Thermochem Hess’s Law and Enthalpy of Formation Sections 5.6 and 5.7

Hess’s Law Enthalpy is a state function – independent of path it takes to get there –Distance vs. displacement (physics again!).  H is well known for many reactions, and it is inconvenient to directly measure  H for every reaction in which we are interested. However, we can estimate  H for any reaction using  H values that are already known and the properties of enthalpy.

Hess’s Law Hess’s law states that “If a reaction is carried out in a series of steps,  H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

Hess’s Law Because  H is a “state function”, the total enthalpy change depends only on the initial state of the reactants and the final state of the products. It doesn’t matter if a reaction happens in 1 step or 2 (or 3 or infinity) if you start and end at the same place the overall  H is the same.

Calculation of  H Imagine this reaction as occurring in 3 steps which add up to the overall equation C 3 H 8 (g)  3 C (graphite) + 4 H 2 (g)  H 1 = kJ 3 C (graphite) + 3 O 2 (g)  3 CO 2 (g)  H 2 = -1181kJ 4 H 2 (g) + 2 O 2 (g)  4 H 2 O (l)  H 3 = kJ C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Then  H =  H 1 +  H 2 +  H 3 = – = C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l)

Using Hess’s Law IF you are given an overall equation and a set of equations to arrange so they add up to the overall equation, you need to reverse them or multiply them to get the overall equation. EXAMPLE: Given this information: 2 CO(g)  2 C(s) + O 2 (g)  H 1 = 221 kJ CO(g) + 2 H 2 (g)  CH 3 OH(g)  H 2 =  91 kJ Find  H for 2 C(s) + O 2 (g) + 4 H 2 (g)  2 CH 3 OH(g)

To get the final equation, reverse the first equation and change sign of  H 1 : 2 C (s) + O 2 (g)  2 CO (g) -  H 1 =  221 kJ Double the second equation and double  H 2 2[CO (g) + 2 H 2 (g)  CH 3 OH (g) ] 2  H 2 = 2(  91 kJ) 2 C (s) + 4 H 2 (g) + O 2 (g)  2 CH 3 OH (g)  H rxn = -  H  H 2  H rxn =  221 kJ + 2(  91 kJ) =  403 kJ

Enthalpies of Formation The enthalpy (or heat) of formation of a substance,  H f, is the heat released or absorbed when making a 1 mole of a compound or element “from scratch” “From scratch” means from its elements in their naturally occurring elemental forms and phases at 25C. By definition the heat of formation of a pure element in its normal form and phase at 25C is zero (e.g. O 2 (g) and H 2 (g) but NOT O (g) or H 2 (l)) EXAMPLES OF FORMATION REACTIONS: C(s) + 1/2O 2 (g)  CO(g),  H =  kJ H 2 (g) + 1/2O 2 (g)  H 2 O (l)  H = -285 kJ What is the formation reaction for C 2 H 6 (g)?

Standard Enthalpies of Formation Standard enthalpies of formation,  H f, are measured under standard conditions (25°C and 1.00 atm pressure). An element in its standard form at these conditions has an enthalpy of formation of zero. Watch states carefully when looking up these values! H 2 O (g) has a different from H 2 O(l). Appendix C in your book p.1123 (or Google) 

Calculation of  H using H f We can use heat of formation to calculate the enthalpy change  H for any reaction using:  H =  n  H f(products) -  m  H f(reactants) where n and m are the stoichiometric coefficients. Basically you add up all of the heats of formation of each product times its coefficient in the equation. Do the same for the reactants. Subtract the reactant sum from the product sum and that is  H! 

C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Calculation of  H  H= [3( kJ) + 4( kJ)] - [1( kJ) + 5(0 kJ)] = [( kJ) + ( kJ)] - [( kJ) + (0 kJ)] = ( kJ) - ( kJ) = kJ What is  H for this reaction? CH 4 (g) + 2O 2 (g)  2 H 2 O (g) + CO 2 (g)

C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (l) Calculation of  H  H= [3( kJ) + 4( kJ)] - [1( kJ) + 5(0 kJ)] = [( kJ) + ( kJ)] - [( kJ) + (0 kJ)] = ( kJ) - ( kJ) = kJ What is  H for this reaction? CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2 H 2 O (g)  H= [( kJ) + 2( kJ)] - [1(-79.8 kJ) + 2(0 kJ)] = ( kJ) - (-79.8kJ) = kJ