ÌâÄ¿ÄÚÈÝ

6£®´×ËáÈÜÒºÖдæÔÚµçÀëƽºâCH3COOH?H++CH3COO-£¬ÏÂÁÐÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®CH3COOHÈÜÒºÖмÓÉÙÁ¿µÄCH3COONa¹ÌÌ壬ƽºâÕýÏòÒƶ¯
B£®0.10 mol/LµÄCH3COOHÈÜÒºÖмÓˮϡÊÍ£¬ÈÜÒºÖÐc£¨OH-£©¼õС
C£®´×ËáÈÜÒºÖÐÀë×ÓŨ¶ÈµÄ¹ØϵÂú×㣺c£¨H+£©=c£¨OH-£©+c£¨CH3COO-£©
D£®³£ÎÂÏÂpH=2µÄCH3COOHÈÜÒºÓëpH=12µÄNaOHÈÜÒºµÈÌå»ý»ìºÏºóÈÜÒºµÄpH£¾7

·ÖÎö ´×ËáÈÜÒºÖдæÔÚµçÀëƽºâCH3COOH?H++CH3COO-£¬ËµÃ÷´×ËáÊÇÈõËᣬÈÜÒºÖдæÔÚµçÀëƽºâ£¬Ó°ÏìµçÀëƽºâÒƶ¯µÄÒòËØÓÐζȡ¢Àë×ÓŨ¶ÈµÈ½áºÏÓ°ÏìµçÀëƽºâµÄÒòËؽâ´ð¸ÃÌ⣮

½â´ð ½â£ºA¡¢CH3COOHÈÜÒºÖмÓÉÙÁ¿µÄCH3COONa¹ÌÌ壬´×Ëá¸ùÀë×ÓŨ¶ÈÔö´ó£¬´×ËáµÄµçÀëƽºâÄæÏòÒƶ¯£¬¹ÊA´íÎó£»
B¡¢´×ËáÈÜÒº¼ÓˮϡÊÍ£¬ÈÜÒºËáÐÔ¼õÈõ£¬ÇâÑõ¸ùÀë×ÓŨ¶ÈÔö´ó£¬¹ÊB´íÎó£»
C¡¢¾ÝµçºÉÊغ㣬´×ËáÈÜÒºÖÐÀë×ÓŨ¶È¹ØϵÂú×ãc£¨H+£©=c£¨OH-£©+c£¨CH3COO-£©£¬¹ÊCÕýÈ·£»
D¡¢´×ËáÊÇÈõËᣬpH=2µÄ´×ËáÈÜÒºÖд×ËáŨ¶ÈÔ¶Ô¶´óÓÚ0.01mol/L£¬ËùÒÔpH=2µÄCH3COOHÈÜÒºÓëpH=12µÄNaOHÈÜÒºµÈÌå»ý»ìºÏ£¬ÈÜÒºÏÔËáÐÔ£¬pH£¼7£¬¹ÊD´íÎó£»
¹ÊÑ¡C£®

µãÆÀ ±¾Ì⿼²éÁËÈõµç½âÖʵçÀ룬Ã÷È·Ó°ÏìƽºâÒƶ¯µÄÒòËؼ´¿É½â´ð£¬Ò×´íÑ¡ÏîÊÇB£¬×¢Òâ¼ÓˮϡÊÍ´Ù½ø´×ËáµçÀ룬µ«ÈÜÒºÖÐÇâÀë×ÓŨ¶È½µµÍ£¬ÇâÑõ¸ùÀë×ÓŨ¶ÈÔö´ó£¬ÎªÒ×´íµã£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
1£®¹¤ÒµÉÏÓÃNH3ºÍCO2·´Ó¦ºÏ³ÉÄòËØ£®
2NH3£¨g£©+CO2£¨g£©?CO£¨NH2£©2£¨g£©+H2O£¨g£©¡÷H1=-536.1kJ/mol

