Either way, you get to the finish. Hess’s Law The enthalpy change of a physical or chemical process is independent of the pathway of the process and the number of intermediate steps in the process. Finish Start Either way, you get to the finish.

Hess’s law allows you to determine the energy of chemical reaction without directly measuring it. The enthalpy change of a chemical process is equal to the sum of the enthalpy changes of all the individual steps that make up the process.

and.. the H values must be treated accordingly. Determine the heat of reaction for the reaction: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ Hint: The three reactions must be algebraically manipulated to add up to the desired reaction. and.. the H values must be treated accordingly.

Found in more than one place, SKIP IT (its hard). Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) Using the following sets of reactions: N2(g) + O2(g)  2NO(g) H = 180.6 kJ N2(g) + 3H2(g)  2NH3(g) H = -91.8 kJ 2H2(g) + O2(g)  2H2O(g) H = -483.7 kJ 4NH3  2N2 + 6H2 H = +183.6 kJ NH3: Reverse and x 2 Found in more than one place, SKIP IT (its hard). O2 : NO: x2 2N2 + 2O2  4NO H = 361.2 kJ H2O: x3 6H2 + 3O2  6H2O H = -1451.1 kJ

Found in more than one place, SKIP IT. Goal: 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g) 4NH3  2N2 + 6H2 H = +183.6 kJ NH3: Reverse and x2 Found in more than one place, SKIP IT. O2 : NO: x2 2N2 + 2O2  4NO H = 361.2 kJ H2O: x3 6H2 + 3O2  6H2O H = -1451.1 kJ Cancel terms and take sum. + 5O2 + 6H2O H = -906.3 kJ 4NH3  4NO Is the reaction endothermic or exothermic?

Consult your neighbor if necessary. Determine the heat of reaction for the reaction: C2H4(g) + H2(g)  C2H6(g) Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ Consult your neighbor if necessary.

Determine the heat of reaction for the reaction: Goal: C2H4(g) + H2(g)  C2H6(g) H = ? Use the following reactions: C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ C2H6(g) + 7/2O2(g)  2CO2(g) + 3H2O(l) H = -1550 kJ H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H4(g) :use 1 as is C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(l) H = -1401 kJ H2(g) :# 3 as is H2(g) + 1/2O2(g)  H2O(l) H = -286 kJ C2H6(g) : rev #2 2CO2(g) + 3H2O(l)  C2H6(g) + 7/2O2(g) H = +1550 kJ C2H4(g) + H2(g)  C2H6(g) H = -137 kJ

Homework: Page 317 #1-3

Where do the reference equations come from? Many can be determined through calorimetry. ie. combustion reactions Another type of reference equation is called standard enthalpies of formation.

Standard Enthalpies of formation: The standard enthalpy of formation is a measure of the energy released or consumed when one mole of a substance is created under standard conditions from its pure elements. The symbol of the standard enthalpy of formation is ΔHof. Δ = A change in enthalpy o = A degree signifies that it's a standard enthalpy change. f = The f indicates that the substance is formed from its elements An important point to be made about the standard enthalpy of formation is that when a pure element is in its standard form its standard enthalpy formation is zero.

ΔHoreaction=∑nΔHof(products)−∑nΔHof(Reactants) The equation for the standard enthalpy change of formation, shown below, is commonly used: ΔHoreaction=∑nΔHof(products)−∑nΔHof(Reactants) This equation essentially states that the standard enthalpy change of formation is equal to the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants.

ΔHreactiono = ΔHfo[C] - (ΔHfo[A] + ΔHfo[B]) Given a simple chemical equation with the variables A, B and C representing different compounds: A+B⇋C and the standard enthalpy of formation values: ΔHfo[A] = 433 KJ/mol ΔHfo[B] = -256 KJ/mol ΔHfo[C] = 523 KJ/mol the equation for the standard enthalpy change of formation is as follows:  ΔHreactiono = ΔHfo[C] - (ΔHfo[A] + ΔHfo[B]) ΔHreactiono = (1 mol)(523 kJ/mol) - ((1 mol)(433 kJ/mol) + (1 mol)(-256 kJ/mol)) Because there is one mole each of A, B and C,  the standard enthalpy of formation of each reactant and product is multiplied by 1 mole, which eliminates the mol denominator: ΔHreactiono = 346 kJ The result is 346 kJ, which is the standard enthalpy change of formation for the creation of variable "C".

Calculate the ΔHreactiono for the formation of NO2(g). NO2(g) is formed from the combination of NO(g) and O2(g) in the following reaction: 2NO(g)+O2(g)⇋2NO2(g) To find the ΔHreactiono, use the formula for the standard enthalpy change of formation: ΔHoreaction=∑ΔHof(products)−∑ΔHof(Reactants) The relevant standard enthalpy of formation values from the data table are as follows: O2(g): 0 kJ/mol NO(g): 90.25 kJ/mol NO2(g): 33.18 kJ/mol Plugging these values into the formula above gives the following: ΔHreactiono= (2 mol)(33.18 kJ/mol) - ((2 mol)(90.25 kJ/mol) + (1 mol)(0 kJ/mol)) ΔHreactiono =-114.14kJ

Homework: Page 324 #1-6

Thermodynamic Quantities of Selected Substances @ 298.15 K