ÌâÄ¿ÄÚÈÝ

ijʵÑéС×éÓÃ0.50 mol/L NaOHÈÜÒººÍ0.50 mol/L H2SO4ÈÜÒº½øÐÐÖкÍÈȵIJⶨ¡£
¢ñ.ÅäÖÆ0.50 mol/L NaOHÈÜÒº
£¨1£©ÈôʵÑéÖдóԼҪʹÓÃ245 mL NaOHÈÜÒº£¬ÔòÖÁÉÙÐèÒª³ÆÁ¿NaOH¹ÌÌå      g¡£
£¨2£©´ÓÏÂͼÖÐÑ¡Ôñ³ÆÁ¿NaOH¹ÌÌåËùÐèÒªµÄÒÇÆ÷(ÌîÐòºÅ)        ¡£
Ãû³Æ
ÍÐÅÌÌìƽ(´øíÀÂë)
СÉÕ±­
ÛáÛöǯ
²£Á§°ô
Ò©³×
Á¿Í²
ÒÇÆ÷






ÐòºÅ
a
b
c
d
e
f
 
¢ò.²â¶¨ÖкÍÈȵÄʵÑé×°ÖÃÈçͼËùʾ

£¨1£©Ð´³öÏ¡ÁòËáºÍÏ¡ÇâÑõ»¯ÄÆÈÜÒº·´Ó¦±íʾÖкÍÈȵÄÈÈ»¯Ñ§·½³Ìʽ(ÖкÍÈÈÊýֵΪ57.3 kJ¡¤mol£­1)£º_______________________________________¡£
£¨2£©È¡50 mL NaOHÈÜÒººÍ30 mLÁòËá½øÐÐʵÑ飬ʵÑéÊý¾ÝÈçÏÂ±í¡£
¢ÙÇëÌîдϱíÖеĿհףº
ζÈ
ʵÑé´ÎÊý
³¬Ê¼Î¶Èt1/¡æ
ÖÕֹζÈt2/¡æ
ƽ¾ùζȲî
(t2£­t1)/¡æ
H2SO4
NaOH
ƽ¾ùÖµ
1
26.2
26.0
26.1
30.1
 
2
27.0
27.4
27.2
33.3
3
25.9
25.9
25.9
29.8
4
26.4
26.2
26.3
30.4
 
¢Ú¸ù¾ÝÉÏÊöʵÑéÊý¾Ý¼ÆËã³öµÄÖкÍÈÈΪ53.5 kJ/mol£¬ÕâÓëÖкÍÈȵÄÀíÂÛÖµ57.3 kJ/molÓÐÆ«²î£¬²úÉúÆ«²îµÄÔ­Òò¿ÉÄÜÊÇ(Ìî×Öĸ)______¡£
a£®ÊµÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û²î
b£®ÔÚÁ¿È¡NaOHÈÜÒºµÄÌå»ýʱÑöÊÓ¶ÁÊý
c£®·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÏ¡ÁòËáµÄСÉÕ±­ÖÐ
d£®ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨H2SO4ÈÜÒºµÄζÈ
¢ñ£¨1£©5.0 £¨2£©abe 
¢ò£¨1£©1/2H2SO4(aq)+NaOH(aq)=1/2Na2SO4(aq)+H2O(l)  ¦¤H="-57.3KJ/mol" £¨2£© ¢Ù   4.0 ¢Úa c d

