10£®ÒԸߴ¿H2ΪȼÁϵÄÖÊ×Ó½»»»Ä¤È¼Áϵç³Ø¾ßÓÐÄÜÁ¿Ð§Âʸߡ¢ÎÞÎÛȾµÈÓŵ㣬µ«È¼ÁÏÖÐÈô»ìÓÐCO½«ÏÔÖøËõ¶Ìµç³ØÊÙÃü£®ÒÔ¼×´¼ÎªÔ­ÁÏÖÆÈ¡¸ß´¿H2ÊÇÖØÒªÑо¿·½Ïò£®
£¨1£©¼×´¼ÔÚ´ß»¯¼Á×÷ÓÃÏÂÁѽâ¿ÉµÃµ½H2£¬ÇâÔªËØÀûÓÃÂÊ´ï100%£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽΪCH3OH $\frac{\underline{\;¸ßÎÂ\;}}{\;}$CO+2H2£¬¸Ã·½·¨µÄȱµãÊDzúÎïH2ÖÐCOº¬Á¿×î¸ß£®
£¨2£©¼×´¼Ë®ÕôÆøÖØÕûÖÆÇâÖ÷Òª·¢ÉúÒÔÏÂÁ½¸ö·´Ó¦£º
Ö÷·´Ó¦£ºCH3OH£¨g£©+H2O£¨g£©¨TCO2£¨g£©+3H2£¨g£©¡÷H=+49kJ•mol-1
¸±·´Ó¦£ºH2£¨g£©+CO2£¨g£©¨TCO£¨g£©+H2O£¨g£©¡÷H=+41kJ•mol-1
¢Ù¼ÈÄܼӿ췴ӦËÙÂÊÓÖÄÜÌá¸ßCH3OHƽºâת»¯ÂʵÄÒ»ÖÖ´ëÊ©ÊÇÉý¸ßζȣ®
¢Ú·ÖÎöÊʵ±Ôö´óË®´¼±È£¨nH2O£ºnCH3OH£©¶Ô¼×´¼Ë®ÕôÆøÖØÕûÖÆÇâµÄºÃ´¦Ìá¸ß¼×´¼µÄÀûÓÃÂÊ£¬ÓÐÀûÓÚÒÖÖÆCOµÄÉú³É£®
¢ÛijζÈÏ£¬½«nH2O£ºnCH3OH=1£º1µÄÔ­ÁÏÆø³äÈëºãÈÝÃܱÕÈÝÆ÷ÖУ¬³õʼѹǿΪp1£¬·´Ó¦´ïµ½Æ½ºâʱ×ÜѹǿΪp2£¬Ôòƽºâʱ¼×´¼µÄת»¯ÂÊΪ£¨$\frac{{P}_{2}}{{P}_{1}}$-1£©¡Á100%£®£¨ºöÂÔ¸±·´Ó¦£©
¢Ü¹¤ÒµÉú²úÖУ¬µ¥Î»Ê±¼äÄÚ£¬µ¥Î»Ìå»ýµÄ´ß»¯¼ÁËù´¦ÀíµÄÆøÌåÌå»ý½Ð×ö¿ÕËÙ[µ¥Î»Îªm3/£¨m3´ß»¯¼Á•h£©£¬¼ò»¯Îªh-1]£®Ò»¶¨Ìõ¼þÏ£¬¼×´¼µÄת»¯ÂÊÓëζȡ¢¿ÕËٵĹØϵÈçͼ£®¿ÕËÙÔ½´ó£¬¼×´¼µÄת»¯ÂÊÊÜζÈÓ°ÏìÔ½´ó£®ÆäËûÌõ¼þÏàͬ£¬±È½Ï230¡æʱ1200h-1ºÍ300h-1Á½ÖÖ¿ÕËÙÏÂÏàͬʱ¼äÄÚH2µÄ²úÁ¿£¬Ç°ÕßԼΪºóÕßµÄ3.2±¶£®£¨ºöÂÔ¸±·´Ó¦£¬±£Áô2λÓÐЧÊý×Ö£©
£¨3£©¼×´¼Ë®ÕôÆøÖØÕûÖÆÇâÏûºÄ´óÁ¿ÈÈÄÜ£¬¿Æѧ¼ÒÌá³öÔÚÔ­ÁÏÆøÖвôÈëÒ»¶¨Á¿ÑõÆø£¬ÀíÂÛÉÏ¿ÉʵÏÖ¼×´¼Ë®ÕôÆø×ÔÈÈÖØÕûÖÆÇ⣮
