1.2.5 Hess’s Law- the equation

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

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

1.2.5 Hess’s Law- the equation