Hess’ Law PreLab
I.
Energy
A.
Energy = the capacity to do work or produce heat
1. Conservation of Energy = energy can’t be created or destroyed; it can
only change forms
2. Potential Energy = due to position or composition
a. Water behind a dam has energy that can be converted to work
b. Attractive and Repulsive forces (gravity) govern this type
c. Gasoline burns to make heat: forces holding atoms together
3. Kinetic Energy = due to motion of the object
4. Heat = q = energy transferred between objects at different temperatures
a. Temperature = KE of particles in random motion
b. Heat is not a substance, but our language often treats it that way
B. Chemical Energy
1. Mechanical Energy = energy of the movement of objects
2. Chemical Energy = energy of the change in chemical bonds
a. CH4(g) + 2O2(g) -------> CO2(g) + 2H2O(g) + heat
b. Heat energy is liberated by rearranging the chemical bonds
3. Dividing up the Universe
a. System = the specific reactants and products we are investigating
b. Surroundings = everything else in the Universe
4. Heat flow in chemical reactions (heat = q = DH = Enthalpy)
a. Exothermic = energy flows out of the system as heat (-DH)
i. Products have a lower PE than the reactants
ii. Heat can be viewed as a product
iii. Heat released results in an increase in KE of surrounding particles
b. Endothermic = energy flows into the system (+DH)
i. N2(g) + O2(g) + heat -------> 2NO(g)
ii. Heat can be viewed as a reactant
iii. Heat absorbed results in less KE of the surrounding particles
iv. Products have more PE than the reactants
II.
Hess’s Law
A. Enthalpy is a State Function
1. Path doesn’t matter
2. As long as we know reactants and products, steps don’t matter for DH
3. Example:
N2(g) + O2(g) -------> 2NO(g)
DH1 = 180 kJ
2NO(g) + O2(g) -------> 2NO2(g) DH2 = -112 kJ
N2(g) + 2O2(g) -------> 2NO2(g) DHtotal = 68 kJ
4.
Hess’s Law
a. You may sum steps in order to find overall DH
b. The DH for the reverse reaction will simply change signs (+/-)
c. If you multiply a reaction, you must multiply DH the same
5. Explanation
a. Sign of DH depends on direction of heat flow. Heat flow is reversed if
the overall reaction is reversed.
Xe(g) + 2F2(g) -------> XeF4(s) + 251kJ (DH = -251kJ)
XeF4(s) + 251kJ -------> Xe(g) + 2F2(g)
(DH = +251kJ)
b. DH is an extensive property = depends on the amount of substance
(an intensive property = depends only on identity of the substance)
Xe(g) + 2F2(g) -------> XeF4(s) + 251kJ (DH = -251kJ)
2Xe(g) + 4F2(g) -------> 2XeF4(s) + 502kJ (DH = -502kJ)
B. Examples
1. Calculate DH for the conversion of graphite to diamond using the known
DH’s for the combustion of graphite and diamond.
a. Cg(s) + O2(g) -------> CO2(g)
DH = -394kJ
Cd(s) + O2(g) -------> CO2(g)
DH = -396 kJ
b. Cg(s) + O2(g) -------> CO2(g)
CO2(g) -------> Cd(s) + O2(g)
Cg(s) -------> Cd(s)
DH = -394kJ
(reverse) DH = +396 kJ
DH = +2 kJ
2. Today’s Experiment: Find DH for oxidation of Mg
a. Reactions
1. Mg(s) + 2HCl(aq) -------> Mg2+(aq) + H2(g) + 2Cl-(aq)
2. Mg2+(aq) + H2O(l) + 2Cl-(aq) -------> MgO(s) + 2HCl(aq)
3. H2(g) + 1/2O2(g) -------> H2O(l)
Total: Mg(s) + 1/2O2(g) -------> MgO(s)
b. Accepted value for this reaction is DH = -602 kJ/mol
DH1 = exp
DH2 = exp
DH3 = exp
DHtotal = ?
3. Procedure
a. Run reaction #1 in our calorimeter
Do twice
b. Calculate DH1 = (4.18 J/goC)(mtotal)(DT) = ____J
c. Convert to kJ/mol: (DH1 x 1kJ/1000J)/(mol of Mg)
d. Do the same thing for the reverse of Reaction #2 Do twice
e. Remember to change the sign of the DH2 that you get
f. The DH3 is found in your text Appendix = -286 kJ/mol
g. Calculate DHtotal = DH1 + DH2 + DH3
Avg Dev =
4. Deviation from the mean = |DHi – DHavg|
(Dev1 + Dev2)/2
Incident: Explosion of Nitric Acid Waste
Incident: Explosion of Liquid Nitrogen Tank