ÒÑÖª£ºCH3OH£¨g£©+$\frac{1}{2}$O2£¨g£©¨TCO2£¨g£©+2H2£¨g£©¡÷H=-193kJ•mol-1
Ôò5CH3OH£¨g£©+4H2O£¨g£©+$\frac{1}{2}$O2£¨g£©¨T5CO2£¨g£©+14H2£¨g£©µÄ¡÷H=+3 kJ•mol-1£®

·ÖÎö £¨1£©¼×´¼ÔÚ´ß»¯¼Á×÷ÓÃÏÂÁѽâ¿ÉµÃµ½H2£¬ÇâÔªËØÀûÓÃÂÊ´ï100%£¬·´Ó¦Éú³ÉCOÓëÇâÆø£»µÃµ½»ìºÏÆøÌåÖÐCO²»Ò×·ÖÀë³ýÈ¥£»
£¨2£©¢ÙÕý·´Ó¦ÎªÆøÌåÌå»ýÔö´óµÄÎüÈÈ·´Ó¦£¬Éý¸ßζÈÓÐÀûÓÚƽºâÕýÏòÒƶ¯£¬ÇÒÔö´ó·´Ó¦ËÙÂÊ£»
¢ÚÔö´óË®´¼±È£¬¼×´¼µÄÀûÓÃÂÊÔö´ó£»
¢ÛºãκãÈÝÏ£¬ÆøÌåµÄѹǿ֮±ÈµÈÓÚÆäÎïÖʵÄÁ¿Ö®±È£¬¼ÆËãƽºâʱÆøÌåµÄ×ÜÎïÖʵÄÁ¿£¬ÔÙÀûÓòîÁ¿·¨¼ÆËã²Î¼Ó·´Ó¦¼×´¼µÄÎïÖʵÄÁ¿£¬½ø¶ø¼ÆËã¼×´¼µÄת»¯ÂÊ£»
¢ÜÓÉͼ¿ÉÖª£¬¿ÕËÙÔ½´ó£¬¼×´¼µÄת»¯ÂÊÊÜζÈÓ°ÏìÔ½´ó£»
ÓÉͼ¿ÉÖª£¬230¡æʱ1200h-1ºÍ300h-1Á½ÖÖ¿ÕËÙϼ״¼µÄת»¯ÂÊ·Ö±ðΪ75%¡¢95%£¬ÇâÆøµÄ²úÁ¿Ö®±ÈµÈÓڲμӷ´Ó¦¼×´¼µÄÌå»ýÖ®±È£»
£¨3£©ÒÑÖª£º¢Ù£®CH3OH£¨g£©+H2O£¨g£©?CO2£¨g£©+3H2£¨g£©£©¡÷H=+49 kJ•mol-1
¢Ú£®CH3OH£¨g£©+$\frac{1}{2}$O2£¨g£©?CO2£¨g£©+2H2£¨g£©¡÷H=-193kJ•mol-1
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ù¡Á4+¢Ú¿ÉµÃ£º5CH3OH£¨g£©+4H2O£¨g£©+$\frac{1}{2}$O2£¨g£©?5CO2£¨g£©+14H2£¨g£©£®

½â´ð ½â£º£¨1£©¼×´¼ÔÚ´ß»¯¼Á×÷ÓÃÏÂÁѽâ¿ÉµÃµ½H2£¬ÇâÔªËØÀûÓÃÂÊ´ï100%£¬·´Ó¦Éú³ÉCOÓëÇâÆø£¬·´Ó¦·½³ÌʽΪ£ºCH3OH $\frac{\underline{\;¸ßÎÂ\;}}{\;}$CO+2H2£»È±µãÊǵõ½»ìºÏÆøÌåÖÐCO²»Ò×·ÖÀë³ýÈ¥£¬²úÎïH2ÖÐCOº¬Á¿×î¸ß£¬
¹Ê´ð°¸Îª£ºCH3OH $\frac{\underline{\;¸ßÎÂ\;}}{\;}$CO+2H2£»²úÎïH2ÖÐCOº¬Á¿×î¸ß£»