£¨1£©¸Ã¹ý³Ìʵ¼ÊÉÏ·ÖÁ½²½½øÐУ¬µÚÒ»²½²úÉú°±»ù¼×Ëá泥¬µÚ¶þ²½ÊÇ°±»ù¼×Ëáï§ÍÑË®Éú³ÉÄòËØ£®Çëд³öµÚÒ»²½µÄ·´Ó¦·½³Ìʽ2NH3£¨g£©+CO2£¨g£©?NH2COONH4£®
£¨2£©ÔÚÒ»¶¨Ìõ¼þϺϳÉÄòËØ£¬µ±°±Ì¼±È$\frac{n£¨N{H}_{3}£©}{n£¨C{O}_{2}£©}$=4ʱ£¬CO2µÄת»¯ÂÊËæʱ¼äµÄ±ä»¯¹ØϵÈçͼ1Ëùʾ£¬ÏÂÁÐ˵·¨´íÎóµÄÊÇBD£¨Ìî±êºÅ£©£®
A£®¸Ã·´Ó¦ÔÚ60minʱ´ïµ½Æ½ºâ״̬£»
B£®NH3µÄƽºâת»¯ÂÊΪ70%£»
C£®Ôö¼Ó°±Ì¼±È¿É½øÒ»²½Ìá¸ßCO2µÄƽºâת»¯ÂÊ£»
D£®MµãµÄÄæ·´Ó¦ËÙÂÊvÄ棨CO2£©´óÓÚNµãµÄÕý·´Ó¦ËÙÂÊvÕý£¨CO2£©£»
£¨3£©·´Ó¦2NH3£¨g£©+CO2£¨g£©?CO£¨NH2£©2£¨g£©+H2O£¨g£©£¬ÔÚ200¡æʱ¿É×Ô·¢ÏòÕý·´Ó¦½øÐУ¬ÔòÉý¸ßζȣ¬Æ½ºâ³£ÊýK¼õС£¨Ìî¡°Ôö´ó¡±¡¢¡°¼õС¡±¡¢¡°²»±ä¡±£©£®
£¨4£©ÈôÒ»¶¨Ìõ¼þϵõ½µÄÄòËØΪ¹ÌÌ壬2NH3£¨g£©+CO2£¨g£©?CO£¨NH2£©2£¨s£©+H2O£¨g£©£¬¿ªÊ¼ÔÚÒ»ÃܱÕÈÝÆ÷ÖгäÈë0.1mol NH3¡¢0.65mol CO2¡¢0.65mol H2O£¨g£©£¬±£³ÖζȲ»±ä£¬´ïµ½Æ½ºâʱ£¬»ìºÏÆøÌåµÄƽ¾ùĦ¶ûÖÊÁ¿ÓëÆðʼÏà±È±äС£¨Ìî¡°±ä´ó¡±¡¢¡°±äС¡±¡¢¡°²»±ä¡±£©£®
£¨5£©ÄòËØÔÚÒ»¶¨Î¶ȺÍѹÁ¦Ï»Ჿ·Ö±äΪNH4CNO£¨ÇèËá泥©£¬ÒÑÖª25¡æ°±Ë®µÄµçÀë³£ÊýΪ1.7¡Á10-5£¬ÇèËáHCNOµÄµçÀë³£ÊýΪ3.3¡Á10-4£®ÇèËáï§Ë®ÈÜÒºÖи÷Àë×ÓµÄŨ¶È´óС˳ÐòΪc£¨CNO-£©£¾c£¨NH4+£©£¾c£¨H+£©£¾c£¨OH-£©£®
£¨6£©µç½âÄòËØ[CO£¨NH2£©2]µÄ¼îÐÔÈÜÒºÖÆÇâµÄ×°ÖÃʾÒâͼÈçͼ2£¨µç½â³ØÖиôĤ½ö×èÖ¹ÆøÌåͨ¹ý£¬a¡¢b¼«¾ùΪ¶èÐԵ缫£©£®µç½âʱ£¬a¼«µÄµç¼«·´Ó¦Ê½ÎªCO£¨NH2£©2-6e-+8OH-=CO32-+N2¡ü+6H2O£¬ÈôÔÚb¼«²úÉú±ê¿öÏÂ224mLÇâÆø£¬ÔòÏûºÄÄòËØ0.2g£®

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

¾«Ó¢¼Ò½ÌÍø