Hess’s Law and Standard Enthalpies of Formation Unit 10 Lesson 4 .

Hess’s Law H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested. However, we can estimate H using published H values and the properties of enthalpy.

Hess’s Law The heat released or absorbed in a chemical process is the same whether the process takes place in one or several steps If two or more chemical equations can be added together to produce an overall equation, the sum of the enthalpy equals the enthalpy change of the overall equation. This is called the Heat of Summation, ∆H

Hess’s Law Steps Read through the whole question Plan a Strategy Evaluate the given equations. Rearrange and manipulate the equations so that they will produce the overall equation. (don’t forget to reverse the sign of ∆H if you reverse the reaction) Add the enthalpy terms.

H2O(g) + C(s) → CO(g) + H2(g) Example 1 Use these equations to calculate the molar enthalpy change which produces hydrogen gas. C(s) + ½ O2(g) → CO(g) ∆H = -110.5kJ H2(g) + ½ O2(g) → H2O(g) ∆H = -241.8kJ The above needs to be reversed to match our overall equation: H2O(g) → H2(g) + ½ O2(g) ∆H = +241.8kJ

H2O(g) + C(s) → CO(g) + H2(g) Use these equations to calculate the molar enthalpy change which produces hydrogen gas. C(s) + ½ O2(g) → CO(g) ∆H = -110.5kJ H2O(g) → H2(g) + ½ O2(g) ∆H = +241.8kJ _+____________________________________ C(s) + H2O(g) → H2(g) + CO(g) ∆H=+131.3kJ

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas in the reaction above. C4H10(g) + 6 ½ O2(g) → 4CO2(g) + 5H2O(g) ∆H= -2657.4kJ/mol C(s) + O2(g) → CO2(g) ∆H= -393.5kJ/mol H2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas in the reaction above. C4H10(g) + 6 ½ O2(g) → 4CO2(g) + 5H2O(g) ∆H= -2657.4kJ Needs to be switched because C4H10 is a product 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ The sign must change since we switched the equation

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas in the reaction above. 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ C(s) + O2(g) → CO2(g) ∆H= -393.5kJ Needs to be multiplied by 4 due to the 4C(s) in the equation 4C(s) + 4O2(g) → 4CO2(g) ∆H= 4(-393.5kJ) Distribute the 4

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas in the reaction above. 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol 4C(s) + 4O2(g) → 4CO2(g) ∆H= 4(-393.5kJ/mol) H2(g) + ½O2(g) → H2O(g) ∆H= -241.8kJ/mol Needs to be multiplied by 5 due to the 5H2 (g) in the equation

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas in the reaction above. 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol 4C(s) + 4O2(g) → 4CO2(g) ∆H= 4(-393.5kJ/mol) 5H2(g) + 2½O2(g) → 5H2O(g) ∆H= 5(-241.8kJ/mol)

Example 2 4C(s) + 5H2(g) → C4H10(g) Use these equations to calculate the molar enthalpy change which produces butane gas in the reaction above. 5H2O(g) + 4CO2(g) → 6 ½ O2(g) + C4H10(g) ∆H= +2657.4kJ/mol 4C(s) + 4O2(g) → 4CO2(g) ∆H= 4(-393.5kJ/mol) 5H2(g) + 2½O2(g) → 5H2O(g) ∆H= 5(-241.8kJ/mol) __+__________________________________________________ 4C(s) + 5H2(g) → C4H10(g) ∆H = -125.6kJ

Standard Enthalpy of Formation Standard Enthalpy (ΔH°f) – The change in enthalpy that accompanies the formation of one mole of a compound from its constituent elements. In other words, how much energy is gained or released when hydrogen combines with oxygen to form one mole of water 1H2 + 1/2O2  1H2O Works the same for any compound

Calculating ΔH rxn From ΔH°f The enthalpy of a reaction can be calculated from the heat of formation (DHf) of the substances in the reaction. H = nHf°products – mHf° reactants where n and m are the stoichiometric coefficients DHf = 0 kJ/mol for any element (including diatomics) DHf for compounds can be looked up in a chart. Values can be positive or negative, be careful with signs. “Sum of reactants” “Sum of Products” **This equation is NOT in your ref. pack!

Example Problem (DHf CO = -110.5 kJ, DHf CO2 = -393.5 kJ) What is the heat of reaction for the reaction 2CO + O2  2CO2 (DHf CO = -110.5 kJ, DHf CO2 = -393.5 kJ) DHrxn = Σ n DHf(products) – Σ m DHf(reactants) DHrxn = (2(-393.5 kJ)) – (2(-110.5 kJ) + 0 kJ) DHrxn = -787.0 kJ + 221.0 kJ DHrxn = -566.0 kJ Since there are 2 CO2’s Since there are 2 CO’s