£¨2£©¢ÙÕý·´Ó¦ÎªÆøÌåÌå»ýÔö´óµÄÎüÈÈ·´Ó¦£¬Éý¸ßζÈÔö´ó·´Ó¦ËÙÂÊ£¬ÓÐÀûÓÚƽºâÕýÏòÒƶ¯£¬¼×´¼µÄת»¯ÂÊÔö´ó£¬¹Ê´ð°¸Îª£ºÉý¸ßζȣ»
¢ÚÔö´óË®´¼±È£¨nH2O£ºnCH3OH£©£¬ÓÐÀûÓÚ·´Ó¦ÕýÏò½øÐУ¬Ìá¸ß¼×´¼µÄÀûÓÃÂÊ£¬ÓÐÀûÓÚÒÖÖÆCOµÄÉú³É£¬¹Ê´ð°¸Îª£ºÌá¸ß¼×´¼µÄÀûÓÃÂÊ£¬ÓÐÀûÓÚÒÖÖÆCOµÄÉú³É£»
¢ÛÉèÆðʼ n£¨H2O£©=n£¨CH3OH£©=1mol£¬ºãκãÈÝÏ£¬ÆøÌåµÄѹǿ֮±ÈµÈÓÚÆäÎïÖʵÄÁ¿Ö®±È£¬Æ½ºâʱÆøÌåµÄ×ÜÎïÖʵÄÁ¿2mol¡Á$\frac{{P}_{2}}{{P}_{1}}$£¬
CH3OH£¨g£©+H2O£¨g£©?CO2£¨g£©+3H2£¨g£©ÎïÖʵÄÁ¿Ôö´ó
1                                         2
£¨ $\frac{{P}_{2}}{{P}_{1}}$-1£©mol                       2mol¡Á$\frac{{P}_{2}}{{P}_{1}}$-2mol=2£¨$\frac{{P}_{2}}{{P}_{1}}$-1£©mol
¹Ê¼×´¼µÄת»¯ÂÊΪ$\frac{£¨\frac{{P}_{2}}{{P}_{1}}-1£©}{1mol}$¡Á100%=£¨$\frac{{P}_{2}}{{P}_{1}}$-1£©¡Á100%£¬
¹Ê´ð°¸Îª£º£¨$\frac{{P}_{2}}{{P}_{1}}$-1£©¡Á100%£»
¢ÜÓÉͼ¿ÉÖª£¬¿ÕËÙÔ½´ó£¬¼×´¼µÄת»¯ÂÊÊÜζÈÓ°ÏìÔ½´ó£»
ÓÉͼ¿ÉÖª£¬230¡æʱ1200h-1ºÍ300h-1Á½ÖÖ¿ÕËÙϼ״¼µÄת»¯ÂÊ·Ö±ðΪ75%¡¢95%£¬ÇâÆøµÄ²úÁ¿Ö®±ÈµÈÓڲμӷ´Ó¦¼×´¼µÄÌå»ýÖ®±È£¬¹ÊÏàͬʱ¼äÄÚH2µÄ²úÁ¿£¬Ç°ÕßԼΪºóÕßµÄ$\frac{1200¡Á75%}{300¡Á95%}$¡Ö3.2±¶£¬
¹Ê´ð°¸Îª£º´ó£»3.2£»
£¨3£©ÒÑÖª£º¢Ù£®CH3OH£¨g£©+H2O£¨g£©?CO2£¨g£©+3H2£¨g£©£©¡÷H=+49 kJ•mol-1
¢Ú£®CH3OH£¨g£©+$\frac{1}{2}$O2£¨g£©?CO2£¨g£©+2H2£¨g£©¡÷H=-193kJ•mol-1