ÊÔÌâ·ÖÎö£º¢ñ£¨1£©ÈÝÁ¿Æ¿Óë245 mL NaOHÈÜÒº½Ó½üµÄ¹æ¸ñÊÇ250 mLµÄ¡£ÈÜÒºÓоùÒ»ÐÔ£¬¸÷´¦µÄŨ¶ÈÏàͬ¡£n(NaOH)="V¡¤C=0.25L¡Á0.50" mol/L="0.125" mol.m(NaOH)="n¡¤M=0.125" mol¡Á40g/mol=5.0g.£¨2£©³ÆÁ¿¹ÌÌåÒ©Æ·ÒªÓÃÌìƽ¡¢Ò©³×¡£ÓÉÓÚÇâÑõ»¯ÄÆÓи¯Ê´ÐÔËùÒÔ²»ÄÜÖ±½ÓÓÃÌìƽÀ´³ÆÁ¿£¬Òª·ÅÔÚÉÕ±­ÖгÆÁ¿¡£Òò´Ë³ÆÁ¿NaOH¹ÌÌåËùÐèÒªµÄÒÇÆ÷´úÂëÊÇa b e¢ò£¨1£©ÖкÍÈÈÊÇËá¼î·¢Éú·´Ó¦²úÉú1Ħ¶ûµÄˮʱ·Å³öµÄÈÈ¡£Ç¿ËáÓëÇ¿¼î·¢ÉúÖкͷ´Ó¦±íʾÖкÍÈȵÄÈÈ»¯Ñ§·½³ÌʽΪ£º1/2H2SO4(aq)+NaOH(aq)=1/2Na2SO4(aq)+H2O(l)  ¦¤H="-57.3KJ/mol" £¨2£©¢Ù·ÖÎöÉÏÊöËÄ×éÊý¾Ý¿ÉÒÔ¿´³ö£ºµÚ¶þ×éÊý¾ÝÎó²îÌ«´ó£¬ÒªÈ¥µô¡£Çó³öÆäÓàÈý×éÊý¾ÝµÄζȲîµÄƽ¾ùÖµ¾ÍÊÇƽ¾ùζȲ{£¨30.1-26.1£©+£¨29.8-25.9£©+£¨30.4-26.3£©}¡Â3=4.0¡£¢ÚÖкÍÈȵÄʵÑéֵΪ53.5 kJ/mol±ÈÖкÍÈȵÄÀíÂÛÖµ57.3 kJ/molС£¬¿ÉÄܵÄÔ­ÒòÊÇa£®ÊµÑé×°Öñ£Î¡¢¸ôÈÈЧ¹û²î¡£c£®·Ö¶à´Î°ÑNaOHÈÜÒºµ¹ÈëÊ¢ÓÐÏ¡ÁòËáµÄСÉÕ±­ÖÐd£®ÓÃζȼƲⶨNaOHÈÜÒºÆðʼζȺóÖ±½Ó²â¶¨H2SO4ÈÜÒºµÄζȡ£¶ÔÓÚb£®ÔÚÁ¿È¡NaOHÈÜÒºµÄÌå»ýʱÑöÊÓ¶ÁÊý£¬ÔòÇâÑõ»¯ÄƵÄÌå»ýƫС£¬²úÉúµÄË®µÄÎïÖʵÄÁ¿Æ«Ð¡ÖкÍÈȵÄÊýÖµ¾ÍÆ«´ó¡£ÓëÌâ¸ÉÏàÎ¥±³¡£¹Ê²â¶¨µÄÖкÍÈÈƫСµÄÔ­ÒòÊÇa c d¡£
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿
ΪÁ˲ⶨº¬ÓÐH2C2O4¡¤2H2O¡¢KHC2O4ºÍK2SO4µÄÊÔÑùÖи÷ÎïÖʵÄÖÊÁ¿·ÖÊý£¬½øÐÐÈçÏÂʵÑ飺
¢Ù³ÆÈ¡6.0 gÊÔÑù£¬¼ÓË®Èܽ⣬Åä³É250 mLÊÔÑùÈÜÒº¡£
¢ÚÓÃËáʽµÎ¶¨¹ÜÁ¿È¡25.00 mLÊÔÑùÈÜÒº·ÅÈë׶ÐÎÆ¿ÖУ¬²¢¼ÓÈë2¡«3µÎ·Ó̪ÊÔÒº£¬ÓÃ0.2500 mol/L NaOHÈÜÒºµÎ¶¨£¬ÏûºÄNaOHÈÜÒº20.00 mL¡£
¢ÛÔÙÈ¡25.00 mLÊÔÑùÈÜÒº·ÅÈëÁíһ׶ÐÎÆ¿ÖУ¬ÓÃ0.1000 mol/LµÄËáÐÔ¸ßÃÌËá¼ØÈÜÒºµÎ¶¨£¬ÏûºÄ¸ßÃÌËá¼ØÈÜÒº16.00 mL¡£
»Ø´ðÏÂÁÐÎÊÌ⣺
£¨1£©ÒÑÖª£º0.10 mol/L KHC2O4ÈÜÒºpHԼΪ3£¬ÆäÖк¬Ì¼ÔªËصÄÁ£×ÓŨ¶ÈÓÉ´óµ½Ð¡µÄ˳ÐòΪ        ¡£
£¨2£©²½Öè¢ÙËùÐèÒªµÄ²£Á§ÒÇÆ÷ÓÐÉÕ±­¡¢²£Á§°ô¡¢                              ¡£
£¨3£©Íê³É²¢ÅäƽÏÂÁÐÀë×Ó·½³Ìʽ£º   C2O42-+   MnO4-+   H+=   CO2+   Mn2++             
£¨4£©µÎ¶¨Ê±±ßµÎ±ßÒ¡¶¯×¶ÐÎÆ¿£¬ÑÛ¾¦Ó¦¹Û²ì                            ¡£
£¨5£©²½Öè¢ÚÖÐÁ¿È¡ÊÔÑùÈÜҺʱ£¬ËáʽµÎ¶¨¹ÜÓÃÕôÁóˮϴ¹ýºóûÓÐÈóÏ´£¬Ôò²âµÃµÄH2C2O4¡¤2H2OµÄÖÊÁ¿·ÖÊý       ¡££¨Ìî¡°Æ«´ó¡±¡¢¡°Æ«Ð¡¡±»ò¡°ÎÞÓ°Ï족£©
£¨6£©²½Öè¢ÛÖÐÅжϵζ¨ÖÕµãµÄ·½·¨ÊÇ                                 ¡£
£¨7£©ÊÔÑùÖÐH2C2O4¡¤2H2OµÄÖÊÁ¿·ÖÊýΪ             ¡£

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

¾«Ó¢¼Ò½ÌÍø