ÌâÄ¿ÄÚÈÝ

ËÄ´¨ÓзḻµÄÌìÈ»Æø×ÊÔ´¡£ÒÔÌìÈ»ÆøΪԭÁϺϳÉÄòËصÄÖ÷Òª²½ÖèÈçÏÂͼËùʾ£¨Í¼ÖÐijЩת»¯²½Öè¼°Éú³ÉÎïδÁгö£©£º

ÇëÌîдÏÂÁпհףº

£¨1£©ÒÑÖª0.5 mol¼×ÍéºÍ0.5 molË®ÕôÆøÔÚt ¡æ£¬p k Paʱ£¬ÍêÈ«·´Ó¦Éú³ÉÒ»Ñõ»¯Ì¼ºÍÇâÆø£¨ºÏ³ÉÆø£©£¬ÎüÊÕÁËa kJÈÈÁ¿¡£¸Ã·´Ó¦µÄÈÈ»¯Ñ§·½³ÌʽÊÇ£º       _____________    

£¨2£©Ôںϳɰ±µÄʵ¼ÊÉú²ú¹ý³ÌÖУ¬³£²ÉÈ¡µÄ´ëÊ©Ö®Ò»ÊÇ£º½«Éú³ÉµÄ°±´Ó»ìºÏÆøÌåÖм´Ê¹·ÖÀë³öÀ´£¬²¢½«·ÖÀë³ö°±ºóµÄµªÆøºÍÇâÆøÑ­»·ÀûÓã¬Í¬Ê±²¹³äµªÆøºÍÇâÆø¡£ÇëÓ¦Óû¯Ñ§·´Ó¦ËÙÂʺͻ¯Ñ§Æ½ºâµÄ¹Ûµã˵Ã÷²ÉÈ¡¸Ã´ëÊ©µÄÀíÓÉ£º            _____________________ 

£¨3£©µ±¼×ÍéºÏ³É°±ÆøµÄת»¯ÂÊΪ75£¥Ê±£¬ÒÔ5.60¡Á107 L¼×ÍéΪԭÁÏÄܹ»ºÏ³É                       L °±Æø¡££¨¼ÙÉèÌå»ý¾ùÔÚ±ê×¼×´¿öϲⶨ£©

£¨4£©ÒÑÖªÄòËصĽṹ¼òʽΪ£¬Çëд³öÁ½ÖÖº¬ÓÐ̼ÑõË«¼üµÄÄòËصÄͬ·ÖÒì¹¹ÌåµÄ½á¹¹¼òʽ£º¢Ù                      £¬¢Ú                        ¡£

 

¡¾´ð°¸¡¿

£¨1£©CH4(g)+H2O(g)CO(g)+3H2(g)£»¡÷H£½2a kJ/mol

£¨2£©Ôö´óµªÆøºÍÇâÆøµÄŨ¶ÈÓÐÀûÓÚÔö´ó·´Ó¦ËÙÂÊ£»¼õС°±ÆøµÄŨ¶È£¬Ôö´óµªÆøºÍÇâÆøµÄŨ¶È¶¼ÓÐÀûÓÚƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯¡£

£¨3£©1.12¡Á108

£¨4£©

¡¾½âÎö¡¿£¨1£©¸ù¾ÝÌâÄ¿Ëùʾд³ö»¯Ñ§·´Ó¦·½³Ìʽ²¢Åäƽ£ºCH4+H2O¡úCO+3H2£¬¶øÇÒÿÓÐ1molµÄ¼×ÍéºÍË®·´Ó¦¿ÉÎüÊÕ2a kJµÄÈÈÁ¿£¬ÔÙ¿¼ÂǸ÷ÎïÖÊ״̬£¬Ôò¿ÉµÃÈÈ»¯Ñ§·½³Ìʽ

£¨2£©Ôö´óµªÆøºÍÇâÆøµÄŨ¶ÈÓÐÀûÓÚÔö´ó·´Ó¦ËÙÂÊ£»¼õС°±ÆøµÄŨ¶È£¬Ôö´óµªÆøºÍÇâÆøµÄŨ¶È¶¼ÓÐÀûÓÚƽºâÏòÕý·´Ó¦·½ÏòÒƶ¯

