Standard Enthalpy Change of Reaction

– Define and apply the terms standard enthalpy change of formation (ΔH f ˚) and standard enthalpy of combustion (ΔH c ˚).  Q – What are the “standard” conditions?  A – 298K (25˚C), 1.00E5 Pa (100 kPa) or around room temp and pressure

– Define and apply the terms standard enthalpy change of formation (ΔH f ˚) and standard enthalpy of combustion (ΔH c ˚).  ΔH f ˚= the enthalpy change that occurs when 1 mol of a substance is formed from its elements in their standard states. (see table 11 of IB data booklet) Gives a measure of the stability of a substance relative to its elements Can be used to calculate the enthalpy changes of all reactions, hypothetical or real

Sample problem 1  Q – the ΔH f ˚ of ethanol is given in Table 11 of the IB data booklet. Give the thermo chemical equation which represents the standard enthalpy of formation of eth anol.  A – Start with the chemical equation for the formation of ethanol from its component elements in their standard states.  _C(graphite) + _H 2 (g) + _O 2 (g)  _C 2 H 5 OH(l) ΔH f ˚= -277 kJmol -1 Continue by making the coefficient for ethanol 1 because ΔH f ˚ is per mole  2C(graphite) + 3H 2 (g) + 1/2O 2 (g)  C 2 H 5 OH(l) ΔH f ˚= -277 kJmol -1

Sample problem 2  Q – Which of the following does NOT have a standard heat of formation value of zero at 25˚C and 1.00E5 Pa? Cl 2 (g) I 2 (s) Br 2 (g) Na(s)  A – Elements in their STANDARD states have a zero value. Bromine is a LIQUID in its standard state.

Sample problem 3  Q – Which of the following DOES have a standard heat of formation value of zero at 25˚C and 1.00E5 Pa? H(g) Hg(s) C(diamond) Si(s)  A – Graphite is more stable (but not harder) than diamond so Si is the only choice in its standard state.

Using ΔH f ˚  The following expression is used to predict the standard enthalpy change for an entire reaction. ΔH˚ reaction =  ΔH f ˚ products -  ΔH f ˚ reactants  Sample Problem Calculate the enthalpy change for the reaction C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O(g) -105 zero 3(-394) 4(-286) kJ/mol  ΔH˚ reaction = (3(-394)+4(-286))-(-105)  ΔH˚ reaction = kJ/mol

– Define and apply the terms standard enthalpy change of formation (ΔH f ˚) and standard enthalpy of combustion (ΔH c ˚).  ΔH c ˚= the enthalpy change that occurs when 1 mol of a substance burns completely under standard conditions (see table 12 of IB data booklet) Balance these equations for one mole of the reactant instead of the product.

 Determine the enthalpy change of a reaction using standard enthalpy changes of formation and combustion. ΔH˚ reaction =  ΔH f ˚ products -  ΔH f ˚ reactants ΔH˚ reaction =  ΔH c ˚ reactants -  ΔH c ˚ products  They are “flipped” because formation focuses on the energy used in creation and combustion focuses on the energy used in destruction.

Summing up  So now we have several ways to predict the enthalpy changes of a chemical reaction Using Hess’s Law Average Bond Enthalpies ○ ΔH˚=  bonds broken-  bonds formed Standard Enthalpies of formation ○ ΔH˚ reaction =  ΔH f ˚ products -  ΔH f ˚ reactants Standard Enthalpies of combustion ○ ΔH˚ reaction =  ΔH c ˚ reactants -  ΔH c ˚ products

HW (Due Monday)  Do # 6 from you Hess Lab post lab questions and turn the whole thing in tomorrow.