ÏÂͼÉæ¼°µÄÎïÖÊËùº¬ÔªËØÖУ¬³ýÒ»ÖÖÔªËØÍ⣬ÆäÓà¾ùΪ¶ÌÖÜÆÚÔªËØ¡£ÒÑÖª£ºA¡¢FΪÎÞÉ«ÆøÌåµ¥ÖÊ£¬BΪ¾ßÓд̼¤ÐÔÆøζµÄÆøÌ壬CΪºÚÉ«Ñõ»¯ÎEΪºìÉ«½ðÊôµ¥ÖÊ£¨²¿·Ö·´Ó¦µÄ²úÎïδÁгö£©¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©BµÄµç×ÓʽΪ        £»

£¨2£©Ð´³öBºÍC·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                            £»

£¨3£©Ð´³öEÓëGµÄÏ¡ÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£¬²¢Óõ¥ÏßÇűê³öµç×ÓתÒƵķ½ÏòºÍÊýÄ¿£º

                                                      £»

£¨4£©J¡¢K¾ùÊÇͬÖÖ½ðÊôµÄÂÈ»¯ÎÇÒKΪ°×É«³Áµí¡£Ð´³öSO2»¹Ô­JÉú³ÉKµÄÀë×Ó·½³Ìʽ                                        £»

£¨5£©Æû³µÎ²ÆøÖг£º¬ÓÐDºÍCO £¬¶þÕßÔÚ´ß»¯¼Á×÷ÓÃÏ¿ÉÒԴ󲿷Öת»¯ÎªÁ½ÖÖ¶Ô¿ÕÆøÎÞÎÛȾµÄÎïÖÊ£¬ÒÑÖª£º    F(g) + A(g) = 2D (g)   ¡÷H = +180.5KJ/mol

2C (s)+ O2 (g)= 2CO(g)  ¡÷H = -221.0 KJ/mol

C (s)+ O2(g) = CO2(g)  ¡÷H = -393.5 KJ/mol

ÔòÉÏÊöβÆøת»¯µÄÈÈ»¯Ñ§·½³ÌʽΪ£º                                   ¡£

 

£¨10·Ö£©£¨Ã¿Ð¡Ìâ¸÷2·Ö£©

£¨1£©NH3£»(ÂÔ)  £¨2£©3CuO + 2NH3 ===3Cu + N2 +3H2O(¼ÓÈÈ)

£¨3£© 3Cu +8H++ 2NO3- = 3Cu2+ + 2NO¡ü + 4H2O ¡¡

£¨4£©2Cu2+ +2Cl-+ SO2 +2H2O= 2CuCl¡ý+ 4H+ + SO42-

£¨5£©2NO(g)+2CO(g)==N2(g)+2CO2(g) ¡÷H=¡ª746.5KJ/mol

½âÎö:

 

Á·Ï°²áϵÁдð°¸
Ïà¹ØÏ°Ìâ

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

ÏÂͼÉæ¼°µÄÎïÖÊËùº¬ÔªËØÖУ¬³ýÒ»ÖÖÔªËØÍ⣬ÆäÓà¾ùΪ¶ÌÖÜÆÚÔªËØ£®ÒÑÖª£ºA¡¢FΪÎÞÉ«ÆøÌåµ¥ÖÊ£¬BΪ¾ßÓд̼¤ÐÔÆøζµÄÆøÌ壬CΪºÚÉ«Ñõ»¯ÎEΪºìÉ«½ðÊôµ¥ÖÊ£¨²¿·Ö·´Ó¦µÄ²úÎïδÁгö£©£®Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©BµÄµç×ÓʽΪ
£»
£¨2£©Ð´³öBºÍC·´Ó¦µÄ»¯Ñ§·½³Ìʽ
3CuO+2NH3
 ¡÷ 
.
 
3Cu+N2+3H2O
3CuO+2NH3
 ¡÷ 
.
 
3Cu+N2+3H2O
£»
ɾ³ý´Ë¿Õ
ɾ³ý´Ë¿Õ

£¨3£©Ð´³öEÓëGµÄÏ¡ÈÜÒºµÄÀë×Ó·½³Ìʽ£¬²¢±ê³öµç×ÓתÒÆÊýÄ¿£º
£»
£¨4£©J¡¢K¾ùÊÇͬÖÖ½ðÊôµÄÂÈ»¯ÎÇÒKΪ°×É«³Áµí£®Ð´³öSO2»¹Ô­JÉú³ÉKµÄÀë×Ó·½³Ìʽ
2Cu2++2Cl-+SO2+2H2O=2CuCl¡ý+4H++SO42-
2Cu2++2Cl-+SO2+2H2O=2CuCl¡ý+4H++SO42-
£»
£¨5£©Æû³µÎ²ÆøÖг£º¬ÓÐDºÍCO£¬¶þÕßÔÚ´ß»¯¼Á×÷ÓÃÏ¿ÉÒԴ󲿷Öת»¯ÎªÁ½ÖÖ¶Ô¿ÕÆøÎÞÎÛȾµÄÎïÖÊ£¬ÒÑÖª£ºF£¨g£©+A£¨g£©=2D £¨g£©¡÷H=+180.5KJ/mol
2C £¨s£©+O2 £¨g£©=2CO£¨g£©¡÷H=-221.0KJ/mol
C £¨s£©+O2£¨g£©=CO2£¨g£©¡÷H=-393.5KJ/mol
ÔòÉÏÊöβÆøת»¯µÄÈÈ»¯Ñ§·½³ÌʽΪ£º
2NO£¨g£©+2CO£¨g£©=N2£¨g£©+2CO2£¨g£©¡÷H=-746.5KJ/mol
2NO£¨g£©+2CO£¨g£©=N2£¨g£©+2CO2£¨g£©¡÷H=-746.5KJ/mol
£®

