for the reaction..
PCl3(g)+Cl2(g)---->PCl5(g), ∆H=-x kJ

If the ∆H°PCl3 is -y kj, what is ∆H°PCl5?

1) (x-y) kJ
2) (y-x) kj
3) -(x+y) kj
4) (x+y) kj

-(x+y)
 
  • -8
let us assume that the delta H PCl5 be Z
delta H=delta H Product - delta H Reactant
delta H= -x (given)
delta Hreactant = -y

-x=-z-(-y)
-x=-z+y
-x-y=z
-(x+y)=z

Ans:  -(x+y)=delta H PCl5
 
  • 19
Then what about pcl3. Its enthalpy be should also included in reactants. Isn't it?
  • -6
See when we write X + Y —> XY (delta H = - x J), it means that on reaction of X and Y, x Joules were removed to give XY. Hence we can write it as X + Y - x Joules —> XY. In the question given, PCl3 has - y kJ and PCl5 has z (assume) and delta H is - x kJ. Reaction is: PCl3 + Cl2 - x kJ —> PCl5. So substitute energies instead of reactants as: - x kJ - y kJ = z kJ which will give you (-x -y) kJ or —(x+y) kJ
  • -2
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