ÌâÄ¿ÄÚÈÝ

10£®Ñо¿µªÑõ»¯ÎïÓëÐü¸¡ÔÚ´óÆøÖк£ÑÎÁ£×ÓµÄÏ໥×÷ÓÃʱ£¬Éæ¼°ÈçÏ·´Ó¦£º
2NO2£¨g£©+NaCl£¨s£©?NaNO3£¨s£©+ClNO£¨g£©¡¡ K1¡÷H£¼0£¨I£©
2NO£¨g£©+Cl2£¨g£©?2ClNO£¨g£©¡¡¡¡K2¡÷H£¼0¡¡£¨II£©
£¨1£©4NO2£¨g£©+2NaCl£¨s£©?2NaNO3£¨s£©+2NO£¨g£©+Cl2£¨g£©µÄƽºâ³£ÊýK=$\frac{{{K}^{2}}_{1}}{{K}_{2}}$£¨ÓÃK1¡¢K2±íʾ£©£®
£¨2£©ÎªÑо¿²»Í¬Ìõ¼þ¶Ô·´Ó¦£¨II£©µÄÓ°Ï죬ÔÚºãÎÂÌõ¼þÏ£¬Ïò 2LºãÈÝÃܱÕÈÝÆ÷ÖмÓÈë0.2mol NOºÍ0.1mol Cl2£¬10minʱ·´Ó¦£¨II£©´ïµ½Æ½ºâ£®²âµÃ10minÄÚV£¨ClNO£©=7.5¡Á10-3mol•L-1•min-1£¬Ôòƽºâºón£¨Cl2£©=0.025mol£¬NOµÄת»¯ÂʧÑ1=75%£®ÆäËüÌõ¼þ±£³Ö²»±ä£¬·´Ó¦£¨II£©ÔÚºãѹÌõ¼þϽøÐУ¬Æ½ºâʱNOµÄת»¯ÂʧÑ2£¾§Ñ1£¨Ìî¡°£¾¡±¡°£¼¡±»ò¡°=¡±£©£¬Æ½ºâ³£ÊýK2²»±ä£¨Ìî¡°Ôö´ó¡±¡°¼õС¡±»ò¡°²»±ä¡±£©£®
ÈôҪʹK2¼õС£¬¿É²ÉÓõĴëÊ©ÊÇÉý¸ßζȣ®
£¨3£©ÊµÑéÊÒ¿ÉÓÃNaOHÈÜÒºÎüÊÕNO2£¬·´Ó¦Îª2NO2+2NaOH=NaNO3+NaNO2+H2O£®º¬0.2mol NaOHµÄË®ÈÜÒºÓë0.2mol NO2Ç¡ºÃÍêÈ«·´Ó¦µÃ 1LÈÜÒºA£¬ÈÜÒºBΪ0.1mol•L?1µÄCH3COONaÈÜÒº£¬ÔòÁ½ÈÜÒºÖÐc£¨NO3?£©¡¢c£¨NO2-£©ºÍc£¨CH3COO?£©ÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨NO3?£©£¾c£¨NO2-£©£¾c£¨CH3COO?£©£®£¨ÒÑÖªHNO2µÄµçÀë³£ÊýKa=7.1¡Á10-4mol•L?1£¬CH3COOHµÄµçÀë³£ÊýKa=1.7¡Á10-5mol•L?1£¬¿ÉʹÈÜÒºAºÍÈÜÒºBµÄpHÏàµÈµÄ·½·¨ÊÇBC£®
¡¡¡¡A£®ÏòÈÜÒºAÖмÓÊÊÁ¿Ë®¡¡¡¡¡¡¡¡¡¡¡¡ B£®ÏòÈÜÒºAÖмÓÊÊÁ¿NaOH
C£®ÏòÈÜÒºBÖмÓÊÊÁ¿Ë®¡¡¡¡¡¡¡¡¡¡¡¡ D..ÏòÈÜÒºBÖмÓÊÊÁ¿NaOH
