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If the reaction above for the formation of water were reversed, the sign of H would be reversed That would indicate that it would take 4836 kJ of energy to decompose 2 mol of water This would then become an endothermic process H is dependent upon the state of matter The enthalpy change would be different for the formation of liquid water instead of gaseous water H can also indicate whether a reaction will be spontaneous A negative (exothermic) value of H is associated with a spontaneous reaction However, in many reactions this is not the case There is another factor to consider in predicting a reaction s spontaneity We will cover this other factor a little later in this chapter Enthalpies of reaction can be measured using a calorimeter However, they can also be calculated in other ways Hess s law states that if a reaction occurs in a series of steps, then the enthalpy change for the overall reaction is simply the sum of the enthalpy changes of the individual steps If, in adding the equations of the steps together, it is necessary to reverse one of the given reactions, then the sign of H must also be reversed Also, particular attention must be used if the reaction stoichiometry has to be adjusted The value of an individual H may need to be adjusted It doesn t matter whether the steps used are the actual steps in the mechanism of the reaction because Hreaction ( Hrxn) is a state function, a function that doesn t depend on the pathway, but only on the initial and final states Let s see how Hess s law can be applied, given the following information: C(s) + O2(g) CO2(g) H2(g) + (1/2)O2 (g) H2O(1) C2H2(g) + (5/2)O2(g) 2 CO2(g) + H2O(1) find the enthalpy change for: 2C(s) + H2(g) C2H2(g) Answer: 2[C(s) + O2(g) CO2(g)] H2(g) + (1/2) O2(g) H2O(1) 2 CO2(g) + H2O(1) C2H2(g) + (5/2) O2(g) 2C(s) + H2(g) C2H2(g) 2 ( 3935 kJ) 2858 kJ ( 12998 kJ) 2270 kJ H = 3935 kJ H = 2858 kJ H = 12998 kJ
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Enthalpies of reaction can also be calculated from individual enthalpies of formation (or heats of formation), Hf , for the reactants and products Because the temperature, pressure, and state of the substance will cause these enthalpies to vary, it is common to use a standard state convention For gases, the standard state is 1 atm pressure For a substance in an aqueous solution, the standard state is 1 molar concentration And for a pure substance (compound or element), the standard state is the most stable form at 1 atm pressure and 25 C A degree symbol to the right of the H indicates a standard state, H The standard enthalpy of formation of a substance ( Hf ) is the change in enthalpy when 1 mol of the substance is formed from its elements when all substances are in their standard states These values are then tabulated and can be used in determining H rxn Hf of an element in its standard state is zero Hf rxn can be determined from the tabulated Hf of the individual reactants and products It is the sum of the Hf of the products minus the sum of the Hf of the reactants: H rxn = Hf products Hf reactants
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128 U Step 4 Review the Knowledge You Need to Score High
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In using this equation be sure to consider the number of moles of each, because Hf for the individual compounds refer to the formation of 1 mol For example, let s use standard enthalpies of formation to calculate Hrxn for: 6 H2O(g) + 4 NO(g) 5 O2(g) + 4 NH3(g) Answer: H rxn = {5[ H f O2 ( g )]+ 4[ H f NH3 ( g )]} {[6 H f H 2O(g)]+ 4[ H f NO(g)]} Using tabulated standard enthalpies of formation gives: Hrxn = [5(000 kJ) + 4( 4619 kJ)] [6( 24185 kJ) + 4(9037)] = 90468 kJ People commonly forget to subtract all the reactants from the products The values of Hf will be given to you on the AP exam, or you will be asked to stop before putting the numbers into the problem An alternative means of estimating the heat of reaction is to take the sum of the average bond energies of the reactant molecules and subtract the sum of the average bond energies of the product molecules
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