Bond & Group Enthalpies. The standard entahlpy change associated with breaking the bond in a diatomic molecule in the gas phase: N N (g) -------> 2 N (g) D H [N N] = + 946 kJ / mole of N N bonds
Bond & Group Enthalpies
The standard entahlpy change associated with breaking the bond in a diatomic molecule in the gas phase:
N N (g) -------> 2 N (g)
DH [N N] = + 946 kJ / mole of N N bonds
is a measure of the strength of that bond (in this case the N N triple bond) and is refered to as a bond enthalpy or sometimes less rigorously as a bond energy.
At 298 K what strictly speaking is the bond energy of a N N triple bond?
Note that energy had to be transferred into the N N bond in order to break it and bond breaking is always an endothermic process.
Is bond formation endothermic or exothermic?
The standard enthalpy of formation of oxygen atoms in the gas phase is + 247.5 kJ / mole of oxygen atoms. What is the bond enthalpy of an O = O double bond?
Defining bond enthalpies in polyatomic molecules is more ambiguous! Consider the N-H single bond energies in NH3. Spectroscopic studies have allowed us to determine the enthalpies of the following three bond breaking processes:
NH3 (g) -----> NH2 (g) + H (g) DH = + 460 kJ/mole
NH2 (g) -----> NH (g) + H (g) DH = + 399 kJ/mole
NH (g) -----> N (g) + H (g) DH = + 314 kJ/mole
Which one of these three values should we use for the N-H bond enthalpy? What is actually done is to use the average of these three values as measure of the average N-H bond enthalpy in NH3.
DHN-H average = + 391 kJ/mole of N-H bonds
The utility of bond enthalpies is that at a certain level of approximation they are transferable from one compound to another. For example we can use the N-H bond enthalpy that we already know to calculate the N=N double bond enthalpy in the compound diimine: The enthalpy change for completely dissociating diimine into constituent atoms in the gas phase is:
N=N (g) -------> 2 N (g) + 2 H (g) DH= +1,200 kJ/mole
DH= +1,200 kJ/mole = (1 mol) DH [N=N] + (2 mol) DH [N-H]
= (1 mol) DH [N=N] + (2 mol) (+391 kJ/mole)
from which we calculate the N=N double bond enthalpy as:
DH [N=N] = + 418 kJ/mole
In this manner tables of bond enthalpies can be constructed:
DH [N - N]+ 391 kJ/mole DH [N = N]+ 418 kJ/mole DH [N N]+ 946 kJ/mole
How are bond order (i.e., whether a bond is single, double, triple, etc.) and bond enthalpy related?
A partial table of single bond enthalpies is shown below:
H-H436 kJ/moleH-F563 kJ/mole C-C348 kJ/moleH-Cl432 kJ/mole N-N161 kJ/moleH-Br366 kJ/mole O-O139 kJ/moleH-I299 kJ/mole F-F153 kJ/moleC-N292 kJ/mole Cl-Cl243 kJ/moleC-O351 kJ/mole Br-Br193 kJ/moleC-S259 kJ/mole I-I151 kJ/moleC-F441 kJ/mole C-H413 kJ/moleC-Cl328 kJ/mole N-H391 kJ/moleC-Br276 kJ/mole S-S213 kJ/moleC-I240 kJ/mole O-H463 kJ/moleS-H339 kJ/mole
A partial table of multiple bond enthalpies is shown below:
C=C615 kJ/moleC=S477 kJ/mole C=N615 kJ/moleN=N418 kJ/mole O=O495 kJ/moleC=O728 kJ/mole C=O686 kJ/mole (formaldehyde) C=O715 kJ/mole (other aldehydes) CC946 kJ/mole NN946 kJ/mole CN891 kJ/mole (nitriles)CN866 (HCN)
Note that bond enthalpies are associated with a process in which the reactants and products are in the gas phase. If other phases are present, then other enthalpy changes have to be included in the overall enthalpy change.
Use bond enthalpies to estimate the standard enthalpy of formation of CO2 (g). In addition to data that can be found in the preceeding table of bond enthalpies you will need the heat of sublimation of graphite which is +713.0 kJ/mole.
Bond enthalpies can be used to estimate the heats of reactions that might otherwise be difficult to measure. For example we can use bond enthalpies to estimate the standard heat of formation of hydrogen azide, HN3 (g), an explosive compound. We will proceed by breaking all the bonds in the reactants (an endothermic process) to form a collection of atoms in the gas phase that we will then reassemble into the products (an exothermic process):
1/2 H2 (g) + 3/2 N2 (g) --------> N=N=N (g)
- 2 DH [N=N] - 1 DH [N-H] = -2 (+ 418 kJ) - 1 (391 kJ) = -1,227 kJ
1/2 DH [H-H] = 1/2 (+ 436 kJ) = + 218 kJ
3/2 DH [N N] = 3/2 (+ 946 kJ) = + 1,419 kJ
[ H (g) + 3 N (g) ]
DHof [HN3 (g)] = (+ 218 kJ) + (+1,429 kJ) + (-1,227 kJ) = + 410 kJ
Use bond enthalpies to estimate the heat of isomerization of cyclopropane to propene in the gas phase:
--------> CH3CH=CH2 (g)
The experimental value is -32.88 kJ. Why do you think the experimental value and the value calculated from bond enthalpies are so different? Could you use the value you calculated and the experimental value to estimate the strain enthalpy in cyclopropane?
The standard enthalpy of formation of ozone, O3 (g), is +142.7 kJ/mole. Estimate the strength of an oxygen oxygen partial double bond, O O.
DHof, - CH - = - 6.19 kJ/mol
DHof, - CH2 - = - 20.7 kJ/mol
DHof, - CH3= - 42.17 kJ/mol
Group enthalpies provide an alternative and slightly more reliable way to estimate heats of formation and ultimately heats of reaction. Standardgroup enthalpiesof formationare defined as the contribution that a specific chemical group makes to the standard enthalpy of formation for some compound in the gas phase.
For example consider 2-methylbutane in the gas phase:
The standard enthalpy of formation of gaseous 2-methylbutane would then be calculated as:
DHof, 2-methylbutane (g) = 3 DHof, - CH3 + 1 DHof, - CH2 + 1 DHof, - CH
= (3 mols) (- 42.17 kJ / mol) + 1 (- 20.7 kJ) + 1 (- 6.19 kJ)
= - 153.0 kJ
Use the on-line NIST Web Book to compare this result with a literature value.
Tables of standard group enthalpies of formation for compounds in the gas phase can be found in Sydney W. Benson’s Thermochemical Kinetics, McGraw-Hill, New York (1976):
DHof, 298.2 K (kJ / mol)
C(H)3(C) => - CH3- 42.17C(H)2(C)2- 20.7C(H)(C)3- 6.19C(C)4+ 8.16C(Cl)(H)2(C)- 65.7C(Br)(H)2(C)- 22C(I)(H)2(C)+ 37C(Cl)(H)(C)2- 60.2C(Br)(H)(C)2- 9.7C(Cl)(C)3- 53.1
Use Benson’s standard group enthalpies of formation to estimate the enthalpy change for the gas phase isomerization of 1-chloro-2-methylpropane to 2-chlorobutane:
What value would an estimate based on bond enthalpies have given?