£¨4£©ÔÚ25¡æÏ£¬½«a mol•L-1µÄ°±Ë®Óë0.01mol•L-1µÄÑÎËáµÈÌå»ý»ìºÏ£¬·´Ó¦Ê±ÈÜÒºÖÐc£¨NH${\;}_{4}^{+}$£©=c£¨Cl-£©£®ÔòÈÜÒºÏÔÖУ¨Ìî¡°Ëᡱ¡°¼î¡±»ò¡°ÖС±£©ÐÔ£»Óú¬aµÄ´úÊýʽ±íʾNH3•H2OµÄµçÀë³£ÊýKb=$\frac{1{0}^{-9}}{a-0.01}$£®
£¨5£©Ë®ÈÜÒºÖеÄÐÐΪÊÇÖÐѧ»¯Ñ§µÄÖØÒªÄÚÈÝ£®ÒÑÖªÏÂÁÐÎïÖʵĵçÀë³£ÊýÖµ£º
HClO£ºKa=3¡Á10-8HCN£ºKa=4.9¡Á10-10H2CO3£ºKa1=4.3¡Á10-7Ka2=5.6¡Á10-11
84Ïû¶¾Òº£¨ÓÐЧ³É·ÝΪNaClO£©ÖÐͨÈëÉÙÁ¿µÄCO2£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪNaClO+CO2+H2O=NaHCO3+HClO£®

·ÖÎö £¨1£©2NO2£¨g£©+NaCl£¨s£©?NaNO3£¨s£©+ClNO£¨g£© K1¡÷H£¼0 £¨I£©
2NO£¨g£©+Cl2£¨g£©?2ClNO£¨g£© K2¡÷H£¼0 £¨II£©
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢ñ¡Á2-¢ò¿ÉµÃ£º4NO2£¨g£©+2NaCl£¨s£©?2NaNO3£¨s£©+2NO£¨g£©+Cl2£¨g£©£¬Ôò¸Ã·´Ó¦Æ½ºâ³£ÊýΪ¢ñµÄƽºâ³£Êýƽ·½Óë¢òµÄƽºâ³£ÊýµÄÉÌ£»
£¨2£©²âµÃ10minÄÚv£¨ClNO£©=7.5¡Á10-3mol•L-1•min-1£¬Ôò¡÷n£¨ClNO£©=7.5¡Á10-3mol•L-1•min-1¡Á10min¡Á2L=0.15mol£¬ÓÉ·½³Ìʽ¼ÆËã²Î¼Ó·´Ó¦NO¡¢ÂÈÆøµÄÎïÖʵÄÁ¿£¬½ø¶ø¼ÆËãÆ½ºâʱÂÈÆøµÄÎïÖʵÄÁ¿¡¢NOµÄת»¯ÂÊ£»
ÆäËüÌõ¼þ±£³Ö²»±ä£¬·´Ó¦£¨II£©ÔÚºãѹÌõ¼þϽøÐУ¬ÕýÏòÒÆ¶¯Ê±ÎïÖʵÄÁ¿¼õС£¬Ï൱ÓÚѹǿÔö´ó£»
ƽºâ³£ÊýÖ»ÊÜζÈÓ°Ï죬ζȲ»±ä£¬Æ½ºâ³£Êý²»±ä£¬Õý·´Ó¦Îª·ÅÈÈ·´Ó¦£¬½µµÍÎÂ¶ÈÆ½ºâÕýÏòÒÆ¶¯£¬Æ½ºâ³£ÊýÔö´ó£»
£¨3£©0.2mol NaOHµÄË®ÈÜÒºÓë0.2mol NO2Ç¡ºÃÍêÈ«·´Ó¦µÃ1LÈÜÒºA£¬ÓÉ2NO2+2NaOH¨TNaNO3+NaNO2+H2O£¬µÃµ½ÈÜÒºAÖÐNaNO3ÎïÖʵÄÁ¿Å¨¶ÈΪ0£®mol/L£¬NaNO2ÎïÖʵÄÁ¿Îª0.1mol/L£»
