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Lecture 3: Hess’ Law
• Reading: Zumdahl 9.5
• Outline– Definition of Hess’ Law– Using Hess’ Law (examples)
First Law of Thermodynamics
First Law: Energy of the Universe is Constant
E = q + w
q = heat. Transferred between two bodies
w = work. Force acting over a distance (F x d)
Definition of Enthalpy
• Thermodynamic Definition of Enthalpy (H):
H = E + PV
E = energy of the system
P = pressure of the system
V = volume of the system
Changes in Enthalpy• Consider the following expression for a chemical process:
H = Hproducts - Hreactants
If H >0, then qp >0. The reaction is endothermic
If H <0, then qp <0. The reaction is exothermic
Heat Capacity, energy and enthalpy
Ideal Monatomic Gas
• Cv = 3/2R
• Cp = Cv + R = 5/2 R
Polyatomic Gas
• Cv > 3/2R
• Cp > 5/2 R
• E = q + w
• w = -PextV (for now)
• E = nCvT = qV
• H = nCpT = qP
Thermodynamic State Functions
• Thermodynamic State Functions: Thermodynamic properties that are dependent on the state of the system only. (Example: E and H)
• Other variables will be dependent on pathway (Example: q and w). These are NOT state functions. The pathway from one state to the other must be defined.
Hess’ Law Defined
• From lecture 3: Enthalpy is a state function. As such, H for going from some initial state to some final state is pathway independent.
• Hess’ Law: H for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick any pathway to calculate H for a reaction.
Hess’ Law: An Example
Using Hess’ Law
• When calculating H for a chemical reaction as a single step, we can use combinations of reactions as “pathways” to determine H for our “single step” reaction.
2NO2 (g)
N2 (g) + 2O2 (g)
q
2NO2 (g)N2 (g) + 2O2 (g)
Example (cont.)
• Our reaction of interest is:
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
• This reaction can also be carried out in two steps:
N2 (g) + O2 (g) 2NO(g) H = 180 kJ2NO (g) + O2 (g) 2NO2(g) H = -112 kJ
Example (cont.)
• If we take the previous two reactions and add them, we get the original reaction of interest:
N2 (g) + O2 (g) 2NO(g) H = 180 kJ 2NO (g) + O2 (g) 2NO2(g) H = -112 kJ
N2 (g) + 2O2 (g) 2NO2(g) H = 68 kJ
Changes in Enthalpy• Consider the following expression for a chemical process:
H = Hproducts - Hreactants
If H >0, then qp >0. The reaction is endothermic
If H <0, then qp <0. The reaction is exothermic
Example (cont.)
• Note the important things about this example, the sum of H for the two reaction steps is equal to the H for the reaction of interest.
• We can combine reactions of known H to determine the H for the “combined” reaction.
Hess’ Law: Details
• Once can always reverse the direction of a reaction when making a combined reaction. When you do this, the sign of H changes.
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
2NO2(g) N2(g) + 2O2(g) H = -68 kJ
Details (cont.)• The magnitude of H is directly proportional to
the quantities involved (it is an “extensive” quantity).
• As such, if the coefficients of a reaction are multiplied by a constant, the value of H is also multiplied by the same integer.
N2(g) + 2O2(g) 2NO2(g) H = 68 kJ
N2(g) + 4O2(g) 4NO2(g) H = 136 kJ
Using Hess’ Law
• When trying to combine reactions to form a reaction of interest, one usually works backwards from the reaction of interest.
• Example:
What is H for the following reaction?
3C (gr) + 4H2 (g) C3H8 (g)
Example (cont.)
3C (gr) + 4H2 (g) C3H8 (g) H = ?
• You’re given the following reactions:
C (gr) + O2 (g) CO2 (g) H = -394 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220 kJ
H2 (g) + 1/2O2 (g) H2O (l) H = -286 kJ
Example (cont.)
• Step 1. Only reaction 1 has C (gr). Therefore, we will multiply by 3 to get the correct amount of C (gr) with respect to our final equation.
C (gr) + O2 (g) CO2 (g) H = -394 kJ
3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ
Initial:
Final:
Example (cont.)
• Step 2. To get C3H8 on the product side of the reaction, we need to reverse reaction 2.
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220 kJ
C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) H = -2220 kJInitial:
Final:
Example (cont.)
• Step 3: Add two “new” reactions together to see what is left:
3C (gr) + 3O2 (g) 3CO2 (g) H = -1182 kJ
3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) H = +2220 kJ2
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ
Example (cont.)
• Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ
H2 (g) + 1/2O2 (g) H2O (l) H = -286 kJ
3C (gr) + 4H2 (g) C3H8 (g)
Need to multiply second reaction by 4
Example (cont.)
• Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ
4H2 (g) + 2O2 (g) 4H2O (l) H = -1144 kJ
3C (gr) + 4H2 (g) C3H8 (g)
Example (cont.)• Step 4 (cont.):
3C (gr) + 4H2O (l) C3H8 (g) + 2O2 H = +1038 kJ
4H2 (g) + 2O2 (g) 4H2O (l) H = -1144 kJ
3C (gr) + 4H2 (g) C3H8 (g) H = -106 kJ
Changes in Enthalpy• Consider the following expression for a chemical process:
H = Hproducts - Hreactants
If H >0, then qp >0. The reaction is endothermic
If H <0, then qp <0. The reaction is exothermic
Another Example
• Calculate H for the following reaction:
H2(g) + Cl2(g) 2HCl(g)
Given the following:
NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ
N2 (g) + 3H2 (g) 2NH3 (g) H = -92 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = -629 kJ
Another Example (cont.)
• Step 1: Only the first reaction contains the product of interest (HCl). Therefore, reverse the reaction and multiply by 2 to get stoichiometry correct.
NH3 (g) + HCl (g) NH4Cl(s) H = -176 kJ
2NH4Cl(s) 2NH3 (g) + 2HCl (g) H = 352 kJ
Another Example (cont.)
• Step 2. Need Cl2 as a reactant, therefore, add reaction 3 to result from step 1 and see what is left.
2NH4Cl(s) 2NH3 (g) + 2HCl (g) H = 352 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) H = -629 kJ
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g)
H = -277 kJ
Another Example (cont.)• Step 3. Use remaining known reaction in
combination with the result from Step 2 to get final reaction.
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) H = -277 kJ
( N2 (g) + 3H2(g) 2NH3(g) H = -92 kJ)
H2(g) + Cl2(g) 2HCl(g) H = ?
Need to take middle reaction and reverse it
Another Example (cont.)• Step 3. Use remaining known reaction in
combination with the result from Step 2 to get final reaction.
N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) H = -277 kJ
2NH3(g) 3H2 (g) + N2 (g) H = +92 kJ
H2(g) + Cl2(g) 2HCl(g) H = -185 kJ
1
Changes in Enthalpy• Consider the following expression for a chemical process:
H = Hproducts - Hreactants
If H >0, then qp >0. The reaction is endothermic
If H <0, then qp <0. The reaction is exothermic