Heats of Formation of Solid Binary Compounds. 

 Table (continued). 



33 



Equation. 



M = 



^X 10000 

 no. of atoms 

 of halogen 



Heat of formation. 



N 

 M 



Constant. 



Calculated 

 =Mx2-5. 



N= found. 



( Pb 2 4- 2Br 2 =2PbBr 2 4- heat 1 

 [^ 5 +(2X-00404)= ^ + . j 



rSn 2 + 2Br 2 =2SnBr 2 4- heat 1 

 j ^+(2x00404)=^ +, \ 



C Cu 2 4- 2C1 2 =2CuCl 2 4- heat "1 



1 i^4+ (2x004fi0) =7-fr + • I 



f Sn Q + 2I 2 =2SnI 2 4- heat ] 



1 4- 2 - 2 4- x > 

 1503^ 383 589 T 



f Pb 2 4- 2I 2 =2PbT 2 4- heat "] 

 1 605 383 656 J 



17-8 

 15-8 



18-4 



95 



9-6 



44 j 

 39 | 



46 < 



24 



-■{ 



36-3 (B) 

 362 (B) 



38-5 (B) 

 32-8 (F&S) 



35-7 (B) 

 344 (B) 



32-9 (B) 

 31-0 (B) 

 25-8 (T) 

 305 (A) 

 29-5 (F & S) 



27-0 (B) 



25-2 (B) 

 264 (B) 

 23-2 (F & S) 

 269 



| 2-2 (38-5) 

 12-3(35-7) 



► 1-8 (32-9) 



J 

 2-8 (27-0) 



I 2-6 (25-2) 



|"Sn 2 + 4Br 2 =2SnB ri 4- heat "I 

 { 5 L 3 4-(4X00404)= 3-| + . } 

 r Sn 2 + 4I 2 =2SnI 4 + heat 1 



1 4- -*- = -i + * r 



1 503^ 383 419 T J 



144 

 9-5 



36 

 24 



28-8 (B) 

 20-0 (B) 



2-0 (28-8) 

 21 (20-0) 



|"A1 2 + 3C1 2 =A1 2 C1 6 4- heat j 

 {_1_+(3X 00460)= ^ + x j 

 f Al 2 4- 3Br 2 = Al 2 Br 6 4- heat j 

 {_1_+(3X -00404)= 1. + . } 



r Sb 4 4- 6C1 2 =4SbCl 3 4- heat j 

 (^+(6X00460)= 3 -| + . } 



fAl 2 4- 3I 2 =A1 2 I 6 4- heat] 



1 4- a = i- + * r 



I 873^ 383 398 T J 

 fP 4 + 6I 2 -4PI 3 4- heat] 



1 X 4- 6 ~ 4 4- x \ 

 1317 383 328 T J 



21-2 

 17-6 

 14-5 

 10-8 

 55 



53 | 

 44 

 36 J 



27 

 14 



53-6 (B) 

 53-6 (T) 



44-2 (B) 



30-4 (B) 

 305 (A) 



28-8 



23-4 (B) 

 3-5 (B) 



1 2-5 (53-6) 

 2-5 (44-2) 



y 2-1 (30-5) 

 2-2 (23 4) 

 0'6 (3-5) 



Phil. Mag. S. 5. Vol. 11. No. 65. Jan. 1881. 



D 