HNO2µÄµçÀë³£ÊýKa=7.1¡Á10-4mol•L-1£¬CH3COOHµÄµçÀë³£ÊýKa=1.7¡Á10-5mol•L-1£¬ËµÃ÷CH3COOHËáÐÔСÓÚHNO2µÄËáÐÔ£¬ÈÜÒºÖд×Ëá¸ùÀë×ÓË®½â³Ì¶ÈСÓÚÑÇÏõËá¸ùÀë×ÓË®½â³Ì¶È£¬ÈÜÒºB¼îÐÔ´óÓÚAÈÜÒº£»
£¨4£©ÔÚ25¡æÏ£¬Æ½ºâʱÈÜÒºÖÐc£¨NH4+£©=c£¨Cl-£©=0.005mol/L£¬¸ù¾ÝÎïÁÏÊØºãµÃc£¨NH3£®H2O£©=£¨0.5a-0.005£©mol/L£¬¸ù¾ÝµçºÉÊØºãµÃc£¨H+£©=c£¨OH-£©=10-7mol/L£¬ÈÜÒº³ÊÖÐÐÔ£¬NH3•H2OµÄµçÀë³£ÊýKb=$\frac{c£¨O{H}^{-}£©c£¨N{{H}_{4}}^{+}£©}{c£¨N{H}_{3}£®{H}_{2}O£©}$£»
£¨5£©ÓÉKa¿ÉÖªËáÐÔ£¬·¢ÉúÇ¿ËáÖÆÈ¡ÈõËáµÄ·´Ó¦£®

½â´ð ½â£º£¨1£©2NO2£¨g£©+NaCl£¨s£©?NaNO3£¨s£©+ClNO£¨g£© K1¡÷H£¼0 £¨I£©
2NO£¨g£©+Cl2£¨g£©?2ClNO£¨g£© K2¡÷H£¼0 £¨II£©
¸ù¾Ý¸Ç˹¶¨ÂÉ£¬¢ñ¡Á2-¢ò¿ÉµÃ£º4NO2£¨g£©+2NaCl£¨s£©?2NaNO3£¨s£©+2NO£¨g£©+Cl2£¨g£©£¬Ôò¸Ã·´Ó¦Æ½ºâ³£ÊýK=$\frac{{{K}^{2}}_{1}}{{K}_{2}}$£¬
¹Ê´ð°¸Îª£º$\frac{{{K}^{2}}_{1}}{{K}_{2}}$£»
£¨2£©²âµÃ10minÄÚv£¨ClNO£©=7.5¡Á10-3mol•L-1•min-1£¬Ôò¡÷n£¨ClNO£©=7.5¡Á10-3mol•L-1•min-1¡Á10min¡Á2L=0.15mol£¬
ÓÉ·½³Ìʽ¿ÉÖª£¬²Î¼Ó·´Ó¦ÂÈÆøµÄÎïÖʵÄÁ¿Îª0.15mol¡Á$\frac{1}{2}$=0.075mol£¬¹ÊƽºâʱÂÈÆøµÄÎïÖʵÄÁ¿Îª0.1mol-0.075mol=0.025mol£»
²Î¼Ó·´Ó¦NOÎïÖʵÄÁ¿Îª0.15mol£¬ÔòNOµÄת»¯ÂÊΪ$\frac{0.15mol}{0.2mol}$¡Á100%=75%£»
