Hess' Law
Hess' Law states that the enthalpy of a reaction is independent of whether the reaction occurs in one or several steps. If this was not the case energy would not be conserved. But because it is supported by expereimental data, it allows us to algebraically add equations and their accompanying DH's to obtain the DH for the desired or target equation.
Keep the following rules in mind:
Example 1 Carbon disulfide is a very flammable solvent. It burns according to the following equation:
CS2(l) + 3 O2(g) --> CO2(g) + 2 SO2(g)
Calculate DH for the above reaction using the following data:
(1) C(s) + 2 S(s) --> CS2(l) DH = 88 kJ
(2) C(s) + O2(g) --> CO2(g) DH = -394 kJ
(3) S(s) + O2(g) --> SO2(g) DH = -297 kJ
Solution Equation (1) is the only one with CS2(l), but it has the compound as a product. Our target equation has CS2(l) on the left hand side. So we reverse equation (1) and change the sign of DH.
reverse(1) CS2(l) --> C(s) + 2 S(s) DH = - 88 kJ
Equation (2) is the only one with CO2(g), and,
compared to the target equation, it has the correct amount on the desired side
of the equation. So we keep equation (2) as is.
(2) C(s)
+ O2(g) --> CO2(g) DH = -394 kJ
Equation (3) is the only one with SO2(g). It 's
on the correct side of the equation, but the target equation needs twice as
much. So we multiply equation (3) by 2.
Eq(3) *2 2S(s) + 2O2(g)
--> 2SO2(g) DH = 2(-297 kJ) = -594 kJ
Now if we add the above
three equations:
CS2(l) + C(s) + O2(g) + 2S(s) + 2O2(g) --> C(s)
+ 2 S(s) +
CO2(g) + 2SO2(g) DH =- 88 -394-594 = -1076 kJ
Notice that C(s) and 2S(s) cancel. We are left
with the target equation
CS2(l) + 3 O2(g) -->
CO2(g) + 2 SO2(g) and of course
the DH which we needed.
Answer: DH = -1076 kJ