Hess’ Law PreLab

I. EnergyA. 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 = H = Enthalpy)

a. Exothermic = energy flows out of the system as heat (-H)

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 (+H)

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 LawA. Enthalpy is a State Function

1. Path doesn’t matter2. As long as we know reactants and products, steps don’t matter for H3. Example:

N2(g) + O2(g) -------> 2NO(g) H1 = 180 kJ

2NO(g) + O2(g) -------> 2NO2(g) H2 = -112 kJ

N2(g) + 2O2(g) -------> 2NO2(g) Htotal = 68 kJ

4. Hess’s Law a. You may sum steps in order to find overall Hb. The H for the reverse reaction will simply change signs (+/-)c. If you multiply a reaction, you must multiply H the same

5. Explanation

a. Sign of H depends on direction of heat flow. Heat flow is reversed if the overall reaction is reversed.

Xe(g) + 2F2(g) -------> XeF4(s) + 251kJ (H = -251kJ)

XeF4(s) + 251kJ -------> Xe(g) + 2F2(g) (H = +251kJ)

b. H 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 (H = -251kJ)

2Xe(g) + 4F2(g) -------> 2XeF4(s) + 502kJ (H = -502kJ)

B. Examples

1. Calculate H for the conversion of graphite to diamond using the known H’s for the combustion of graphite and diamond.

a. Cg(s) + O2(g) -------> CO2(g) H = -394kJ

Cd(s) + O2(g) -------> CO2(g) H = -396 kJ

b. Cg(s) + O2(g) -------> CO2(g) H = -394kJ

CO2(g) -------> Cd(s) + O2(g) (reverse) H = +396 kJ

Cg(s) -------> Cd(s) H = +2 kJ

2. Today’s Experiment: Find H for oxidation of Mg

a. Reactions

1. Mg(s) + 2HCl(aq) -------> Mg2+(aq) + H2(g) + 2Cl-(aq) H1 = exp

2. Mg2+(aq) + H2O(l) + 2Cl-(aq) -------> MgO(s) + 2HCl(aq) H2 = exp

3. H2(g) + 1/2O2(g) -------> H2O(l) H3 = expTotal: Mg(s) + 1/2O2(g) -------> MgO(s) Htotal = ?

b. Accepted value for this reaction is H = -602 kJ/mol

3. Procedure

a. Run reaction #1 in our calorimeter

b. Calculate H1 = (4.18 J/goC)(mtotal)(T) = ____J

c. Convert to kJ/mol: (H1 x 1kJ/1000J)/(mol of Mg)

d. Do the same thing for the reverse of Reaction #2

e. Remember to change the sign of the H2 that you get

f. The H3 is found in your text Appendix = -286 kJ/mol

g. Calculate Htotal = H1 + H2 + H3

4. Deviation from the mean = |Hi – Havg|

Do twice

Do twice

Avg Dev =

(Dev1 + Dev2)/2

Incident: Explosion of Nitric Acid Waste

Incident: Explosion of Liquid Nitrogen Tank