ÆäËüÌõ¼þ±£³Ö²»±ä£¬·´Ó¦£¨II£©ÔÚºãѹÌõ¼þϽøÐУ¬ÕýÏòÒÆ¶¯Ê±ÎïÖʵÄÁ¿¼õС£¬Ï൱ÓÚѹǿÔö´ó£¬ÕýÏò½øÐеij̶ÈÔö´ó£¬ÔòƽºâʱNOµÄת»¯ÂʧÑ2£¾§Ñ1£¬Æ½ºâ³£ÊýÖ»ÊÜζÈÓ°Ï죬ζȲ»±ä£¬Æ½ºâ³£Êý²»±ä£¬Õý·´Ó¦Îª·ÅÈÈ·´Ó¦£¬½µµÍÎÂ¶ÈÆ½ºâÕýÏòÒÆ¶¯£¬Æ½ºâ³£ÊýÔö´ó£¬
¹Ê´ð°¸Îª£º0.025£»75%£»£¾£»²»±ä£»Éý¸ßζȣ»
£¨3£©0.2mol NaOHµÄË®ÈÜÒºÓë0.2mol NO2Ç¡ºÃÍêÈ«·´Ó¦µÃ1LÈÜÒºA£¬ÓÉ2NO2+2NaOH¨TNaNO3+NaNO2+H2O£¬µÃµ½ÈÜÒºAÖÐNaNO3ÎïÖʵÄÁ¿Å¨¶ÈΪ0£®mol/L£¬NaNO2ÎïÖʵÄÁ¿Îª0.1mol/L£¬ÈÜÒºBΪ0.1mol•L-1µÄCH3COONaÈÜÒº£»
HNO2µÄµçÀë³£ÊýKa=7.1¡Á10-4mol•L-1£¬CH3COOHµÄµçÀë³£ÊýKa=1.7¡Á10-5mol•L-1£¬ËµÃ÷CH3COOHËáÐÔСÓÚHNO2µÄËáÐÔ£¬ÈÜÒºÖд×Ëá¸ùÀë×ÓË®½â³Ì¶ÈСÓÚÑÇÏõËá¸ùÀë×ÓË®½â³Ì¶È£¬ÈÜÒºB¼îÐÔ´óÓÚAÈÜÒº£¬Á½ÈÜÒºÖÐc£¨NO3-£©¡¢c£¨NO2-£©ºÍc£¨CH3COO-£©ÓÉ´óµ½Ð¡µÄ˳ÐòΪc£¨NO3-£©£¾c£¨NO2-£©£¾c£¨CH3COO-£©£¬
A£®ÈÜÒºB¼îÐÔ´óÓÚAÈÜÒº£¬ÏòÈÜÒºAÖмÓÊÊÁ¿Ë®£¬Ï¡ÊÍÈÜÒº£¬¼îÐÔ¼õÈõ£¬²»Äܵ÷½ÚÈÜÒºpHÏàµÈ£¬¹ÊA´íÎó£»
B£®ÏòÈÜÒºAÖмÓÊÊÁ¿NaOH£¬Ôö´ó¼îÐÔ£¬¿ÉÒÔµ÷½ÚÈÜÒºpHÏàµÈ£¬¹ÊBÕýÈ·£»
C£®ÏòÈÜÒºBÖмÓÊÊÁ¿Ë®£¬Ï¡ÊÍÈÜÒº¼îÐÔ¼õÈõ£¬¿ÉÒÔµ÷½ÚÈÜÒºpHÏàµÈ£¬¹ÊCÕýÈ·£»
D£®ÈÜÒºB¼îÐÔ´óÓÚAÈÜÒº£¬ÏòÈÜÒºBÖмÓÊÊÁ¿NaOH£¬ÈÜÒºpH¸ü´ó£¬¹ÊD´íÎó£»
¹Ê´ð°¸Îª£ºc£¨NO3-£©£¾c£¨NO2-£©£¾c£¨CH3COO-£©£»BC£»
£¨4£©ÔÚ25¡æÏ£¬Æ½ºâʱÈÜÒºÖÐc£¨NH4+£©=c£¨Cl-£©=0.005mol/L£¬¸ù¾ÝÎïÁÏÊØºãµÃc£¨NH3£®H2O£©=£¨0.5a-0.005£©mol/L£¬¸ù¾ÝµçºÉÊØºãµÃc£¨H+£©=c£¨OH-£©=10-7mol/L£¬ÈÜÒº³ÊÖÐÐÔ£¬NH3•H2OµÄµçÀë³£ÊýKb=$\frac{c£¨O{H}^{-}£©c£¨N{{H}_{4}}^{+}£©}{c£¨N{H}_{3}£®{H}_{2}O£©}$=$\frac{1{0}^{-7}¡Á5¡Á1{0}^{-3}}{0.5a-5¡Á1{0}^{-3}}$=$\frac{1{0}^{-9}}{a-0.01}$£¬