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢Ù¡Á4+¢Ú¿ÉµÃ£º5CH3OH£¨g£©+4H2O£¨g£©+$\frac{1}{2}$O2£¨g£©?5CO2£¨g£©+14H2£¨g£©£¬Ôò£º¡÷H=4¡Á49 kJ•mol-1-193kJ•mol-1=+3kJ•mol-1£¬
¹Ê´ð°¸Îª£º+3£®

µãÆÀ ±¾Ì⿼²é»¯Ñ§Æ½ºâ¼ÆËãÓëÓ°ÏìÒòËØ¡¢·´Ó¦ÈȼÆËãµÈ£¬²àÖØ¿¼²éѧÉú·ÖÎö¼ÆËãÄÜÁ¦£¬×¢Òâ¸Ç˹¶¨ÂÉÔÚ·´Ó¦ÈȼÆËãÖÐÓ¦Óã®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

7£®ÓÃËáʽµÎ¶¨¹Ü׼ȷÒÆÈ¡10.00mLijδ֪Ũ¶ÈµÄÑÎËáÈÜÓÚÒ»½à¾»µÄ׶ÐÎÆ¿ÖУ¬È»ºóÓÃ0.20mol•L -1µÄÇâÑõ»¯ÄÆÈÜÒº£¨Ö¸Ê¾¼ÁΪ·Ó̪£©£®µÎ¶¨½á¹ûÈçÏ£º
NaOHÆðʼ¶ÁÊýNaOHÖÕµã¶ÁÊý
µÚÒ»´Î      0.50mL    18.60mL
µÚ¶þ´Î      0.70mL    19.00mL
£¨1£©¸ù¾ÝÒÔÉÏÊý¾Ý¿ÉÒÔ¼ÆËã³öÑÎËáµÄÎïÖʵÄÁ¿Å¨¶ÈΪ0.364mol•L-1£®
£¨2£©´ïµ½µÎ¶¨ÖÕµãµÄ±êÖ¾ÊÇÎÞÉ«±äΪºìÉ«£¬ÇÒ30sÄÚ²»±äÉ«£®
£¨3£©ÒÔϲÙ×÷Ôì³É²â¶¨½á¹ûÆ«¸ßµÄÔ­Òò¿ÉÄÜÊÇAD£®
A£®Î´Óñê×¼ÒºÈóÏ´¼îʽµÎ¶¨¹Ü
B£®µÎ¶¨ÖÕµã¶ÁÊýʱ£¬¸©Êӵζ¨¹ÜµÄ¿Ì¶È£¬ÆäËü²Ù×÷¾ùÕýÈ·
C£®Ê¢×°Î´ÖªÒºµÄ׶ÐÎÆ¿ÓÃÕôÁóˮϴ¹ý£¬Î´Óôý²âÒºÈóÏ´
D£®µÎ¶¨µ½ÖÕµã¶ÁÊýʱ·¢Ïֵζ¨¹Ü¼â×ì´¦Ðü¹ÒÒ»µÎÈÜÒº£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

8£®Ð´³öÏÂÁз´Ó¦µÄÀë×Ó·½³Ìʽ£®
£¨1£©Ì¼Ëá±µ£¨BaCO3£©ºÍÑÎËá·´Ó¦£ºBaCO3+2H+=Ba2++H2O+CO2¡ü£®
£¨2£©H2SO4ÈÜÒºÓëBa£¨OH£©2ÈÜÒº·´Ó¦£º2H++SO42-+Ba2++2OH-=BaSO4¡ý+2H2O£®
£¨3£©FeÓëÏ¡ÁòËáÈÜÒº·´Ó¦£ºFe+2H+¨TFe2++H2¡ü£®
£¨4£©Ñõ»¯Ã¾£¨MgO£©ÓëÏ¡ÏõËᣨHNO3£©·´Ó¦£ºMgO+2H+¨TMg2++H2O£®
£¨5£©Ð´³öÒ»¸ö·ûºÏÀë×Ó·´Ó¦µÄ»¯Ñ§·½³Ìʽ£ºCO32-+Ca2+=Ca CO3¡ýNa2CO3+CaCl2=CaCO3¡ý+2NaCl