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º ÌâÐÍ£º

£¨10·Ö£©ÏÂͼÉæ¼°µÄÎïÖÊËùº¬ÔªËØÖУ¬³ýÒ»ÖÖÔªËØÍ⣬ÆäÓà¾ùΪ¶ÌÖÜÆÚÔªËØ¡£ÒÑÖª£ºA¡¢FΪÎÞÉ«ÆøÌåµ¥ÖÊ£¬BΪ¾ßÓд̼¤ÐÔÆøζµÄÆøÌ壬CΪºÚÉ«Ñõ»¯ÎEΪºìÉ«½ðÊôµ¥ÖÊ£¨²¿·Ö·´Ó¦µÄ²úÎïδÁгö£©¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©BµÄµç×ÓʽΪ­­­­­        £»

£¨2£©Ð´³öBºÍC·´Ó¦µÄ»¯Ñ§·½³Ìʽ                             £»

£¨3£©Ð´³öEÓëGµÄÏ¡ÈÜÒºµÄÀë×Ó·½³Ìʽ£¬Ð´³öµç×ÓתÒÆÊýÄ¿£º           £¬       £»

£¨4£©Æû³µÎ²ÆøÖг£º¬ÓÐDºÍCO£¬¶þÕßÔÚ´ß»¯¼Á×÷ÓÃÏ¿ÉÒԴ󲿷Öת»¯ÎªÁ½ÖÖ¶Ô¿ÕÆøÎÞÎÛȾµÄÎïÖÊ£¬ÒÑÖª£ºF(g) + A(g) = 2D (g) £»¡÷H = +180.5KJ/mol

2C (s)+ O2 (g)= 2CO(g) £»¡÷H = -221.0 KJ/mol

C (s)+ O2(g) = CO2(g)£»¡÷H = -393.5 KJ/mol

ÔòÉÏÊöβÆøת»¯µÄÈÈ»¯Ñ§·½³ÌʽΪ£º                         ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêºÓ±±Ê¡Õý¶¨ÖÐѧ¸ß¶þµÚ¶þѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

£¨10·Ö£©ÏÂͼÉæ¼°µÄÎïÖÊËùº¬ÔªËØÖУ¬³ýÒ»ÖÖÔªËØÍ⣬ÆäÓà¾ùΪ¶ÌÖÜÆÚÔªËØ¡£ÒÑÖª£ºA¡¢FΪÎÞÉ«ÆøÌåµ¥ÖÊ£¬BΪ¾ßÓд̼¤ÐÔÆøζµÄÆøÌ壬CΪºÚÉ«Ñõ»¯ÎEΪºìÉ«½ðÊôµ¥ÖÊ£¨²¿·Ö·´Ó¦µÄ²úÎïδÁгö£©¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©BµÄµç×ÓʽΪ­­­­­        £»
£¨2£©Ð´³öBºÍC·´Ó¦µÄ»¯Ñ§·½³Ìʽ                             £»
£¨3£©Ð´³öEÓëGµÄÏ¡ÈÜÒºµÄÀë×Ó·½³Ìʽ£¬Ð´³öµç×ÓתÒÆÊýÄ¿£º           £¬       £»
£¨4£©Æû³µÎ²ÆøÖг£º¬ÓÐDºÍCO£¬¶þÕßÔÚ´ß»¯¼Á×÷ÓÃÏ¿ÉÒԴ󲿷Öת»¯ÎªÁ½ÖÖ¶Ô¿ÕÆøÎÞÎÛȾµÄÎïÖÊ£¬ÒÑÖª£ºF(g) + A(g) =" 2D" (g) £»¡÷H =" +180.5KJ/mol"
2C (s)+ O2 (g)=" 2CO(g)" £»¡÷H =" -221.0" KJ/mol
C (s)+ O2(g) = CO2(g)£»¡÷H =" -393.5" KJ/mol
ÔòÉÏÊöβÆøת»¯µÄÈÈ»¯Ñ§·½³ÌʽΪ£º                         ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2012½ìºÚÁú½­Ê¡¼Ñľ˹һÖиßÈýÉÏѧÆÚµÚËĴε÷Ñп¼ÊÔ»¯Ñ§ÊÔ¾í ÌâÐÍ£ºÌî¿ÕÌâ

