Lecture 4: Hess’s Law

Reading: Zumdahl 9.5

Outline:

Definition of Hess’ Law

Using Hess’ Law (examples)

Q: What is Hess’s Law ?

• Recall (lecture 3) Enthalpy is a state function. As such, DH for going from some initial state to some final state is pathway independent.

• Hess’s Law: DH for a process involving the transformation of reactants into products is not dependent on pathway.

(This means we can calculate DH for a reaction by a single step, or by multiple steps)

Using Hess’s Law

When calculating DH for a chemical reaction as a single step, we can use combinations of reactions as “pathways” to determine DH for our “single step” reaction.

2NO2 (g)

N2 (g) + 2O2 (g)

q

2NO2 (g)N2 (g) + 2O2 (g)

The reaction of interest is:

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

This reaction can also be carried out in two steps:

N2 (g) + O2 (g) 2NO(g) DH = +180 kJ

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

• If we take the previous two reactions and add them, we get the original reaction of interest:

N2 (g) + O2 (g) 2NO(g) DH = +180 kJ

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

N2 (g) + 2O2 (g) 2NO2(g) DH = + 68 kJ

Note: the important things about this example is that the sum of DH for the two reaction steps is equal to the DH for the reaction of interest.

Big idea: We can combine reactions of known

DH to determine the DH for the overall

“combined” reaction.

Hess’s Law: An Important Detail

One can always reverse the direction of a

reaction when making a combined reaction.

When you do this, the sign of DH changes.

N2(g) + 2O2(g) 2NO2(g) DH = +68 kJ

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

One more detail:• The magnitude of DH is directly proportional to the

quantities involved. (This means DH is an “extensive” quantity).

• So, if the coefficients of a reaction are multiplied by a number, the value of DH is also multiplied by the same number.

N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ

2N2(g) + 4O2(g) 4NO2(g) DH = 136 kJ

Using Hess’s Law: tips

• When trying to combine reactions to form a reaction of interest, it is usually best to work backwards from the reaction of interest.

• Example:

What is DH for the following reaction?

3C (gr) + 4H2 (g) C3H8 (g)

3C (gr) + 4H2 (g) C3H8 (g) DH = ?

You’re given the following reactions:

C (gr) + O2 (g) CO2 (g) DH = -394 kJ

C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJ

H2 (g) + 1/2O2 (g) H2O (l) DH = -286 kJ

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.

(x3) C (gr) + O2 (g) CO2 (g) DH = -394 kJ

3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ

Step 2: To get C3H8 on the product side of the reaction, we need to reverse reaction 2, and change

the sign of DH.

3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ

C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJ

Step 3: Add two “new” reactions together to see what

remains:

3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ

3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ2

3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ

• Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction:

3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ

H2 (g) + 1/2O2 (g) H2O (l) DH = -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 DH = +1038 kJ

4H2 (g) + 2O2 (g) 4H2O (l) DH = -1144 kJ

3C (gr) + 4H2 (g) C3H8 (g)

3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ

4H2 (g) + 2O2 (g) 4H2O (l) DH = -1144 kJ

3C (gr) + 4H2 (g) C3H8 (g) DH = -106 kJ

Which is the one step reaction of interest

Another Example:

• Calculate DH for the following reaction:

H2(g) + Cl2(g) 2HCl(g)

Given the following:

NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ

N2 (g) + 3H2 (g) 2NH3 (g) DH = - 92 kJ

N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = - 629 kJ

Step 1: Only the first reaction contains the product of interest (HCl), but as a reactant. Therefore, reverse this reaction and multiply by 2 to get stoichiometry correct.

NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ

2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = +352 kJ

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) DH = 352 kJ

N2(g) + 4H2(g) + Cl2(g) 2NH4Cl(s) DH = -629 kJ

N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g)

DH = -277 kJ

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) DH = -277 kJ

? ( N2 (g) + 3H2(g) 2NH3(g) DH = -92 kJ)

H2(g) + Cl2(g) 2HCl(g) DH = ?

Key: need to reverse the middle reaction

• 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) DH = -277 kJ

2NH3(g) 3H2 (g) + N2 (g) DH = +92 kJ

H2(g) + Cl2(g) 2HCl(g) DH = -185 kJ

1

This is the desired reaction and resultant DH!