£¨6£©Ð´³öNaHCO3µÄµçÀë·½³ÌʽNaHCO3¨TNa++HCO3-£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÌî¿ÕÌâ

5£®Ð´³öÏÂÁÐÀë×Ó·½³ÌʽÄܱíʾµÄÒ»¸ö»¯Ñ§·½³Ìʽ£º
£¨1£©OH-+H+¨TH2ONaOH+HCl=NaCl+H2O
£¨2£©Cu2++2OH-¨TCu£¨OH£©2CuSO4+2NaOH¨TCu£¨OH£©2¡ý+Na2SO4£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

5£®ÀûÓÃËùѧ»¯Ñ§ÖªÊ¶»Ø´ðÎÊÌâ

¢ñ¡¢¼×´¼ÊÇÖØÒªµÄ»¯¹¤Ô­ÁÏ£¬ÓֿɳÆΪȼÁÏ£®¹¤ÒµÉÏÀûÓúϳÉÆø£¨Ö÷Òª³É·ÖΪCO¡¢CO2ºÍH2£©ÔÚ´ß»¯¼ÁµÄ×÷ÓÃϺϳɼ״¼£¬·¢ÉúµÄÖ÷·´Ó¦ÈçÏ£º
¢ÙCO£¨g£©+2H2£¨g£©¨TCH3OH£¨g£©¡÷H=£¿
¢ÚCO2£¨g£©+3H2£¨g£©¨TCH3OH£¨g£©+H2O£¨g£©¡÷H=-58kJ/mol
¢ÛCO2£¨g£©+H2£¨g£©¨TCO£¨g£©+H2O£¨g£©¡÷H=+41kJ/mol
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑÖª·´Ó¦¢ÙÖеÄÏà¹ØµÄ»¯Ñ§¼ü¼üÄÜÊý¾ÝÈçÏ£º
»¯Ñ§¼üH-HC-OC
$\frac{\underline{\;¡û\;}}{\;}$O
H-OC-H
E/£¨kJ•mol-1£©4363431076465x
Ôòx=413kJ/mol£®
£¨2£©ÈôT¡æʱ½«6molCO2ºÍ8molH2³äÈë2LÃܱÕÈÝÆ÷Öз¢Éú·´Ó¦¢Ú£¬²âµÃH2µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯Èçͼ1ÖÐ״̬¢ñ£¨Í¼1ÖÐʵÏߣ©Ëùʾ£®Í¼1ÖÐÊý¾ÝA£¨1£¬6£©´ú±íÔÚ1minʱH2µÄÎïÖʵÄÁ¿ÊÇ6mol£®
¢ÙT¡æʱ£¬×´Ì¬¢ñÌõ¼þÏ£¬Æ½ºâ³£ÊýK=0.5£»
¢ÚÆäËûÌõ¼þ²»±ä£¬½ö¸Ä±äζÈʱ£¬²âµÃH2µÄÎïÖʵÄÁ¿Ëæʱ¼ä±ä»¯ÈçͼÖÐ״̬¢óËùʾ£¬Ôò״̬¢ó¶ÔÓ¦µÄζȣ¾£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©T¡æ£»
¢ÛÒ»¶¨Î¶ÈÏ£¬´Ë·´Ó¦ÔÚºãÈÝÈÝÆ÷ÖнøÐУ¬ÄÜÅжϸ÷´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬ÒÀ¾ÝµÄÊÇbc£®