¹Ê´ð°¸Îª£ºÖУ»$\frac{1{0}^{-9}}{a-0.01}$£»
£¨5£©ÓÉKa¿É֪̼ËáµÄËáÐÔ´óÓÚHClOµÄËáÐÔ£¬Ôò84Ïû¶¾Òº£¨ÓÐЧ³É·ÝΪNaClO£©ÖÐͨÈëÉÙÁ¿µÄCO2£¬¸Ã·´Ó¦µÄ»¯Ñ§·½³ÌʽΪNaClO+CO2+H2O=NaHCO3+HClO£¬
¹Ê´ð°¸Îª£ºNaClO+CO2+H2O=NaHCO3+HClO£®

µãÆÀ ±¾Ì⿼²é»¯Ñ§Æ½ºâµÄ¼ÆË㣬Ϊ¸ßƵ¿¼µã£¬°ÑÎÕ»¯Ñ§Æ½ºâÒÆ¶¯¡¢ìÊ±ä¼ÆËã¡¢µçÀëÆ½ºâ³£Êý¼ÆËãΪ½â´ðµÄ¹Ø¼ü£¬²àÖØ·ÖÎöÓë¼ÆËãÄÜÁ¦µÄ¿¼²é£¬×ÛºÏÐÔ½ÏÇ¿£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
18£®X¡¢Y¡¢Z¡¢WΪËÄÖÖ¶ÌÖÜÆÚÔªËØ£¬ÆäÖÐYÔªËØÔ­×ÓºËÍâ×îÍâ²ãµç×ÓÊýÊÇÆäµç×Ó²ãÊýµÄ3±¶£¬ËüÃÇÔÚÖÜÆÚ±íÖеÄÏà¶ÔλÖÃÈç±íËùʾ£º
XY
ZW
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©WλÓÚÖÜÆÚ±íÖеÚÈýÖÜÆÚ£¬µÚ¢÷A×壻
£¨2£©X¿ÉÐγÉ˫ԭ×Ó·Ö×Ó£¬Æä·Ö×ӵĵç×ÓʽÊÇ£»YºÍÇâÔªËØÐγɵÄ10µç×Ó΢Á£Öг£¼û+1¼ÛÑôÀë×ÓΪH3O+£¨Ìѧʽ£¬ÏÂͬ£©£»ZºÍÇâÔªËØÐγɵÄ18µç×Ó΢Á£Öг£¼ûµÄ-1¼ÛÒõÀë×ÓµÄË®½â·½³ÌʽHS-+H2O?H2S+OH-£®
£¨3£©¹¤ÒµÉϽ«¸ÉÔïµÄWµ¥ÖÊͨÈëÈÛÈÚµÄZµ¥ÖÊÖпÉÖÆµÃ»¯ºÏÎïZ2W2£¬¸ÃÎïÖÊ¿ÉÓëË®·´Ó¦Éú³ÉÒ»ÖÖÄÜʹƷºìÈÜÒºÍÊÉ«µÄÆøÌ壬0.2mol¸ÃÎïÖʲμӷ´Ó¦Ê±×ªÒÆ0.3molµç×Ó£¬ÆäÖÐÖ»ÓÐÒ»ÖÖÔªËØ»¯ºÏ¼Û·¢Éú¸Ä±ä£¬Ð´³öZ2W2ÓëË®·´Ó¦µÄ»¯Ñ§·½³Ìʽ2S2Cl2+2H2O¨T3S¡ý+SO2¡ü+4HCl£®
