1.2.5 hess’s law- 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)

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 ΔHo

f of the products and subtract ΔHo

f of all of the reactants.

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

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

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)

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.

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 - ∑ΔHreactantsC6H6 (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

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

ΔH = -3267.4 kJ   

1. Forgetting to multiply ΔH values by the appropriate coefficient.

2. Using the wrong value of ΔH for water:ΔHf° for H2O(l) = -285.8 kJ/mol;ΔHf° for H2O(g) = -241.8 kJ/mol

3. Solving for ΔH as "Reactants - Products" instead of "Products – Reactants".

4. Accidentally changing the sign for ΔH.

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

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˚)

1.2.5- practice problems- hand in when finished

Next Class: Hess’s Law Lab