a£® 2¸öC=O¶ÏÁѵÄͬʱÓÐ2¸öH-OÉú³É     b£®ÈÝÆ÷ÖлìºÏÆøÌåƽ¾ùĦ¶ûÖÊÁ¿²»±ä
c£®vÄ棨H2£©=3vÕý£¨CH3OH£©               d£®¼×´¼ºÍË®ÕôÆøµÄÌå»ý±È±£³Ö²»±ä
¢ò¡¢¶þÑõ»¯Ì¼µÄ»ØÊÕÀûÓÃÊÇ»·±£ÁìÓòÑо¿µÄÈȵã¿ÎÌ⣮
£¨1£©CO2¾­¹ý´ß»¯Ç⻯ºÏ³ÉµÍ̼ϩÌþ£®ÔÚ2LºãÈÝÃܱÕÈÝÆ÷ÖгäÈë2moI CO2ºÍnmol H2£¬ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º2C02£¨g£©+6H2£¨g£©?CH2=CH2£¨g£©+4H20£¨g£©£¬¡÷H=-128kJ/mol£®CO2µÄת»¯ÂÊÓëζȡ¢Í¶ÁϱÈ[X=$\frac{n£¨{H}_{2}£©}{n£¨C{O}_{2}£©}$]µÄ¹ØϵÈçͼ2Ëùʾ£®
¢ÙX2£¾  X1£¨Ìî¡°£¾¡±¡¢¡°£¼¡±»ò¡°=¡±£©
¢ÚÔÚ500Kʱ£¬ÈôBµãµÄͶÁϱÈΪ3.5£¬ÇÒ´Ó·´Ó¦¿ªÊ¼µ½BµãÐèÒª10min£¬Ôòv£¨H2£©=0.225mol/£¨L£®min£©£®
£¨2£©ÒÔÏ¡ÁòËáΪµç½âÖÊÈÜÒº£¬ÀûÓÃÌ«ÑôÄܽ«CO2ת»¯ÎªµÍ̼ϩÌþ£¬¹¤×÷Ô­Àíͼ3ÈçÏ£¬Ôò×ó²à²úÉúÒÒÏ©µÄµç¼«·´Ó¦Ê½Îª2CO2+12e-+12H+=CH2=CH2+4H2O£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

15£®2SO2£¨g£©+2O2£¨g£©?2SO3£¨g£©ÊÇÉú²úÁòËáµÄÖ÷Òª·´Ó¦Ö®Ò»£®Ï±íÊÇÔ­ÁÏÆø°´V£¨SO2£©£ºV£¨O2£©£ºV£¨N2£©=7£º11£º82ͶÁÏ£¬ÔÚ1.01¡Á105Paʱ£¬²»Í¬Î¶ÈÏÂSO2µÄƽºâת»¯ÂÊ£®
ζÈ/400500600
SO2ת»¯ÂÊ/%99.293.573.7
£¨1£©¸Ã·´Ó¦ÊÇ·ÅÈÈ·´Ó¦£¨Ìî¡°·ÅÈÈ¡±»ò¡°ÎüÈÈ¡±£©£®
£¨2£©400¡æ£¬1.01¡Á105Paʱ£¬½«º¬10 mol SO2µÄÔ­ÁÏÆøͨÈëÒ»ÃܱÕÈÝÆ÷ÖнøÐз´Ó¦£¬Æ½ºâʱSO2µÄÎïÖʵÄÁ¿ÊÇ0.08mol£®
£¨3£©ÁòË᳧βÆø£¨Ö÷Òª³É·ÖSO2¡¢O2ºÍN2£©ÖеÍŨ¶ÈSO2µÄÎüÊÕÓкܶ෽·¨£®