£¨4£©½«0.20molZY2ºÍ0.10molO2³äÈëÒ»¸ö¹Ì¶¨ÈÝ»ýΪ5LµÄÃܱÕÈÝÆ÷ÖУ¬ÔÚÒ»¶¨Î¶Ȳ¢Óд߻¯¼Á´æÔÚÏ£¬½øÐз´Ó¦£¬¾­°ë·ÖÖӴﵽƽºâ£¬²âµÃÈÜÒºÖк¬YZ30.18mol£¬Ôòv£¨O2£©=0.036mol/£¨L•min£©£»ÈôζȲ»±ä£¬¼ÌÐøÍ¨Èë0.20molYZ2ºÍ0.10molO2£¬ÔòƽºâÏòÕý·´Ó¦·½ÏòÒÆ¶¯£¨Ìî¡°ÏòÕý·´Ó¦·½Ïò¡±¡¢¡°ÏòÄæ·´Ó¦·½Ïò¡±»ò¡°²»¡±£©£¬Ôٴδﵽƽºâºó£¬0.36mol£¼n£¨YZ3£©£¼0.40mol£®
15£®µªÊǵØÇòÉϺ¬Á¿·á¸»µÄÒ»ÖÖÔªËØ£¬µª¼°Æä»¯ºÏÎïÔÚ¹¤Å©ÒµÉú²ú¡¢Éú»îÖÐÓÐ×ÅÖØÒª×÷Ó㬺ϳɰ±¹¤ÒµÔÚ¹úÃñÉú²úÖÐÓÐÖØÒªÒâÒ壮ÒÔÏÂÊǹØÓںϳɰ±µÄÓйØÎÊÌ⣬Çë»Ø´ð£º
£¨1£©ÈôÔÚÒ»ÈÝ»ýΪ2LµÄÃܱÕÈÝÆ÷ÖмÓÈë0.2molµÄN2ºÍ0.6molµÄH2ÔÚÒ»¶¨Ìõ¼þÏ·¢Éú·´Ó¦£º
N2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H£¼0£¬ÈôÔÚ5·ÖÖÓʱ·´Ó¦´ïµ½Æ½ºâ£¬´Ëʱ²âµÃNH3µÄÎïÖʵÄÁ¿Îª0.2mol£®Ôòǰ5·ÖÖӵį½¾ù·´Ó¦ËÙÂÊv£¨N2£©=0.01mol/£¨L•min£©£®Æ½ºâʱH2µÄת»¯ÂÊΪ50%
£¨2£©ÈôÔÚ0.5LµÄÃܱÕÈÝÆ÷ÖУ¬Ò»¶¨Á¿µÄµªÆøºÍÇâÆø½øÐÐÈçÏ·´Ó¦£º
N2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H£¼0£¬Æä»¯Ñ§Æ½ºâ³£ÊýKÓëζÈTµÄ¹ØÏµÈç±íËùʾ£º
T/¡æ  200   300    400
KK1K3  0.5
ÇëÍê³ÉÏÂÁÐÎÊÌ⣺
¢ÙÊԱȽÏK1¡¢K2µÄ´óС£¬K1£¾K2£¨Ìî¡°£¼¡±¡°£¾¡±»ò¡°=¡±£©£»
¢ÚÏÂÁи÷ÏîÄÜ×÷ΪÅжϸ÷´Ó¦´ïµ½»¯Ñ§Æ½ºâ״̬µÄÒÀ¾ÝÊÇC£¨ÌîÐòºÅ×Öĸ£©
A£®ÈÝÆ÷ÄÚN2¡¢H2¡¢NH3µÄÎïÖʵÄÁ¿Å¨¶ÈÖ®±ÈΪ1£º3£º2
B£®v£¨H2£©Õý=3v£¨H2£©Äæ
C£®ÈÝÆ÷ÄÚѹǿ±£³Ö²»±ä
D£®»ìºÏÆøÌåµÄÃܶȱ£³Ö²»±ä
¢Û400¡æ£¬µ±²âµÃNH3¡¢N2ºÍH2ÎïÖʵÄÁ¿·Ö±ðΪ3mol¡¢2molºÍ 1molʱ£¬Ôòv£¨N2£©Õý£¾v£¨N2£©Äæ
£¨Ìî¡°£¼¡±¡°£¾¡±»ò¡°=¡±£©

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

¾«Ó¢¼Ò½ÌÍø