1.2.5 Hess’s Law- the equation

The equation There is another way to calculate enthalpy changes based on the principal of Hess's Law. If you are not given the intermediate reactions then: ΔH for a reaction may be calculated using ΔHf values and the equation: ΔH = ∑ΔH (products) – ∑ΔH (reactants)

The Equation You may not be familiar with the ∑ symbol. It stands for "summation" or "the sum of". To find ΔHo for the reaction, add together all ΔHof of the products and subtract ΔHof of all of the reactants.

Use a Table of Thermochemical Data to locate ΔHf values for all reactants and products

Things to watch out for: The physical state is important (s,l,g,aq) The balancing coefficients in the equation, as you must multiply the ΔHf values by the coefficients. be very careful with + and - values. you should begin by writing all the ΔHf values directly below all participants in the equation

Example: Using a Table of Thermochemical Data, calculate ΔH for the combustion of benzene, C6H6, as shown by the following reaction: C6H6 (l) + 15/2 O2 (g) → 6 CO2 (g) + 3 H2O (l)

Solution: Remember that ΔH for any pure element = 0. (some exceptions) What are Δ H values for C6H6 (l) , CO2(g), and H2O(l). Remember, these ΔH values are given for 1 mole. In our final reaction; There are 6 mole of CO2, so multiply ΔH by 6. There are 3 mole of H2O, so multiply ΔH by 3.

From the table C6H6 (l) ΔH = +49.0 kJ 6 CO2 (g) ΔH = 6(-393.5 kJ)= -2361kJ 3 H2O (l) ΔH = 3(-285.8 kJ)= -857.4 kJ 15/2 O2 (g) ΔH = 0 kJ

ΔH = ∑ΔHproducts - ∑ΔHreactants Using the formula ΔH = ∑ΔHproducts - ∑ΔHreactants C6H6 (l)+15/2 O2 (g)→ 6 CO2 (g)+3 H2O (l) 49.0 + (15/2 ×0) → (6×-393.5)+(3×-285.8) 49.0 -3218.4 Reactants Products

answer ΔH = ∑ΔHproducts – ∑ΔHreactants ΔH =-3218.4 – (+49.0) ΔH = -3267.4 kJ   

Common Sources of Error Forgetting to multiply ΔH values by the appropriate coefficient. Using the wrong value of ΔH for water: ΔHf° for H2O(l) = -285.8 kJ/mol; ΔHf° for H2O(g) = -241.8 kJ/mol Solving for ΔH as "Reactants - Products" instead of "Products – Reactants". Accidentally changing the sign for ΔH.

Practice Problem What is the standard heat of reaction for the reaction of gaseous carbon monoxide with oxygen to form gaseous carbon dioxide? 2CO(g) + O2 2 CO2 (g) Hint- use your table to find the heat of formation values for CO and CO2

Solution Using your thermochemical data table you find that ∆H˚f CO(g) =-110.5 kJ/mol ∆H˚f CO2(g)= -383.5 kJ/mol ∆H˚f O2(g)= 0 kJ (recall: all elements have a ∆H˚f of 0 kJ)

Find the ∆H˚f of all reactants, (remembering to multiple by the number of moles of each) (2 mol CO X -110kJ/mol) + (1 mol O2 x 0 kJ/mol)= -221.0 kJ Find the ∆H˚f of the products: (2 mol CO2 x -393.5 kJ/mol) = -787.0 kJ Total ∆H˚= (∑products- ∑ reactants) = (-787.0 kJ) – (-221.0 kJ)= -566.0 kJ The reaction is exothermic (-∆H˚)

assignment 1.2.5- practice problems- hand in when finished Next Class: Hess’s Law Lab