How to Calculate the Molar Heat of Decomposition
- 1). Write out the balanced equation for the decomposition reaction. Websites, such as Jefferson Lab (education.jlab.org/elementbalancing/index.html), are available to practice or review writing decomposition reaction equations.
Example: Suppose water is electrolyzed to yield hydrogen and oxygen; what would the balanced reaction be?
2 H2O ----> 2 H2 + O2 - 2). Divide all the coefficients in front of products or reactants in the equation by the same number so that the coefficient for the compound of interest is 1.
Example: For molar heat of decomposition, you want to know the energy released or absorbed when one mole of a compound decomposes. For water, then, you want to know the change associated with one mole of water. The equation, however, has the coefficient 2 in front of the water. Dividing everything on both sides of the equation by 2 to yield the following:
H2O ---> H2 + 1/2 O2
Notice that you have a 1/2 coefficient on the oxygen; while this would normally be inappropriate, in this case it's fine. - 3). Use a website such as the NIST WebBook, (webbook.nist.gov/chemistry/form-ser.html) where you can search for compounds by formula. Type in the formulas for each of the products and reactants that are not an element in its most stable state. Search for thermodynamic gas phase data. Retrieve the molar enthalpy of formation or "fH° for each. An element in its most stable state will have a "fH° of 0.
Example: The "fH° for liquid water is -285.83 kJ/mol per NIST WebBook. Oxygen and hydrogen gas are both elements in their "standard" state, so the "fH° for each will be 0. - 4). Subtract the sum of the "fH° of the decomposition reactants from the sum of the "fH° of the decomposition products.
Example: Water is the only reactant, and its "fH° is -285.83, so you subtract this value from the sum of the "fH° of the products. Both products have a "fH° of 0, so 0 - -285.83 = +285.83 kJ/mol. This is the molar heat of decomposition for water.
