To find H for this reaction:
1/2C --> 1/3D + B
I will use these two:
(#1) A + 2B --> C ........H= -22.04 kJ
#2) 3A --> 2D ...... ......H= 244.73 kJ
equation #2 is the only provider of "D", & to get 1/3 D produced,.... I will need to use 1/6th of equation #2
equation #1 is the only provider of "C",.. to get 1/2 C as a starting material, I will need to use the reverse of 1/2 of #1
"-1/2 of #1" : 1/2 C --> 1/2 A & 1B
"1/6 of #2" : 1/2 A --> 1/3 D
by combining thes two together the 1/2A --> 1/2 A cancel out & give youi what you asked for
1/2C --> 1/3D + B
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Hess's law says that if 1/6th of # 2 & -1/2 of # 1 give you the reaction that you want, then 1/6th of dH of #2 & -1/2 of dH of #1 gives youi the energy you wish
1/6th of dH#2 : (1/6)(244.73) = 40.79
-1/2 if dH of #1 : (- 1/2)(-22.04) = 11.02
your answer = 51.81kJ
