Question

Hess's law of heat summation states that when a chemical equation can be written as the sum of two or more steps, the enthalpy change for the overall equation is equivalent to the sum of the enthalpy changes for the individual steps. Basically, the enthalpy change for the entire chemical change is the same no matter how you go from a reactant to a product. The book describes finding the enthalpy change for the combustion of graphite to carbon monoxide using the equation: 2C(graphite) + O2(g) ----> 2CO(g). To determine the enthalpy change of pure carbon monoxide from graphite and oxygen, you have to apply this law. This is done using 2 steps:

Step 1) 2C(graphite) + 2O2(g) ----> 2CO2(g), which shows the burning of two mol of graphite in 2 mol of oxygen and producing 2 mol of carbon dioxide. Step 2) 2CO2(g) ---> 2CO(g) + O2(g), which shows decomposing carbon dioxide to give 2 mole of carbon monoxide and 1 mol of oxygen. Then you have to find the enthalpy change for each step and so on. Can you give me an example of when to use this law and show the steps to solve the equation?

Answer 1

The enthalpy change for the reaction can be calculated from the given equations using Hess law

2C(graphite) + O2(g) ----> 2CO(g).

Step 1) 2C(graphite) + 2O2(g) ----> 2CO2(g) ; dH1 = -786 KJ

Step 2) 2CO2(g) ---> 2CO(g) + O2(g) ; dH2 = -566 KJ

------------------------------------------------------adding the above equation we get

2C(graphite) + O2(g) ----> 2CO(g). ; dH = dH1+dH2

Since the required equation got by adding the given equations their dH should add hence we get

dH = -786+(-566) = -1352 KJ

Consider the following example

2CH₄(g) ---> C₂H₆(g) + H₂(g)

1) The data given for the three reactions:

H2 + (1/2)O2 ---> H2O	ΔH = -285.8 kJ
CH4 + 2O2 ---> CO2 + 2H2O	ΔH = -890.4 kJ
C2H6 + (7/2)O2 ---> 2CO2 + 3H2O	ΔH = -1559.9 kJ

2) Manipulate the data equations such that to get the required equation

a) flip (puts hydrogen as a product)
b) multiply by 2 (gives us 2CH₄ on the reactant side)
c) flip (puts C₂H₆ as a product)

3) Then we obtain these data equations:

H2O ---> H2 + 1/2O2	ΔH = +285.8 kJ
2CH4 + 4O2 ---> 2CO2 + 4H2O	ΔH = -1780.8 kJ
2CO2 + 3H2O ---> C2H6 + (7/2)O2	ΔH = +1559.9 kJ

4) When you add the three equations above, the 4O₂ will cancel as will the 2CO₂ and the 4H₂O. Adding the three enthalpies yields the answer to the problem, +64.9 kJ.