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Bond Energies. Problems and Concepts Carol Brown Saint Mary’s Hall. An Organic Story.
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Bond Energies Problems and Concepts Carol Brown Saint Mary’s Hall
An Organic Story • From the instant I laid eyes on O‚Chem I felt a bond between us, actually, more like a triple bond. It was an incredibly polarizing attraction. The feeling was out of this orbital. I mean before I met O‚Chem, I was merely a single electron. So it was a pretty radical experience. Then, however, the torsional twisting was taking a strain on our relationship. Sadly, our thermodynamics was uphill; we just didn't have the right chemistry. It was an unstable relationship, or maybe it was just a lack of equilibrium. We weren’t able to rearrange into a more stable configuration. We decided that we were equally at fault, so it ended in a homolytic split.
Breaking a bond is always endothermic! (Breaking up is hard to do.) ∆H is positive Making a bond is always exothermic! ∆H is negative The Basics
The reaction itself will be either exothermic or endothermic depending upon the number and type of bonds broken and formed.
Steps for solving the problems • Draw correct Lewis structures for each substance. • Total the energy needed for breaking bonds. Remember it is a positive number. • Total the energy gained by forming bonds. Remember it is a negative number. • Add the two for the (approximate) ∆Hrxn.
#1. Calculate the enthalpy of reaction for the process H2(g) + Cl2(g) --> 2 HCl(g)
#1. Calculate the enthalpy of reaction for the process H2(g) + Cl2(g) --> 2 HCl(g)-184.7 kJ
#2. Estimate the enthalpy change for the combustion of hydrogen gas. 2 H2(g) + O2(g) --> 2 H2O(g)
#2. Estimate the enthalpy change for the combustion of hydrogen gas. 2 H2(g) + O2(g) --> 2 H2O(g)-483.6 kJ
#3. Predict the enthalpy of reaction of the following reaction:2 C2H6(g) + 7 O2(g) --> 4 CO2(g) + 6 H2O(g)
#3. Predict the enthalpy of reaction of the following reaction:2 C2H6(g) + 7 O2(g) --> 4 CO2(g) + 6 H2O(g)-2759.1 kJ