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6-11 ijζÈÏ£¬Í­·Û¶ÔÇâÆøÎü¸½·þ´ÓLangmuir¹«Ê½£¬Æä¾ßÌåÐÎʽΪ £¬Ê½ÖÐVÊÇÍ­·Û¶ÔÇâÆøµÄÎü¸½Á¿£¨273.15K, ϵÄÌå»ý£©, pÊÇÇâÆøÑ¹Á¦¡£ÒÑÖªÇâ·Ö×Óºá½ØÃæ»ýΪ13.108¡Á10-22m3,Çó1kgÍ­·ÛµÄ±íÃæ»ý¡£

½â£ºV=V¡ÞKp/(1+Kp) ,Çó³ö V¡Þ=1.36 dm3kg-1 ,a0= V¡ÞA0 NA/22.4=48 m2kg-1

6-12 ÔÚ-192.4¡æÊ±£¬Óù轺Îü¸½N2Æø¡£²â¶¨ÔÚ²»Í¬Æ½ºâѹÁ¦Ï£¬Ã¿Ç§¿Ë¹è½ºÎü¸½N2µÄÌå»ý£¨±ê×¼×´¿ö£©ÈçÏ£º p/(kPa)

8.886 13.93 20.62 27.73 33.77 37.30 V/(dm3)

33.55 36.56 39.80 42.61 44.66 45.92

ÒÑÖª-192.4¡æÊ±N2µÄ±¥ºÍÕôÆøÑ¹Îª147.1kPa, N2·Ö×Ó½ØÃæ»ýΪ16.2¡Á10-20m2,ÓÃBET¹«Ê½ÇóËùÓù轺±È±íÃæ»ý¡£

½â£º´¦ÀíÊý¾Ý£¬ÒÔp/[V(p*-p)]¶Ôp/p* ×÷ͼ£¬Ö±ÏßµÄбÂÊ£½0.17¡Á10£­3 dm£­3 £¬½Ø¾à£½28.3¡Á10£­3 dm£­3 £¬V¡Þ£½1/(бÂÊ£«½Ø¾à)£½35.12 dm3£¬a0= V¡ÞA0 NA/22.4=1.53¡Á105m2 kg-1

6-14 º¬Fe2OaŨ¶ÈΪ1.5 kg m-3µÄÈܽº£¬Ï¡ÊÍ10000±¶ºó£¬ÔÚ³¬ÏÔ΢¾µÏ¹۲죬Êý³öÊÓÒ°ÖпÅÁ£Æ½¾ùΪ4.1¸ö£¨ÊÓÒ°Ãæ»ýΪ°ë¾¶4.0¡Á10-5mµÄÔ²£¬ºñ¶È4.0¡Á10-5m£©£¬ÒÑÖªÖʵãµÄÃÜ

33

¶ÈpΪ5.2¡Á103kg m-3,É轺Á£ÎªÇòÐΣ¬ÊÔ¼ÆËã´Ë½ºÁ£Æ½¾ù°ë¾¶¡£

½â£º4¦Ðr3 /3 = cV/(¦ÑN) , Çó³ör= 0.7¡Á10-7m

6-15 Fe(OH)3ÈܽºÓÚ298K ͨµç45·ÖÖÓ£¬½çÃæÒÆ¶¯10mm£®µç³¡Ç¿¶ÈΪ2 Vcm-1£®ÒÑ֪ˮµÄÏà¶ÔµçÈÝÂÊΪ79£¬Õ³¶ÈΪ0.89¡Á10-3 Pa s,ÇóÈܽºµÄ µçÊÆ£¿£¨Õæ¿ÕµçÈÝÂÊ =8.854¡Á10-12 F m-1£©

½â£º =[¦Çu/( D)](d¦Õ/dl)-1=0.0236V

6-16 ÔÚ298Kʱ,ĤÁ½±ßÀë×Ó³õʼŨ¶È·Ö²¼ÈçÏÂ,×ó±ßRClÈÜҺŨ¶ÈΪ0.1mol dm-3,Ìå»ýΪ1dm3,ÓÒ±ßNaClÈÜҺŨ¶ÈΪ0.1mol dm-3,Ìå»ýΪ2dm3,ÎʴﵽĤƽºâºó,ÆäÉøÍ¸Ñ¹Îª¶àÉÙ?(RClΪ¸ß·Ö×Óµç½âÖÊ,¼ÙÉèÍêÈ«µçÀë,´ïµ½Ä¤Æ½ºâǰºó,Á½±ßÈÜÒºÌå»ý²»±ä)¡£