¢ÙÓð±Ë®ÎüÊÕÉÏÊöβÆø£¬ÈôSO2Ó백ˮǡºÃ·´Ó¦µÃµ½¼îÐԵģ¨NH4£©2SO3ÈÜҺʱ£¬ÔòÓйظÃÈÜÒºµÄÏÂÁйØϵÕýÈ·µÄÊÇac£¨ÌîÐòºÅ£©£®
a£® c£¨NH4+£©+c£¨NH3•H2O£©=2[c£¨SO32-£©+c£¨HSO3-£©+c£¨H2SO3£©]
b£® c£¨NH4+£©+c£¨H+£©=c£¨SO32-£©+c£¨HSO3-£©+c£¨OH-£©
c£® c£¨NH4+£©£¾c£¨SO32-£©£¾c£¨OH-£©£¾c£¨H+£©
¢ÚÓà MnO2ÓëË®µÄÐü×ÇÒºÎüÊÕÉÏÊöβÆø²¢Éú²úMnSO4£®
a£® µÃµ½MnSO4µÄ»¯Ñ§·½³ÌʽÊÇH2O+SO2=H2SO3¡¢MnO2+H2SO3=MnSO4+H2O£®
b£®¸ÃÎüÊÕ¹ý³ÌÉú³ÉMnSO4ʱ£¬ÈÜÒºµÄpH±ä»¯Ç÷ÊÆÈçͼ¼×£¬SO2ÎüÊÕÂÊÓëÈÜÒºpHµÄ¹ØϵÈçͼÒÒ£® ͼ¼×ÖÐpH±ä»¯ÊÇÒòΪÎüÊÕÖÐÓв¿·ÖSO2±»ÑõÆøÑõ»¯×ª»¯ÎªH2SO4£¬Éú³ÉH2SO4·´Ó¦µÄ»¯Ñ§·½³ÌʽÊÇ2SO2+O2+2H2O=2H2SO4£»ÓÉͼÒÒ¿ÉÖªpHµÄ½µµÍ²»ÀûÓÚSO2µÄÎüÊÕ£¨Ìî¡°ÓÐÀûÓÚ¡±»ò¡°²»ÀûÓÚ¡±£©£¬Óû¯Ñ§Æ½ºâÒƶ¯Ô­Àí½âÊÍÆäÔ­ÒòÊÇÈÜÒºÖдæÔÚSO2+H2O?H2SO3?H++HSO3-£¬µ±ÈÜÒºÖÐËáÐÔÔöÇ¿£¬Æ½ºâÏò×óÒƶ¯£¬Ê¹SO2ÆøÌå´ÓÌåϵÖÐÒݳö£®£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º½â´ðÌâ

2£®¹¤ÒµºÏ³É°±·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ£ºN2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷=-92.3kJ/mol£®Ò»¶¨Î¶ÈÏ£¬ÏòÌå»ýΪ2LµÄÃܱÕÈÝÆ÷ÖмÓÈë1mol N2ºÍ3mol H2£¬¾­2minºó´ïµ½Æ½ºâ£¬Æ½ºâʱ²âµÃNH3µÄŨ¶ÈΪ0.5mol/L£®
£¨1£©2min ÄÚH2µÄ·´Ó¦ËÙÂÊv£¨H2£©=0.375mol/£¨L•min£©£»
£¨2£©³ä·Ö·´Ó¦²¢´ïµ½Æ½ºâʱ£¬·Å³öµÄÈÈÁ¿Ð¡ÓÚ92.3kJ£¨Ìî¡°´óÓÚ¡±¡¢¡°Ð¡ÓÚ¡±»ò¡°µÈÓÚ¡±£©£®
Ô­ÒòÊÇ¿ÉÄæ·´Ó¦×îÖÕΪ¶¯Ì¬Æ½ºâ£¬Ê¼ÖÕ´ï²»µ½×î´ó·Å³öÈÈÁ¿£®
£¨3£©ÏÂÁÐ˵·¨¿ÉÖ¤Ã÷·´Ó¦´ïµ½Æ½ºâ״̬µÄC¡¢D£®
