Hess’s Law
Hess’s Law
The value of the ∆H for any reaction that can be written in steps equals the sum of the values of ∆H for each of the individual steps.

Thermal equations 	
1                A        +         B        →        AB           ∆H1 
2                AB      +         B        →        AB2         ∆H2 
then,
                      A         +       2 B      → AB2            ∆H  = ∆H1 + ∆H2 	

Another way to state Hess’s law is: If two or more equations with known enthalpy changes can be added together to form a new “target” equation, then their enthalpy changes may be similarly added together to yield the enthalpy change of the target equation.

Hess’s discovery allowed chemists to determine the enthalpy change of a reaction without direct calorimetry, using two familiar rules for chemical equations and enthalpy changes:
•	If a chemical equation is reversed, then the sign of ∆H changes.
•	If the coefficients of a chemical equation are altered by multiplying or dividing by a constant factor, then the ∆H is altered in the same way.

What is the enthalpy change for the formation of one mole of butane (C4H10 ) gas from its elements? The reaction is:
4 C(s) + 5 H2 (g) → C4H10 (g)   ∆H° = ?
The following known equations, determined by calorimetry, are provided:
(1)  C4H10   (g) +  O2  (g) → 4 CO2 (g) + 5 H2O (g)  		∆H°1=  - 657.4 kJ
 (2)  C  (s)   + O2   (g) → CO2 (g)  				∆H°2 = - 393.5 kJ
(3) 2 H2 (g) + O2 (g) → 2 H2O  (g) 				∆H°3 =  - 483.6 kJ

reversing known equation (1), which will require multiplying its ∆H by -1, will make C4H10(g) a product; 
multiplying known equation (2) by 4 will provide the required amount of C(s) reactant; and 
multiplying known equation (3) by 5/2 will provide the required amount of H2(g) reactant. 

Cancellation when the equations are added will determine whether the required amount of O2(g) remains.

-1 x 	(1): 4 CO2 (g) + 5 H2O (g) → C4H10 (g) +  O2 (g)  		∆H°1= -1 x (-657.4 kJ) = 657.4 kJ
4 x  (2)  4 C  (s)   + 4 O2   (g) → 4 CO2 (g)  			∆H°2 = 4 x (- 393.5 kJ) = 1574.0 kJ
 x  (3) 5 H2 (g) +  O2 (g) → 5 H2O  (g) 			∆H°3 =   x (- 483.6 kJ) = 1209.0 kJ

4 CO2 (g) + 5 H2O (g) 4 C  (s)   + 4 O2   (g) 5 H2 (g) +  O2 (g) → C4H10 (g) +  O2 (g) + 4 CO2 (g) + 5 H2O (g)

leaving:
		4 c (S) + 5 H2 (g) → C4H10 (g)

If the known equations can be added together to form the target equation, then their enthalpy changes can be added together. In this case,

∆H°total = (-1) (∆H°1) + 4∆H°2 +  ∆H°3
∆H°total = (+657.4) + (- 1574.0) + (- 1209.0) kJ
∆H°total = -125.6 kJ
The enthalpy change for the formation of one mole of butane is -125.6 kJ.