½â£ºÄ¤Æ½ºâ (0.1+2x)2x=(0.1-x)2, ½â³öx=0.0215 mol dm-3 , ¦¤c =0.129 mol dm-3

¦Ð=¦¤c RT=320kPa

6-17 ijһ´ó·Ö×ÓÈÜÒºÔÚ300Kʱ,²âµÃÓйØÉøÍ¸Ñ¹µÄÊý¾ÝΪ

c /(g dm-3) 0.5 1.00 1.50

(¦Ð/c)/(Pa g -1dm3) 101.3 104.3 106.4

34

ÊÔÇó´Ë´ó·Ö×ÓµÄÊý¾ù·Ö×ÓÁ¿¡£

½â£ºÒÔ(¦Ð/c)¶Ôc×÷ͼ£¬Ö±ÏߵĽؾࣽ98 Pa g -1dm3 £¬ £½RT/½Ø¾à£½2.5¡Á10 4 g mol-1

µÚËÄÆª »¯Ñ§¶¯Á¦Ñ§

µÚÆßÕ »ùÔª·´Ó¦¶¯Á¦Ñ§ Á· ϰ Ìâ

7-2 »ùÔª·´Ó¦,2A(g)+B(g)£½£½E(g),½«2molµÄAÓë1molµÄB·ÅÈë1ÉýÈÝÆ÷ÖлìºÏ²¢·´Ó¦,ÄÇô·´Ó¦ÎïÏûºÄÒ»°ëʱµÄ·´Ó¦ËÙÂÊÓë·´Ó¦ÆðʼËÙÂʼäµÄ±ÈÖµÊǶàÉÙ£¿:

½â£º[A]:[B]= 2:1 , ·´Ó¦ÎïÏûºÄÒ»°ëʱ [A]=0.5[A]0 ,[B]= 0.5[B]0 , r = k[A]2 [B]

r : r0= 1 : 8

7-3 ·´Ó¦aA==D£¬A·´Ó¦µô15/16ËùÐèʱ¼äÇ¡ÊÇ·´Ó¦µô3/4ËùÐèʱ¼äµÄ2±¶£¬Ôò¸Ã·´Ó¦ÊǼ¸¼¶¡£

½â£ºr = k[A]n , n=1ʱ t = ln ([A]0/[A])/k , t (15/16) : t (3/4) = ln16/ ln4 = 2

35

7-4 Ë«·Ö×Ó·´Ó¦2A(g) B(g) + D(g),ÔÚ623K¡¢³õʼŨ¶ÈΪ0.400mol dm£­3ʱ,°ëË¥ÆÚΪ105s,ÇëÇó³ö

(1) ·´Ó¦ËÙÂʳ£Êýk

(2) A(g)·´Ó¦µô90%ËùÐèʱ¼äΪ¶àÉÙ?

(3) Èô·´Ó¦µÄ»î»¯ÄÜΪ140 kJ mol-1, 573KʱµÄ×î´ó·´Ó¦ËÙÂÊΪ¶àÉÙ?

½â£º(1) r = k[A]2 , t 0.5= 1/(2 k[A]0) , k = 0.012dm3mol-1s-1

(2) 1/[A] ¨C 1/[A]0 =2 k t , t = 945 s

(3) ln(k/k¡¯)=(Ea/R)(1/T ¡¯-1/T) , 573Kʱk = 0.00223dm3mol-1s-1,

×î´ó·´Ó¦ËÙÂÊrmax = k[A]02=3.6¡Á10£­4 moldm-3s-1.

7-5 500KÊ±ÆøÏà»ùÔª·´Ó¦A + B = C£¬ µ±AºÍBµÄ³õʼŨ¶È½ÔΪ0.20 mol dm-3ʱ£¬³õʼËÙÂÊΪ5.0¡Á10-2 mol dm-3 s-1

(1) Çó·´Ó¦µÄËÙÂÊϵÊýk£»

(2) µ±·´Ó¦ÎïA¡¢BµÄ³õʼ·Öѹ¾ùΪ50 kPa£¨¿ªÊ¼ÎÞC£©£¬Ìåϵ×ÜѹΪ75 kPaʱËùÐèʱ¼äΪ¶àÉÙ£¿

½â£º(1) r0 = k[A]0 [B]0 , k =1.25 dm3 mol-1 s-1

(2) p0(A) = p0(B) , r = kp p (A) 2 , p =2 p0(A) - p (A) , p (A)= p0(A)/ 2 , kp = k/(RT) ,

36

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