A£®µ¥Î»Ê±¼äÄÚ£¬¶Ï¿ª1mol N¡ÔN£¬Í¬Ê±¶Ï¿ª3mol H-H
B£®µ¥Î»Ê±¼äÄÚ£¬ÐγÉ1mol N¡ÔN£¬Í¬Ê±ÐγÉ3mol N-H
C£®µ¥Î»Ê±¼äÄÚ£¬¶Ï¿ª1mol N¡ÔN£¬Í¬Ê±¶Ï¿ª6mol N-H
D£®µ¥Î»Ê±¼äÄÚ£¬ÐγÉ1mol N¡ÔN£¬Í¬Ê±¶Ï¿ª3mol H-H
£¨4£©°±µÄÒ»¸öÖØÒªÓÃ;ÊÇÓÃÓÚÖƱ¸»ð¼ý·¢ÉäÔ­ÁÏN2H4£¨ë£©£¬ÒÑÖª£º»ð¼ý·¢ÉäµÄÔ­ÀíÊÇN2H4£¨ë£©ÔÚNO2ÖÐȼÉÕ£¬Éú³ÉN2¡¢Ë®ÕôÆø£®¸ù¾ÝÈçÏ·´Ó¦£º
N2£¨g£©+2O2£¨g£©=2NO2£¨g£©¡÷H1=+67.7kJ/mol
N2H4£¨g£©+O2£¨g£©=N2£¨g£©+2H2O£¨g£©¡÷H2=-534.0kJ/mol
д³öÔÚÏàͬ״̬Ï£¬·¢Éä»ð¼ý·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ2N2H4£¨g£©+N2O4£¨g£©=3N2£¨g£©+4H2O£¨g£©¡÷H=-947.6kJ•mol-1£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

19£®ÏÂÃæÓйر仯¹ý³Ì£¬²»ÊôÓÚÑõ»¯»¹Ô­·´Ó¦µÄÊÇ£¨¡¡¡¡£©
A£®ÉÕ²ËÓùýµÄÌú¹ø£¬¾­·ÅÖó£³öÏÖºì×ØÉ«°ß¼£
B£®ÓÃúÆøÔîȼÉÕÌìÈ»ÆøΪ³´²ËÌṩÈÈÁ¿
C£®Å£Ä̾ÃÖÿÕÆøÖбäÖʸ¯°Ü
D£®Ïò·ÐË®ÖеÎÈëFeCl3±¥ºÍÈÜÒº£¬Êʵ±¼ÓÈÈ£¬ÖƱ¸½ºÌå

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£ºÑ¡ÔñÌâ

20£®³£ÎÂÏ£¬ÏÂÁи÷×éÀë×ÓÔÚÖ¸¶¨ÈÜÒºÖÐÄÜ´óÁ¿¹²´æµÄÊÇ£¨¡¡¡¡£©
A£®c£¨Fe3+£©=0.1 mol•L-1µÄÈÜÒºÖУºK+¡¢ClO-¡¢SO42-¡¢SCN-
B£®$\frac{c£¨{H}^{+}£©}{c£¨O{H}^{-}£©}$=1012µÄÈÜÒºÖУºNH4+¡¢Al3+¡¢NO3-¡¢Cl-
C£®ÓÉË®µçÀë²úÉúµÄc£¨OH-£©=1¡Á10-13mol/LµÄÈÜÒºÖУºCa2+¡¢K+¡¢Cl-¡¢HCO3-
D£®pH=1µÄÈÜÒºÖУºFe2+¡¢NO3-¡¢SO42-¡¢Na+

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