SCH 4U - Hess' Law

Representing Enthalpy Changes

There are four methods:

• include an energy term in the thermochemical equation

CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(g) + 726 kJ

In instances like this, we write the heat term as "726 kJ" because we can see that this amount of energy is released per mole of methanol that undergoes combustion.  If we were to take this value out of the equation for use elsewhere, we would use it as "-726 kJ/mol."

• write a chemical equation and state the enthalpy change

CH₃OH(l) + 3/2O₂(g) → CO₂(g) + 2H₂O(g)   ΔH = -726 kJ

• state the molar enthalpy of a particular reaction

ΔH_c = -726 kJ/mol CH₃OH

In instances like this, we write the heat term as "-726 kJ/mol" because we wouldn't have the knowledge that we have one mol of methanol undergoing combustion unless we're told.

•  draw a potential energy diagram

 

Notice above that a potential energy diagram will look different depending on whether it represents an exothermic or endothermic reaction.  In exothermic, the systems starts with lots of energy and ends with less.  Conversely, the opposite happens for endothermic reactions.
  

Hess’ Law

• It is possible to calculate the enthalpy of a reaction from the tabulated values of other reactions, thus, it is not necessary to do the calorimetry for all reactions.

• Since enthalpy is state function, the enthalpy of a chemical reaction depends only on the amount of matter involved and the initial and final circumstances of the system.  (PS, this means that enthalpy is just like displacement - if we know where we start and we know where we end, it doesn't matter what path we take to get there).
  

• If the reaction can be carried out in one or in many steps, the sum of the enthalpies for all the steps is equivalent to the enthalpy for the single step (ΔH_goal = ΔH₁ + ΔH₂ + ΔH₃ + ···).  This is Hess' Law.
  

 

ex. Calculate the enthalpy of combustion for methane, 

CH₄(g)  +  2O₂(g)  →  CO₂(g)  +  2H₂O(l), 

given CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)   ΔH = -802 kJ

and                    H₂O(l) → H₂O(g)                  ΔH = 44 kJ

Answer:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)   ΔH = -802 kJ

            2H₂O(g) → 2H₂O(l)                  ΔH = -(44 x 2) kJ

_____________________________________________

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)     ΔH = -890 kJ

 

For a bit more guidance, I created a video that takes you through the thought process involved in the solution of #12 from the homework.  Here's another video that takes you through #11.

 

 Homework # 7-15