ÏÂͼÉæ¼°µÄÎïÖÊËùº¬ÔªËØÖУ¬³ýÒ»ÖÖÔªËØÍ⣬ÆäÓà¾ùΪ¶ÌÖÜÆÚÔªËØ¡£ÒÑÖª£ºA¡¢FΪÎÞÉ«ÆøÌåµ¥ÖÊ£¬BΪ¾ßÓд̼¤ÐÔÆøζµÄÆøÌ壬CΪºÚÉ«Ñõ»¯ÎEΪºìÉ«½ðÊôµ¥ÖÊ£¨²¿·Ö·´Ó¦µÄ²úÎïδÁгö£©¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©BµÄµç×ÓʽΪ        £»
£¨2£©Ð´³öBºÍC·´Ó¦µÄ»¯Ñ§·½³Ìʽ                                            £»
£¨3£©Ð´³öEÓëGµÄÏ¡ÈÜÒº·´Ó¦µÄÀë×Ó·½³Ìʽ£¬²¢Óõ¥ÏßÇűê³öµç×ÓתÒƵķ½ÏòºÍÊýÄ¿£º
                                                      £»
£¨4£©J¡¢K¾ùÊÇͬÖÖ½ðÊôµÄÂÈ»¯ÎÇÒKΪ°×É«³Áµí¡£Ð´³öSO2»¹Ô­JÉú³ÉKµÄÀë×Ó·½³Ìʽ                                        £»
£¨5£©Æû³µÎ²ÆøÖг£º¬ÓÐDºÍCO £¬¶þÕßÔÚ´ß»¯¼Á×÷ÓÃÏ¿ÉÒԴ󲿷Öת»¯ÎªÁ½ÖÖ¶Ô¿ÕÆøÎÞÎÛȾµÄÎïÖÊ£¬ÒÑÖª£º    F(g) + A(g) =" 2D" (g)  ¡÷H =" +180.5KJ/mol"
2C (s)+ O2 (g)= 2CO(g) ¡÷H =" -221.0" KJ/mol
C (s)+ O2(g) = CO2(g)  ¡÷H =" -393.5" KJ/mol
ÔòÉÏÊöβÆøת»¯µÄÈÈ»¯Ñ§·½³ÌʽΪ£º                                   ¡£

²é¿´´ð°¸ºÍ½âÎö>>

¿ÆÄ¿£º¸ßÖл¯Ñ§ À´Ô´£º2010ÄêºÓ±±Ê¡¸ß¶þµÚ¶þѧÆÚÆÚÖп¼ÊÔ»¯Ñ§ÊÔÌâ ÌâÐÍ£ºÌî¿ÕÌâ

£¨10·Ö£©ÏÂͼÉæ¼°µÄÎïÖÊËùº¬ÔªËØÖУ¬³ýÒ»ÖÖÔªËØÍ⣬ÆäÓà¾ùΪ¶ÌÖÜÆÚÔªËØ¡£ÒÑÖª£ºA¡¢FΪÎÞÉ«ÆøÌåµ¥ÖÊ£¬BΪ¾ßÓд̼¤ÐÔÆøζµÄÆøÌ壬CΪºÚÉ«Ñõ»¯ÎEΪºìÉ«½ðÊôµ¥ÖÊ£¨²¿·Ö·´Ó¦µÄ²úÎïδÁгö£©¡£Çë»Ø´ðÏÂÁÐÎÊÌ⣺

£¨1£©BµÄµç×ÓʽΪ­­­­­         £»

£¨2£©Ð´³öBºÍC·´Ó¦µÄ»¯Ñ§·½³Ìʽ                              £»

£¨3£©Ð´³öEÓëGµÄÏ¡ÈÜÒºµÄÀë×Ó·½³Ìʽ£¬Ð´³öµç×ÓתÒÆÊýÄ¿£º            £¬        £»

£¨4£©Æû³µÎ²ÆøÖг£º¬ÓÐDºÍCO£¬¶þÕßÔÚ´ß»¯¼Á×÷ÓÃÏ¿ÉÒԴ󲿷Öת»¯ÎªÁ½ÖÖ¶Ô¿ÕÆøÎÞÎÛȾµÄÎïÖÊ£¬ÒÑÖª£ºF(g) + A(g) = 2D (g) £»¡÷H = +180.5KJ/mol

2C (s)+ O2 (g)= 2CO(g) £»¡÷H = -221.0 KJ/mol

C (s)+ O2(g) = CO2(g)£»¡÷H = -393.5 KJ/mol

ÔòÉÏÊöβÆøת»¯µÄÈÈ»¯Ñ§·½³ÌʽΪ£º                          ¡£

 

²é¿´´ð°¸ºÍ½âÎö>>

ͬ²½Á·Ï°²á´ð°¸