ÌâÄ¿ÄÚÈÝ
µª·ÊÊǹ㷺ʹÓõķÊÁÏ£¬Çë½áºÏÓйزÄÁÏ£¬»Ø´ðÎÊÌ⣮
(1)ÁòËáï§ÊÇÅ©´å³£ÓõÄÒ»ÖÖï§Ì¬µª·Ê£®
¢ÙÁòËáï§ÖеªµÄÖÊÁ¿·ÖÊýÊÇ__________£®
¢ÚʵÑé²âµÃijÁòËá立ÊÁÏÖеªÔªËصÄÖÊÁ¿·ÖÊýΪ20£¥£¬Ôò¸ÃÑùÆ·ÖпÉÄÜ»ìÓÐ_____________£®
[¡¡¡¡]
(2)µçÉÁÀ×ÃùÊÇÈÕ³£Éú»îÖÐ˾¿Õ¼û¹ßµÄÏÖÏó£®ÔÚÉÁµç·¢ÉúµÄ¹ý³ÌÖУ¬¿ÕÆøÖеĵªÆøºÍÑõÆøÖ±½Ó»¯ºÏ£®¼ÙÉèij´ÎÀ×µç·Å³öµÄÄÜÁ¿Îª£¬ÒÑÖªºÍ»¯ºÏʱҪÎüÊÕ181kJÄÜÁ¿£¬ÉÁµçʱÓÐ1/1000µÄÄÜÁ¿ÓÃÓÚÕâÒ»·´Ó¦£®
¢Ù´Ë´ÎÉÁµçËù²úÉúµÄNOµÄÎïÖʵÄÁ¿ÊÇ_____mol.
¢Ú´Ë´ÎÀ×µçµÄÉú³ÉÎïÏ൱ÓÚ¸øÍÁÈÀÊ©¼ÓÁË_______ǧ¿ËµÄÄòËØ»¯·Ê¡²ÄòËØ»¯Ñ§Ê½¡³.
½âÎö£º
(1) ¢Ù21£¥£¬¢ÚA(Ìáʾ£ºº¬ÑõµÄÖÊÁ¿·ÖÊýСÓÚ20£¥)(2) ¢ÙÒ»´ÎÉÁµçÓÃÀ´Éú³ÉNOµÄÄÜÁ¿ÊÇÉú³É NOµÄÎïÖʵÄÁ¿Îª¢ÚÀíÂÛÉÏÏ൱ÓÚÄòËØ 41.5molÖÊÁ¿Îª |
£¨11 ·Ö£©¿Æѧ¼ÒÒ»Ö±ÖÂÁ¦Ñо¿³£Î¡¢³£Ñ¹Ï¡°È˹¤¹Ìµª¡±µÄз½·¨¡£ÔøÓÐʵÑ鱨µÀ£ºÔÚ³£Î¡¢³£Ñ¹¡¢¹âÕÕÌõ¼þÏ£¬N2ÔÚ´ß»¯¼Á£¨²ôÓÐÉÙÁ¿Fe2O3µÄTiO2£©±íÃæÓëË®·¢Éú·´Ó¦£¬Éú³ÉµÄÖ÷Òª²úÎïΪNH3£¬·´Ó¦µÄ»¯Ñ§·½³ÌʽÈçÏ£ºN2(g)+ 3H2O(l) 2NH3(g)+ O2(g)¡£»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©½øÒ»²½Ñо¿NH3Éú³ÉÁ¿ÓëζȵĹØϵ£¬²¿·ÖʵÑéÊý¾Ý¼ûÏÂ±í£¨¹âÕÕ¡¢N2ѹÁ¦1.0¡Á105 Pa¡¢·´Ó¦Ê±¼ä3 h£©£¬Ôò¸Ã·´Ó¦µÄÕý·´Ó¦Îª ·´Ó¦£¨Ìî¡°ÎüÈÈ¡±»ò¡°·ÅÈÈ¡±£©
T/K |
303 |
313 |
323 |
NH3Éú³ÉÁ¿/£¨10-6 mol£© |
4.8 |
5.9 |
6.0 |
£¨2£©ÓëÄ¿Ç°¹ã·ºÊ¹ÓõĹ¤ÒµºÏ³É°±·½·¨Ïà±È£¬¸Ã·½·¨Öй̵ª·´Ó¦ËÙÂÊÂý¡£ÇëÌá³ö¿ÉÌá¸ßÆä·´Ó¦ËÙÂÊÇÒÔö´óNH3Éú³ÉÁ¿µÄ½¨Ò飺¡¡¡¡ ¡£
£¨3£©ºÏ³É°±¹¤ÒµÖÐÔÁÏÆøN2¿É´Ó¿ÕÆøÖзÖÀëµÃµ½£¬H2¿ÉÓü×ÍéÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦ÖƵ᣼×ÍéÔÚ¸ßÎÂÏÂÓëË®ÕôÆø·´Ó¦·´Ó¦·½³ÌʽΪ£ºCH4(g)£«H2O(g)£½CO(g)£«3H2(g)¡£²¿·ÖÎïÖʵÄȼÉÕÈÈÊý
¾ÝÈçÏ£º
H2(g) £º¡÷H =£285.8 kJ・mol£1£»
CO(g) £º ¡÷H =£283.0 kJ・mol£1£»
CH4(g) £º¡÷H =£890.3 kJ・mol£1 ¡£
ÒÑÖª1mol H2O(g)ת±äΪ1mol H2O(l)ʱ·Å³ö44.0 kJÈÈÁ¿¡£Ð´³öCH4ºÍH2OÔÚ¸ßÎÂÏ·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ__________________________________¡£
£¨4£©ÓÐÈËÉèÏëÑ°ÇóÊʺϵĴ߻¯¼ÁºÍµç¼«²ÄÁÏ£¬ÒÔN2¡¢H2Ϊµç¼«·´Ó¦ÎÒÔHCl¡ª¡ªNH4ClΪµç½âÖÊÈÜÒºÖƳÉÐÂÐÍȼÁϵç³Ø£¬Çëд³ö¸Ãµç¼«µÄÕý¼«·´Ó¦Ê½
£¨5£©Éú³ÉµÄNH3¿ÉÓÃÓÚÖÆï§Ì¬µª·Ê£¬Èç(NH4)2SO4¡¢NH4Cl£¬ÕâЩ·ÊÁÏÏÔ ÐÔ£¬ÔÒòÊÇ£¨ÓÃÀë×Ó·½³Ìʽ±íʾ£©___________________________£¬Ê¹ÓÃʱ±ÜÃâÓë________________ÎïÖʺÏÊ©¡£