Hess’s Law 5.3 Energetics

Hess’s Law Hess’s Law states that the enthalpy change for any chemical reaction depends only on the difference between the enthalpy of the products and the enthalpy of the reactants. It is independent of the reaction pathways. ΔH1 + ΔH2 = ΔH3 3

Hess’s law Why is this important? Hess’s Law allows us to calculate the enthalpy changes of reactions that cannot be measured directly in the laboratory. For example, although the elements carbon and hydrogen do not combine directly to form propane, the enthalpy change for the reaction: 3C(graphite) + 4H2(g)  C3H8(g) can be calculated from the enthalpy of combustion data of the elements and the compound.

Enthalpies of combustion The enthalpy change of combustion is the enthalpy change when one mole of a substance is completely combusted in oxygen under standard conditions.

Hess’s law 3C(graphite) + 4H2(g) + 5O2(g) C3H8(g) + 5O2(g) ΔH1 ΔH3 3CO2(g) + 4H2O(g) ΔH2 = ΔH1 + ΔH3 The enthalpy of combustion of C ,H2 (ΔH2) and C3H8 (ΔH3 ) can be determined from the enthalpies of combustion table. So, ΔH1 = {(3)(-394) + (4)(-286)} - (1)(-2220) ΔH1 = - 2326 + 2220 = -106 kJ/mol ΔH2

Hess’s law

Hess’s law Example: calculate the enthalpy change that occurs in the following reaction: 2C(graphite) + 3H2(g) +½O2(g)  C2H5OH(l) Since the enthalpy change of formation of ethanol is difficult to measure directly, we will use Hess’s Law and the enthalpy of combustion data to find the answer.

Hess’s law 2C(graphite) + 3H2(g) + 3½O2(g) C2H5OH(l) + 3O2(g) ΔH1 ΔH2 ΔH3 2CO2(g) + 3H2O(g) ΔH2 = ΔH1 + ΔH3 The enthalpy of combustion of C ,H2 (ΔH2) and C2H5OH (ΔH3 ) can be determined from the enthalpies of combustion table. So, ΔH1 = {(2)(-394) + (3)(-286)} - (1)(-1371) ΔH1 = - 1646 + 1371 = -275 kJ/mol

Hess’s Law

Hess’s law Exercise: Calculate the standard enthalpy change when one mole of methane is formed from its elements in their standard states. The standard enthalpies of combustion of carbon, hydrogen, and methane are -393, -286, and -890 kJ/mol, respectively. notes

Hess’s law Expressed in a different way, Hess’s Law states that if a reaction is carried out in a series of steps, Δ H for the reaction will be equal to the sum of the enthalpy changes for the individual steps.

Hess’s Law

Hess’s law

Hess’s law Exercise: Calculate the enthalpy change, ΔHo, for the reaction: S(s) + O2(g)  SO2(g) From the information below: S(s) +3/2 O2(g) SO3(g) ΔHo = -395 kJ SO2(g) + ½ O2(g)  SO3(g) Δho = -98 kJ

Standard enthalpy of formation The standard enthalpy of formation of a substance is the enthalpy change that occurs when one mole of a substance is formed from its elements in their standard state under standard conditions ( 1 atmosphere of pressure and 25oC unless otherwise specified). The standard enthalpy of formation of a substance is given the symbol ΔHof (f stands for formation and o stands for standard state).

Standard enthalpy of formation The standard enthalpy change of formation of an element in its standard state = 0. Therefore, it is important that you know the physical state of the elements under standard conditions. Most elements at 1 atm and 25oC commonly exist as monatomic solids and in one allotropic form with the notable exceptions of:

Standard enthalpy of formation

Thermodynamic data

Standard enthalpy of formation Exercise: Which of the following does not have a standard heat of formation value of zero at 25oC and 1.00 atm? A. Cl2(g) B. I2(s) C. Br2(g) D. Na(s)

Standard enthalpy of formation Determination of the standard enthalpy change for any reaction, ΔHorxn, is done by considering the balanced chemical equation and standard enthalpy of formation data. ΔHorxn for the following generalized chemical equation (where a, b, e, and f are stoichiometric coefficients) a A + b B +…  e E + f F + … can be obtained as such: ΔHorxn= [ e ΔHof, E + f ΔHof, F +…] – [a ΔHof, A + b ΔHof, B +…]

Standard enthalpy of formation The above equation can be generalized as: ΔHorxn= Σ nΔHof, products - Σ nΔHof, reactants Σ (sigma)  “the sum of” n  stoichiometric coefficients

Standard enthalpy of formation Exercise: Calculate the enthalpy change for the reaction: C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g) ΔHof /kJ/mol C3H8 (g) -104 CO2 (g) -394 H2O (g) -286