ÌâÄ¿ÄÚÈÝ

3£®£¨1£©ë£¨N2H4£©ºÍNO2ÊÇÒ»ÖÖË«×é·Ö»ð¼ýÍƽø¼Á£®Á½ÖÖÎïÖÊ»ìºÏ·¢Éú·´Ó¦Éú³ÉN2ºÍH2O£¨g£©£¬ÒÑÖª8gÆøÌåëÂÔÚÉÏÊö·´Ó¦Öзųö142kJÈÈÁ¿£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪ2N2H4£¨g£©+2NO2£¨g£©=3N2£¨g£©+4H2O£¨g£©¡÷H=-1136KL/mol£®
£¨2£©0.3molµÄÆø̬¸ßÄÜȼÁÏÒÒÅðÍ飨B2H6£©ÔÚÑõÆøÖÐȼÉÕ£¬Éú³É¹Ì̬ÈýÑõ»¯¶þÅðºÍҺ̬ˮ£¬·Å³ö649.5kJÈÈÁ¿£¬ÆäÈÈ»¯Ñ§·´Ó¦·½³ÌʽΪB2H6£¨g£©+3O2£¨g£©=B2O3£¨g£©+3H2O£¨l£©£»¡÷H=-2165KJ/mol£»
ÓÖÖªH2O£¨l£©?H2O£¨g£©£»¡÷H=+44kJ/mol£¬Ôò11.2L£¨±ê×¼×´¿ö£©ÒÒÅðÍéÍêȫȼÉÕÉú³ÉÆø̬ˮʱ£¬·Å³öµÄÈÈÁ¿ÊÇ1016.5 kJ£®

·ÖÎö £¨1£©¸ù¾Ý8gëÂȼÉÕ·ÅÈÈ142KJ£¬Çó³ö1molëÂȼÉÕ·ÅÈȶàÉÙ£¬ÔÙÊéдÈÈ»¯Ñ§·½³Ìʽ£¬ÊéдÈÈ»¯Ñ§·½³Ìʽʱ£¬×¢ÒâÎïÖʵÄ״̬¡¢¡÷HµÄÖµÓ뻯ѧ¼ÆÁ¿ÊýµÄ¶ÔÓ¦£»
£¨2£©Ê×ÏȾÝ0.3molÒÒÅðÍéȼÉÕ·ÅÈÈ649.1KJ£¬¼ÆËã³ö1molÒÒÅðÍéȼÉÕ·ÅÈȶàÉÙ£¬ÔÙÊéдÈÈ»¯Ñ§·½³Ìʽ£»µÚ¶þÎÊÖУ¬×¢Òâ11.2LÒÒÅðÍéȼÉÕÉú³ÉË®µÄÎïÖʵÄÁ¿£®

½â´ð ½â£¨1£©8gN2H4µÄÎïÖʵÄÁ¿Îª$\frac{8g}{32g•mo{l}^{-1}}$=0.25mol£¬Ôò1molN2H4·´Ó¦·Å³öµÄÈÈÁ¿Îª$\frac{1mol}{0.25mol}$=568KJ£¬ÆäÈÈ»¯Ñ§·½³ÌʽΪ£º
2N2H4£¨g£©+2NO2£¨g£©=3N2£¨g£©+4H2O£¨g£©¡÷H=-1136KL/mol£¬¹Ê´ð°¸Îª£º2N2H4£¨g£©+2NO2£¨g£©=3N2£¨g£©+4H2O£¨g£©¡÷H=-1136KL/mol£»
£¨2£©0.3molµÄÆø̬ÒÒÅðÍéȼÉշųöµÄÈÈÁ¿Îª649.5KJ£¬Ôò1molÆø̬ÒÒÅðÍéȼÉշųöµÄÈÈÁ¿Îª$\frac{1mol}{0.3mol}$=2165KJ£¬ÔòÆäÈÈ»¯Ñ§·½³ÌʽΪ£º
B2H6£¨g£©+3O2£¨g£©=B2O3£¨g£©+3H2O£¨l£©£»¡÷H=-2165KJ/mol£¬±ê×¼×´¿öÏÂ11.2LÒÒÅðÍéΪ0.5mol£¬ÍêȫȼÉÕÉú³É1.5molҺ̬ˮ·ÅÈÈ2165KJ¡Â2=1082.5KJ£¬ÓÖÖªH2O£¨l£©?H2O£¨g£©£»¡÷H=+44kJ/mol£¬Ôò1.5molҺ̬ˮ±äΪˮÕôÆøÐèÎüÈÈ1.5mol¡Á44KJ/mol=66KJ£¬Ôò·ÅÈÈ1082.5KJ-66KJ=1016.5KJ£¬¹Ê´ð°¸Îª£ºB2H6£¨g£©+3O2£¨g£©=B2O3£¨g£©+3H2O£¨l£©£»¡÷H=-2165KJ/mol£»1016.5£®

µãÆÀ ±¾Ì⿼²éÁËÈÈ»¯Ñ§·½³ÌʽµÄÊéд£¬½áºÏ¸Ç˹¶¨ÂÉ£¬×¢Òâ×¢ÒâÎïÖʵÄ״̬¡¢¡÷HµÄÖµÓ뻯ѧ¼ÆÁ¿ÊýµÄ¶ÔÓ¦£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø