}$=3mol/L£¬Æ½ºâʱµªÆøŨ¶È±ä»¯Á¿Îª¦Á¼×mol/L£¬Ôò£º
               N2£¨g£©+3H2£¨g£©¨T2NH3£¨g£©
ÆðʼŨ¶È£¨mol/L£©£º1      3          0
±ä»¯Å¨¶È£¨mol/L£©£º¦Á¼×     3¦Á¼×       2¦Á¼×
ƽºâŨ¶È£¨mol/L£©£º1-¦Á¼×   3£¨1-¦Á¼×£©  2¦Á¼×
¹Êƽºâ³£ÊýK=$\frac{{c}^{2}£¨N{H}_{3}£©}{c£¨{N}_{2}£©¡Á{c}^{3}£¨{H}_{2}£©}$=$\frac{£¨2{¦Á}_{¼×}£©^{2}}{£¨1-{¦Á}_{¼×}£©¡Á[3£¨1-{¦Á}_{¼×}£©]^{3}}$=$\frac{4{{¦Á}_{¼×}}^{2}}{27£¨1-{¦Á}_{¼×}£©^{4}}$£¬
¼×µÈЧΪÒÒƽºâµÄ»ù´¡ÉÏѹǿÔö´óÒ»±¶£¬Ôö´óѹǿƽºâÕýÏòÒƶ¯£¬N2ת»¯ÂÊÔö´ó£¬¹Ê¦Á¼×£¾¦ÁÒÒ£¬
¹Ê´ð°¸Îª£º$\frac{4{{¦Á}_{¼×}}^{2}}{27£¨1-{¦Á}_{¼×}£©^{4}}$£»£¾£®

µãÆÀ ±¾Ì⿼²é½Ï×ۺϣ¬Éæ¼°ÎïÖʵÄÁ¿»òŨ¶ÈËæʱ¼äµÄ±ä»¯ÇúÏß¡¢»¯Ñ§Æ½ºâÒƶ¯Ô­Àí¡¢»¯Ñ§Æ½ºâµÄ¼ÆË㣬עÖظ߿¼³£¿¼²éµãµÄ¿¼²é£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
19£®¸ßÃÌËá¼ØÊÇÃ̵ÄÖØÒª»¯ºÏÎïºÍ³£ÓõÄÑõ»¯¼Á£®ÒÔÏÂÊǹ¤ÒµÉÏÓÃÈíÃÌ¿óÖƱ¸¸ßÃÌËá¼ØµÄÒ»ÖÖ¹¤ÒÕÁ÷³Ì£®20¡æʱÈܽâ¶È£¨¿Ë/100¿ËË®£©£¬K2CO3£º111£¬KMnO4£º6.34£®
£¨1£©KMnO4Ï¡ÈÜÒºÊÇÒ»ÖÖ³£ÓõÄÏû¶¾¼Á£®ÆäÏû¶¾»úÀíÓëÏÂÁÐBD£¨ÌîÐòºÅ£©ÎïÖÊÏàËÆ£®
A£®75%¾Æ¾«         B£®Ë«ÑõË®          C£®±½·Ó        D£®NaClOÈÜÒº
£¨2£©Ð´³öMnO2¡¢KOHµÄÈÛÈÚ»ìºÏÎïÖÐͨÈë¿ÕÆøʱ·¢ÉúµÄÖ÷Òª·´Ó¦µÄ»¯Ñ§·½³Ìʽ£º2MnO2+4KOH+O2 $\frac{\underline{\;ÈÛÈÚ\;}}{\;}$2K2MnO4+2H2O£®
£¨3£©ÏòK2MnO4ÈÜÒºÖÐͨÈëCO2ÒÔÖƱ¸KMnO4£¬¸Ã·´Ó¦ÖеĻ¹Ô­¼ÁºÍÑõ»¯¼ÁÖ®±ÈÊÇ2£º1£®
£¨4£©ÉÏÊöÁ÷³ÌÖпÉÒÔÑ­»·Ê¹ÓõÄÎïÖÊÓÐʯ»Ò¡¢¶þÑõ»¯Ì¼¡¢KOHºÍMnO2£®
£¨5£©Èô²»¿¼ÂÇÎïÖÊÑ­»·ÓëÖƱ¸¹ý³ÌÖеÄËðʧ£¬Ôò1mol MnO2¿ÉÖƵÃ$\frac{2}{3}$»ò0.67mol KMnO4£®
£¨6£©²Ù×÷¢ñµÄÃû³ÆÊǹýÂË£»²Ù×÷¢òÊDzÉÓÃÕô·¢½á¾§£¨Ìî²Ù×÷²½Ö裩¡¢³ÃÈȹýÂ˵õ½KMnO4´Ö¾§Ì壮
£¨7£©Ò»¶¨Ìå»ýµÄKMnO4ÈÜҺǡºÃÄÜÑõ»¯Ò»¶¨ÖÊÁ¿µÄKHC2O4•H2C2O4•2H2O£®ÈôÓÃc mol/LµÄNaOHÈÜÒºÖкÍÏàͬÖÊÁ¿µÄKHC2O4•H2C2O4•2H2O£¬ËùÐèNaOHÈÜÒºµÄÌå»ýÇ¡ºÃΪKMnO4ÈÜÒºµÄa±¶£¬ÔòKMnO4ÈÜÒºµÄŨ¶ÈΪ$\frac{4ac}{15}$ £¨mol•L-1£©£¨Ìáʾ£º¢ÙH2C2O4ÊǶþÔªÈõËá¢Ú10[KHC2O4•H2C2O4]+8KMnO4+17H2SO4=8MnSO4+9K2SO4+40CO2¡ü+32H2O  £©
3£®Ä³»¯Ñ§ÐËȤС×éÓú¬ÓÐÂÁ¡¢Ìú¡¢Í­µÄºÏ½ðÖÆÈ¡´¿¾»µÄÂÈ»¯ÂÁÈÜÒº¡¢ÂÌ·¯¾§ÌåºÍµ¨·¯¾§Ì壬ÒÔ̽Ë÷¹¤Òµ·ÏÁϵÄÔÙÀûÓã®ÆäʵÑé·½°¸ÈçÏ£º

ÊԻشðÏÂÁÐÎÊÌ⣺
£¨1£©¹ýÂËÓõÄÆ÷²ÄÒÑÓУºÂËÖ½¡¢Ìú¼Ų̈¡¢ÌúȦºÍÉÕ±­£¬»¹Òª²¹³äµÄ²£Á§ÒÇÆ÷ÊÇ£º²£Á§°ô¡¢Â©¶·£®
£¨2£©Ð´³öÏÂÁзûºÅ¶ÔÓ¦ÎïÖʵĻ¯Ñ§Ê½£ºANaAlO2CCO2DAl£¨OH£©3EFeSO4
£¨3£©·´Ó¦¢ÙµÄÀë×Ó·½³ÌʽÊÇ2Al+2OH-+2H2O=2AlO2-+3H2¡ü£»·´Ó¦¢ÚµÄÀë×Ó·½³ÌʽÊÇFe+2H+=Fe2++H2¡ü£»·´Ó¦¢ÛµÄÀë×Ó·½³ÌʽÊÇAl£¨OH£©3+3H+=Al3++3H2O
;¾¶¢òµÄ·½³ÌʽÊÇNaAlO2+CO2+2H2O=Al£¨OH£©3¡ý+NaHCO3£®
£¨4£©ÓÉÂËÒºAÖƵÃAlCl3ÈÜÒºÓÐ;¾¶¢ñºÍ¢òÁ½Ìõ£¬ÄãÈÏΪºÏÀíµÄÊÇ;¾¶¢ò£¬ÀíÓÉÊÇÒòΪÂËÒºAÊÇNaAlO2ºÍNaOHÈÜÒº£¬°´Í¾¾¶IÖ±½ÓÏòAÖмÓÈëÑÎËáµÃµ½µÄAlCl3ÈÜÒºÖк¬ÓдóÁ¿µÄNaClÔÓÖÊ£¬°´Í¾¾¶¢ò£¬Í¨ÈëCO2ÆøÌ壬µÃAl£¨OH£©3³Áµí£¬½«Al£¨OH£©3ÈܽâÓÚÑÎËáÖеõ½µÄÊÇ´¿¾»µÄAlCl3ÈÜÒº£®ËùÒÔ;¾¶¢ò¸üºÏÀí£®
£¨5£©ÈçºÎ¼ìÑéEÖеĽðÊôÑôÀë×Ó£¿È¡ÉÙÐíÈÜÒºEÓëÊÔ¹ÜÖУ¬µÎ¼ÓKSCNÈÜÒº£¬ÈÜÒº²»±äºìÉ«£¬ÔٵμÓÐÂÖÆÂÈË®£¬ÈÜÒº±äΪºìÉ«£¬ËµÃ÷º¬ÓÐFe2+£¬Éæ¼°µÄÀë×Ó·½³Ìʽ£º2Fe2++Cl2=2Fe3++2Cl-£¬Fe3++3SCN-?Fe£¨SCN£©3£®

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

¾«Ó¢¼Ò½ÌÍø