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.

The Basics

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 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