ÌâÄ¿ÄÚÈÝ

£¨13·Ö£©¹¤ÒµºÏ³É°±ÊÇÏõËṤҵÖеÄÖØÒª²½Ö裬ÒÑÖªN2(g) +3H2(g)  2NH3(g) ¦¤H£½£­92.4kJ¡¤mol£­1¡£Çë»Ø´ð£º
(1) µ±ºÏ³É°±·´Ó¦´ïµ½Æ½ºâºó£¬¸Ä±äijһÍâ½çÌõ¼þ£¨²»¸Ä±äN2¡¢H2ºÍNH3µÄÁ¿£©£¬·´Ó¦ËÙÂÊÓëʱ¼äµÄ¹ØϵÈçÓÒͼËùʾ¡£

ͼÖÐt1ʱÒýÆðƽºâÒƶ¯µÄÌõ¼þ¿ÉÄÜÊÇ      __       t3ʱÒýÆðƽºâÒƶ¯µÄÌõ¼þ¿ÉÄÜÊÇ        __   ___  , ÆäÖбíʾƽºâ»ìºÏÎïÖÐNH3µÄº¬Á¿×î¸ßµÄÒ»¶Îʱ¼äÊÇ            ¡£
(2) ζÈΪT ¡æʱ£¬½«2a mol H2ºÍa mol N2·ÅÈë0.5LÃܱÕÈÝÆ÷ÖУ¬³ä·Ö·´Ó¦ºó²âµÃN2µÄת»¯ÂÊΪ50©‡¡£Ôò¸Ã·´Ó¦µÄ»¯Ñ§Æ½ºâ³£ÊýµÄֵΪ           ___________________¡£ÇëÔÚ´ðÌâ¾íÉÏд³ö¾ßÌåµÄ¼ÆËã¹ý³Ì¡£



N2(g)     +     3H2(g)     2NH3(g)
ÆðʼÎïÖʵÄÁ¿Å¨¶È(mol.L-1)      2a              4a              0       
ÎïÖʵÄŨ¶È±ä»¯(mol.L-1)        a                3a            2a               
ƽºâÎïÖʵÄÁ¿Å¨¶È(mol.L-1)     a                 a            2a     £¨3·Ö£©
£¨2·Ö£©

½âÎö

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
£¨2013?¼ÃÄþһģ£©ºÏ³É°±¹¤Òµ¡¢ÁòËṤҵµÄÉú²ú¹¤ÒÕÁ÷³Ì´óÖÂΪ£º

ºÏ³ÉËþºÍ½Ó´¥ÊÒÖеķ´Ó¦·Ö±ðΪ£º
N2£¨g£©+3H2£¨g£©?2NH3£¨g£©¡÷H£¼0£»   2SO2£¨g£©+O2£¨g£©?2SO3£¨g£©¡÷H£¼0
£¨1£©Ð´³öÁ÷³ÌÖÐÉ豸µÄÃû³Æ£ºB
Ñ­»·Ñ¹Ëõ»ú
Ñ­»·Ñ¹Ëõ»ú
£¬X
·ÐÌÚ¯
·ÐÌÚ¯
£®
£¨2£©½øÈëºÏ³ÉËþºÍ½Ó´¥ÊÒÖеÄÆøÌ嶼Ҫ½øÐÐÈÈ´¦Àí£¬×îÀíÏëµÄÈÈ´¦Àí·½·¨ÊÇ
³ä·ÖÀûÓ÷´Ó¦ÖзųöµÄÈÈÁ¿¼ÓÈÈ·´Ó¦Æø
³ä·ÖÀûÓ÷´Ó¦ÖзųöµÄÈÈÁ¿¼ÓÈÈ·´Ó¦Æø
£®
£¨3£©²ÉÓÃÑ­»·²Ù×÷¿ÉÌá¸ßÔ­ÁϵÄÀûÓÃÂÊ£¬ÏÂÁÐÉú²úÖУ¬²ÉÓÃÑ­»·²Ù×÷µÄÊÇ
¢Ù¢Ú¢Û
¢Ù¢Ú¢Û
£¨ÌîÐòºÅ£©£®
¢ÙÁòËṤҵ   ¢ÚºÏ³É°±¹¤Òµ     ¢ÛÏõËṤҵ
£¨4£©¹¤ÒµÉϳ£ÓÃ98.3%µÄŨÁòËáÎüÊÕSO3¶ø²»ÓÃÏ¡ÁòËá»òË®µÄÔ­ÒòÊÇ
ÓÉÓÚÓÃÏ¡ÁòËá»òË®ÎüÊÕSO2ʱÒ×ÐγÉËáÎí£¬²»ÀûÓÚSO2ÎüÊÕ£»
ÓÉÓÚÓÃÏ¡ÁòËá»òË®ÎüÊÕSO2ʱÒ×ÐγÉËáÎí£¬²»ÀûÓÚSO2ÎüÊÕ£»
£®
£¨5£©¹¤ÒµÉú²úÖг£Óð±--Ëá·¨½øÐÐβÆøÍÑÁò£¬ÒÔ´ïµ½Ïû³ýÎÛȾ¡¢·ÏÎïÀûÓõÄÄ¿µÄ£®ÁòËṤҵβÆøÖеÄSO2¾­´¦Àí¿ÉÒԵõ½Ò»ÖÖ»¯·Ê£¬¸Ã·ÊÁϵĻ¯Ñ§Ê½ÊÇ
£¨NH4£©2SO4
£¨NH4£©2SO4
£®

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

¾«Ó¢¼Ò½ÌÍø