ÌâÄ¿ÄÚÈÝ

ͨ¹ýÒÔÏ·´Ó¦¾ù¿É»ñÈ¡H2¡£ÏÂÁÐÓйØËµ·¨ÕýÈ·µÄÊÇ

¢ÙÌ«Ñô¹â´ß»¯·Ö½âË®ÖÆÇ⣺2H2O(l)2H2(g)+O2(g)¦¤H1=571.6kJ/mol

¢Ú½¹Ì¿ÓëË®·´Ó¦ÖÆÇ⣺C(s)+H2O(g)CO(g)+H2(g)¦¤H2=131.3kJ/mol

¢Û¼×ÍéÓëË®·´Ó¦ÖÆÇ⣺CH4(g)+H2O(g)CO(g)+3H2(g)¦¤H3=206.1kJ/mol

A£®·´Ó¦¢ÙÖеçÄÜת»¯Îª»¯Ñ§ÄÜ

B£®·´Ó¦¢ÚΪ·ÅÈÈ·´Ó¦

C£®·´Ó¦¢ÛʹÓô߻¯¼Á£¬¦¤H3¼õС

D£®·´Ó¦CH4(g)C(s)+2H2(g)µÄ¦¤H3=74.8kJ/mol

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
13£®Ìú´¥Ã½ÊÇÖØÒªµÄ´ß»¯¼Á£¬COÒ×ÓëÌú´¥Ã½×÷Óõ¼ÖÂÆäʧȥ´ß»¯»îÐÔ£ºFe+5CO=Fe£¨CO£©5£»³ýÈ¥COµÄ»¯Ñ§·´Ó¦·½³ÌʽΪ£º[Cu£¨NH3£©2]OOCCH3+CO+NH3=[Cu£¨NH3£©3£¨CO£©]OOCCH3£®
Çë»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©C¡¢N¡¢OµÄµçÀëÄÜÓÉ´óµ½Ð¡µÄ˳ÐòΪN£¾O£¾C£¬»ù̬FeÔ­×ӵļ۵ç×ÓÅŲ¼Ê½Îª3d64s2£®
£¨2£©Fe£¨CO£©5ÓÖÃûôÊ»ùÌú£¬³£ÎÂÏÂΪ»ÆÉ«ÓÍ×´ÒºÌ壬ÔòFe£¨CO£©5µÄ¾§ÌåÀàÐÍÊÇ·Ö×Ó¾§Ì壬
Fe£¨CO£©5ÔÚ¿ÕÆøÖÐȼÉÕºóÊ£ÓàµÄ¹ÌÌå³Êºìרɫ£¬ÏàÓ¦µÄ»¯Ñ§·½³ÌʽΪ4Fe£¨CO£©5+13O2$\frac{\underline{\;µãȼ\;}}{\;}$2Fe2O3+20CO2£®
£¨3£©gdºÏÎï[Cu£¨NH3£©2]OOCCH3ÖÐ̼ԭ×ÓµÄÔÓ»¯ÀàÐÍÊÇsp3¡¢sp2£¬ÅäÌåÖÐÌṩ¹Â¶Ôµç×ÓµÄÔ­×ÓÊÇN£®
£¨4£©ÓÃ[Cu£¨NH3£©2]OOCCH3³ýÈ¥COµÄ·´Ó¦ÖУ¬¿Ï¶¨ÓÐbdÐγɣ®
a£®Àë×Ó¼ü    b£®Åäλ¼ü   c£®·Ç¼«ÐÔ¼ü   d£®¦Ä¼ü
£¨5£©µ¥ÖÊÌúµÄ¾§ÌåÔÚ²»Í¬Î¶ÈÏÂÓÐÁ½ÖÖ¶Ñ»ý·½Ê½£¬¾§°û·Ö±ðÈçͼËùʾ£¬ÃæÐÄÁ¢·½¾§°ûºÍÌåÐÄÁ¢·½¾§°ûÖÐʵ¼Êº¬ÓеÄÌúÔ­×Ó¸öÊýÖ®±ÈΪ2£º1£¬ÃæÐÄÁ¢·½¶Ñ»ýÓëÌåÐÄÁ¢·½¶Ñ»ýµÄÁ½ÖÖÌú¾§ÌåµÄÃܶÈÖ®±ÈΪ4$\sqrt{2}$£º3$\sqrt{3}$£¨Ð´³öÒÑ»¯¼òµÄ±ÈÀýʽ¼´¿É£©£®

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

¾«Ó¢¼Ò½ÌÍø