ÌâÄ¿ÄÚÈÝ


¹¤ÒµÉÏÒÔ»ÆÌú¿óΪԭÁÏÉú²úÁòËᣬÆäÖÐÖØÒªµÄÒ»²½ÊÇ´ß»¯Ñõ»¯(Éú²úÖб£³ÖºãκãÈÝÌõ¼þ)£º2SO2(g)£«O2(g)2SO3(g)  ¡÷H£½£­196.6 kJ¡¤mol£­1

(1)Éú²úÖÐΪÌá¸ß·´Ó¦ËÙÂʺÍSO2µÄת»¯ÂÊ£¬ÏÂÁдëÊ©¿ÉÐÐÊÇ             ¡£

A£®Ïò×°ÖÃÖгäÈëO2              B£®Éý¸ßζÈ

C£®Ïò×°ÖÃÖгäÈëN2              D£®Ïò×°ÖÃÖгäÈë¹ýÁ¿µÄSO2

(2)ºãκãѹ£¬Í¨Èë3mol SO2 ºÍ2mol O2 ¼°¹ÌÌå´ß»¯¼Á£¬Æ½ºâʱÈÝÆ÷ÄÚÆøÌåÌå»ýΪÆðʼʱµÄ90%¡£±£³Öͬһ·´Ó¦Î¶ȣ¬ÔÚÏàͬÈÝÆ÷ÖУ¬½«ÆðʼÎïÖʵÄÁ¿¸ÄΪ 5mol SO2(g)¡¢3.5 mol O2(g)¡¢1mol SO3(g)£¬ÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ         

A£®µÚÒ»´Îƽºâʱ·´Ó¦·Å³öµÄÈÈÁ¿Îª294.9kJ 

B£®Á½´ÎƽºâSO2µÄת»¯ÂÊÏàµÈ    

 C£®Á½´ÎƽºâʱµÄO2Ìå»ý·ÖÊýÏàµÈ          

D£®µÚ¶þ´ÎƽºâʱSO3µÄÌå»ý·ÖÊýµÈÓÚ2/9

(3)500 ¡æÊ±½«10 mol SO2ºÍ5.0 mol O2ÖÃÓÚÌå»ýΪ1£ÌµÄºãÈÝÃܱÕÈÝÆ÷ÖУ¬SO2ת»¯ÎªSO3µÄƽºâת»¯ÂÊΪ0.95¡£Ôò500¡æÊ±µÄƽºâ³£ÊýK=           ¡£

(4)550 ¡æ£¬A¡¢B±íʾ²»Í¬Ñ¹Ç¿ÏÂµÄÆ½ºâת»¯ÂÊ(Èçͼ)£¬

ͨ³£¹¤ÒµÉú²úÖвÉÓó£Ñ¹µÄÔ­ÒòÊÇ                    £¬

²¢±È½Ï²»Í¬Ñ¹Ç¿ÏÂµÄÆ½ºâ³£Êý£ºK(0.10 MPa)     K(1.0 MPa)¡£

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø