ÌâÄ¿ÄÚÈÝ
ÏÂÁÐÓйØÈÈ»¯Ñ§·½³ÌʽµÄÐðÊöÕýÈ·µÄÊÇ
A£®ÒÑÖª2H2O(g) £½2H2(g)£«O2(g) ¡÷H£½+483.6 kJ/mol£¬ÔòÇâÆøµÄȼÉÕÈÈΪ
¡÷H£½£241.8kJ/mol
B£®ÒÑÖªC£¨Ê¯Ä«£¬s£©£½ C£¨½ð¸Õʯ£¬s£©¡÷H£¾0£¬Ôò½ð¸Õʯ²»ÈçʯīÎȶ¨
C£®ÒÑÖªÖкÍÈÈΪ¡÷H£½£57£®4kJ/mol £¬ÔòÏ¡´×ËáºÍÏ¡NaOHÈÜÒº·´Ó¦µÄÈÈ»¯Ñ§·½³Ìʽ
Ϊ£ºNaOH(aq)£«CH3COOH(aq)£½CH3COONa(aq)£«H2O(l) ¡÷H£½£57.4kJ/mol
D£®ÒÑÖª2C(s)£«2O2(g)£½2CO2 (g) ¡÷H1£»2C(s)£«O2 (g)£½2CO(g) ¡÷H2£¬Ôò¡÷H1£¾¡÷H2
B
ͼ°ÆäºÏ½ðÔÚÉú²úºÍÉú»îÖж¼Óй㷺µÄÓ¦Óá£Íº¬Á¿ºÜ¸ßµÄͺϽð³Æ×ÏÍ»òºìÍ;ÒÔпΪÖ÷ÒªÌí¼ÓÔªËØµÄÍºÏ½ð³Æ»ÆÍ;ÒÔÄøÎªÖ÷ÒªÌí¼ÓÔªËØµÄÍºÏ½ð³Æ°×Í;»ÆÍÖк¬ÓÐÉÙÁ¿Äø³ÆÄø»ÆÍ¡£
ijѧϰС×éÄâ¶ÔÉÏÊöͼ°ÆäºÏ½ð½øÐÐ̽¾¿¡£
(1)͵ĺËÍâµç×ÓÅŲ¼Ê½:¡¡¡¡¡¡¡¡¡¡¡¡,ÄøÔÚÖÜÆÚ±íÖÐλÓÚ¡¡¡¡¡¡ÖÜÆÚ¡¢
¡¡¡¡¡¡×å¡£
(2)ÏÖÓÐijÖÖͺϽð,¾²â¶¨º¬ÓÐÄøÔªËØ,¶Ô¸ÃͺϽðÀàÐÍ×÷ÈçÏÂ̽¾¿:
¼ÙÉèÒ»:¸ÃͺϽðÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡£
¼ÙÉè¶þ:¸ÃͺϽðÊÇ¡¡¡¡¡¡¡¡¡¡¡¡¡£
(3)Ϊȷ¶¨¸ÃͺϽðµÄ×é³É,Ä㻹ÐèÒª½øÐеÄʵÑé:
| ʵÑé²½Öè | Ô¤ÆÚʵÑéÏÖÏóºÍ½áÂÛ |
| ¢ÙÓÃï±µ¶½«ÍºÏ½ð´ì³É·ÛÄ©,³ÆÈ¡Ò»¶¨ÖÊÁ¿µÄͺϽð·ÛÄ© | |
| ¢Ú | |
| ¢Û |
ÒÑÖª:ÄøºÍÌúÐÔÖÊÏàËÆ,пºÍÂÁÐÔÖÊÏàËÆ¡£
±íÖÐʵÑé²½Öè¿ÉÒÔ²»ÌîÂú,Ò²¿ÉÒÔ×ÔÐÐÌí¼Ó¡£
(4)È¡1.50 g¸ÃͺϽð,Óë×ãÁ¿¼î·´Ó¦²úÉúÆøÌåÌå»ý(ÒѾÕÛËã³É±ê×¼×´¿öϵÄÌå»ý)µÄͼÏñÈçͼËùʾ,¸ÃºÏ½ðº¬Ð¿ÖÊÁ¿·ÖÊý¡¡¡¡¡¡¡¡(СÊýµãºó±£ÁôһλÊý×Ö)¡£
![]()
¸ù¾ÝÓÒϱíÊý¾Ý£¬ÏÂÁÐÅжÏÕýÈ·µÄÊÇ
| µçÀë³£Êý(25¡æ) |
| HF£º Ki = 3.6¡Á10 - 4 |
| H3PO4£º Ki1= 7.5¡Á10 -3£¬ Ki2= 6.2¡Á10 -8£¬ Ki3= 2.2¡Á10 -13 |
A. ÔÚµÈŨ¶ÈµÄNaF¡¢NaH2PO4»ìºÏÈÜÒºÖУ¬
c(Na+)+c(H+)=c(F£)+ c(H2PO4£)+ c(OH£)
B. ÉÙÁ¿H3PO4ºÍNaF·´Ó¦µÄÀë×Ó·½³ÌʽΪ:
H3PO4 +F£= H2PO4£+HF
C. ͬÎÂͬŨ¶Èʱ£¬ÈÜÒºµÄpH£ºNaF£¾NaH2PO4£¾Na2HPO4
D. ½áºÏH+ µÄÄÜÁ¦£ºH2PO4££¾ HPO42££¾F£