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