Bond+Enthalpies

Concepts
Covalently bonded molecules measure bond strength as a function of the energy required to break that bond, or the bond enthalpy. Bond enthalpy is the energy required to break or form one mole of a particular bond of a gaseous substance. Bond enthalpies, such as those seen in the table below, are never an exact value for any specific bond, rather an average. The following table lists bond enthalpies identical to those found in the textbook. It should be noted that these are average values used in estimating the energy required to break/form a bond of a particular type.

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The application of bond enthalpies is to give a quick estimate of whether a reaction will be endothermic or exothermic. Recall, that endothermic reactions have a positive ∆H value whereas exothermic reactions have a negative ∆H value and conceptually, endothermic reactions take in heat from their surroundings whereas exothermic reactions release heat into their surroundings.

To estimate reaction enthalpy, an application of Hess' Law is used. Using the fact that bond breaking is always endothermic and bond formation is always exothermic, the reaction is broken into two phases. Phase 1 is the breaking of the bonds present in the reactants and Phase 2 is the formation of the bonds of the products. **Therefore, the final equation for calculating the enthalpy of reaction** **ΔH = ∑ ΔH(bonds broken) + ∑ ΔH****(bonds formed).** Remember when calculating reaction enthalpy that energy of formation is a negative value.

Calculations
Use the following model as a guide for calculating reaction enthalpies

__1) Write a balanced chemical equation and draw Lewis structures for all molecules__

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__2) Identify the bonds broken/created and the bond enthalpies associated with each phase of reaction__

Phase 1 (Bonds Broken): 2(N=O) --> 2(607) = +1214kJ 2(N-Cl) --> 2(200) = +400kJ

Phase 2 (Bonds Formed): 2(N=O) --> -2(607) = -1214kJ 1(Cl-Cl) --> -1(242) = -242kJ

__3) Calculate **∑**____∆H for both phases of the reaction__

Phase 1 (Bonds Broken):
 * ∑**∆H= 1214 + 400 = 1614kJ

Phase 2 (Bonds Formed):
 * ∑**∆H= -1214 + -242 = -1456kJ

__4) Calculate ∆H of reaction by finding the sum of the **∑**∆H for both phases of reaction__

ΔH = ∑ ΔH(bonds broken) + ∑ ΔH(bonds formed). ΔH --> 1614 + -1456 = 158kJ

Sources: http://www.bustertests.co.uk/studies/energetics-enthalpy-change-calorimetry-hesss-law-bond-enthalpies.php AP Chemistry Textbook And image sources as noted.