Lecture 4: Standard Enthalpies

• Reading: Zumdahl 9.6

• Outline– What is a standard enthalpy?– Defining standard states.

– Using standard enthalpies to determine H°rxn

Hess’ Law

• Enthalpy is a state function. As such, H for going from some initial state to some final state is pathway independent.

H for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick any pathway to calculate H for a reaction.

Definition of H°f

• We saw in the previous lecture the utility of having a series of reactions with known enthalpies.

•What if one could deal with combinations of compounds directly rather than dealing with whole reactions to determine Hrxn?

Definition of H°f (cont.)

• Standard Enthalpy of Formation, H°f:

“The change in enthalpy that accompanies the formation of 1 mole of a compound from its elements with all substances at standard state.”

• We envision taking elements at their standard state, and combining them to make compounds, also at standard state.

What is Standard State?

• Standard State: A precisely defined reference state. It is a common reference point that one can use to compare thermodynamic properties.

• Definitions of Standard State:– For a gas: P = 1 atm (or 100,000 Pascal; 1 atm=101325)– For solutions: 1 M (mol/l).– For liquids and solids: pure liquid or solid– For elements: The form in which the element exists

under conditions of 1 atm and 298 K.

Definitions (cont.)

• Standard elemental states (cont.):– Hydrogen: H2 (g) (not atomic H)– Oxygen: O2 (g)– Carbon: C (gr) … graphite as opposed to diamond

• We will denote the standard state using the subscript “°”. – Example: H°rxn

Importance of Elements

• We will use the elemental forms as a primary reference when examining compounds.

• Pictorially, for chemical reactions we envision taking the reactants to the standard elemental form, then reforming the products.

Example: Combusion of MethaneCH4(g) + 2O2(g) CO2(g) + 2H2O(l)

Elements and H°f• For elements in their

standard state:

H°f = 0

• With respect to the graph, we are envisioning chemical reactions as proceeding through elemental forms. In this way we are comparing H°f for reactants and products to a common reference (zero)

reactants

elements

productsH¡ rxn

q

Tabulated H°f

• In Zumdahl, tables of H°f are provided in Appendix 4.

• The tabulated values represent the H°f for forming the compound from elements at standard conditions.

C (gr) + O2 (g) CO2 (g) Hf° = -394 kJ/mol

Using H°f to determine H°rxn

• Since we have connected the H°f to a common reference point, we can now determine H°rxn by looking at the difference between the H°f for products versus reactants.

• Mathematically:

H°rxn = H°f (products) - H° f (reactants)

How to Calculate H°rxn

• When a reaction is reversed, H changes sign.

• If you multiply a reaction by an integer, H is also multiplied (it is an extensive variable)

• Elements in their standard states are not included in calculations since H°f = 0.

Example

• Determine the H°rxn for the combustion of methanol.

CH3OH (l) + 3/2O2(g) CO2(g) + 2H2O(l)

• H°f in Appendix 4:CH3OH(l) -238.6 kJ/molCO2(g) -393.5 kJ/molH2O(l) -286 kJ/mol

Example (cont.)

CH3OH (l) + 3/2O2(g) CO2(g) + 2H2O(l)

H°rxn = H°f (products) - H° f (reactants)

= H°f(CO2(g)) + 2H°f(H2O(l))

- H°f(CH3OH(l)) = -393.5 kJ - (2 mol)(286 kJ/mol) - (-238.6 kJ) = -728.7 kJ Exothermic!

Another Example

• Using the following reaction:

2ClF3(g) + 2NH3(g) N2(g) + 6HF(g) + Cl2(g)

H°rxn = -1196 kJ

Determine the H°f for ClF3(g)

H°f

NH3(g) -46 kJ/mol HF(g) -271 kJ/mol

Another Example (cont.)

• Given the reaction of interest:

2ClF3(g) + 2NH3(g) N2(g) + 6HF(g) + Cl2(g)

H°rxn = -1196 kJ

H°rxn = H°f (products) - H° f (reactants)

H°rxn = 6H°f(HF(g)) - H°f (NH3(g)) - H° f (ClF3(g))

Another Example (cont.)

H°rxn = 6H°f(HF(g)) - H°f (NH3(g)) - H°f (ClF3(g))

-1196 kJ = (6mol)(-271 kJ/mol)- (2mol)(-46 kJ/mol)

- (2mol)H°f (ClF3(g)) -1196 kJ = -1534 kJ - (2mol)H°f (ClF3(g)) 338 kJ = - (2mol)H°f (ClF3(g)) -169 kJ/mol = H°f (ClF3(g))

Bonus Example

• Problem 9.81 Hvap for water at the normal boiling point (373.2 K) is 40.66 kJ. Given the following heat capacity data, what is Hvap at 340.2 K for 1 mol?

• CP H20(l) = 75 J/mol.K

• CP H20(g) = 36 J/mol.K

Bonus Example (cont.)

H2O(l) H2O(g) H = 40.66 kJ/mol

(373.2 K)

H2O(l) H2O(g)(340.2 K)

H = ?

nCP(H2O(l))T nCP(H2O(g))T

Bonus Example (cont.)

H2O(l) H2O(g) H = 40.66 kJ/mol

(373.2 K)

H2O(l) H2O(g)(340.2 K)

nCP(H2O(l))T nCP(H2O(g))T

Hrxn (340.2K) = H(product) - H(react)

= Hwater(g) (373.2K) + nCP(H2O(g))T - Hwater(l) (373.2K) - nCP(H2O(l))T

Bonus Example (cont.)

Hrxn (340.2K) = kJ/mol + 36 J/mol.K(-33 K) - 75 J/mol.K(-33 K)

Hrxn (340.2K) = kJ/mol + 1.29 kJ/mol = 41.95 kJ/mol

Hrxn (340.2K) = Hwater(g) (373.2K) + nCP(H2O(g))T - Hwater(l) (373.2K) - nCP(H2O(l))T