£¨3£©ÓÉ£ºCH4+H2O¡úCO+3H2¡¢CO+ H2O=CO2+H2¡¢3H2 + N2 2NH3

×ۺϿɵãº3CH4¡«8NH3£¬ËùÒÔÒÔ5.60¡Á107 L¼×ÍéΪԭÁÏÄܹ»ºÏ³É°±ÆøΪ£º

V=5.60¡Á107¡Á75%¡Á8/3=1.12¡Á108L

£¨4£©¿ÉÊʵ±Òƶ¯Ì¼ÑõË«¼üλÖᢱä¸ü¹ÙÄÜÍż´¿É¡£

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ËÄ´¨ÓзḻµÄÌìÈ»Æø×ÊÔ´£®Ä³»¯¹¤³§ÒÔÌìÈ»ÆøΪԭÁϺϳɰ±µÄ¹¤ÒÕÁ÷³ÌʾÒâͼÈçÏ£º
¾«Ó¢¼Ò½ÌÍø
ÒÀ¾ÝÉÏÊöÁ÷³Ì£¬Íê³ÉÏÂÁÐÌî¿Õ£º
£¨1£©Õû¸öÁ÷³ÌÓÐÈý¸öÑ­»·£ºÒ»ÊÇK2CO3£¨aq£©Ñ­»·£¬¶þÊÇN2ºÍH2Ñ­»·£¬µÚÈý¸öÑ­»·Öб»Ñ­»·µÄÎïÖÊÊÇ
 

£¨2£©ÍÑÁò¹ý³ÌÖУ¬ÈôÓÐnmol Fe2O3?H2Oת»¯£¬ÔòÉú³ÉSµÄÎïÖʵÄÁ¿Îª
 
mol£¨Óú¬nµÄ´úÊýʽ±íʾ£©£®
£¨3£©Í¼ÖÐCH4µÄµÚ¶þ´Îת»¯¹ý³ÌÖз¢Éú·´Ó¦µÄ»¯Ñ§·½³ÌʽΪ
 
£®
£¨4£©Ôںϳɰ±µÄʵ¼ÊÉú²ú¹ý³ÌÖУ¬³£²ÉÈ¡µÄ´ëÊ©Ö®Ò»ÊÇ£º½«Éú³ÉµÄ°±´Ó»ìºÏÆøÌåÖм°Ê±·ÖÀë³öÀ´£¬²¢½«·ÖÀë³ö°±ºóµÄµªÆøºÍÇâÆøÑ­»·ÀûÓã¬Í¬Ê±²¹³äµªÆøºÍÇâÆø£®ÇëÔËÓû¯Ñ§·´Ó¦ËÙÂʺͻ¯Ñ§Æ½ºâµÄ¹Ûµã˵Ã÷²ÉÈ¡ÕâЩ´ëÊ©µÄÀíÓÉ£º
 

£¨5£©¸ÄÓùýÁ¿NaOHÈÜÒºÎüÊÕÌìÈ»ÆøÖеÄÁò»¯Ç⣬ÒÔʯī×÷µç¼«µç½âÎüÊÕºóËùµÃÈÜÒº¿É»ØÊÕÁò£¬Æäµç½â×Ü·´Ó¦·½³Ìʽ£¨ºöÂÔÑõÆøµÄÑõ»¯»¹Ô­£©Îª
 

£¨6£©Èô¹¤ÒµÉú²úÖÐÒÔa g°±ÆøºÍ×ãÁ¿µÄ¿ÕÆøΪԭÁÏ£¨²»¿¼ÂÇ¿ÕÆøÖÐN2µÄ·´Ó¦£©×î´óÏ޶ȵÄÖÆÈ¡NH4NO3£¬¾­¹ýһϵÁÐת»¯ºó£¬ÔÙÏò·´Ó¦ºóµÄ»ìºÏÎïÖмÓÈëbgË®£¬µÃµ½ÃܶÈΪd g/mLµÄNH4NO3ÈÜÒº£¬ÀíÂÛÉϸÃÈÜÒºÖÐÈÜÖʵÄÎïÖʵÄÁ¿Å¨¶È¿ÉÄܵÄ×î´óֵΪ
 
mol/L£®

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

¾«Ó¢¼Ò½ÌÍø