hess’ law

Hess’ LawHess’ Law

Reading: p 519-522Reading: p 519-522

OutlineOutline Definition of Hess’ LawDefinition of Hess’ Law Using Hess’ Law (examples)Using Hess’ Law (examples)

First Law of ThermodynamicsFirst Law of Thermodynamics

First Law: Energy of the Universe is First Law: Energy of the Universe is Constant Constant

E = q + wE = q + w

q = heat. Transferred between two bodiesq = heat. Transferred between two bodies

w = work. Force acting over a distance (F x w = work. Force acting over a distance (F x d)d)

Changes in EnthalpyChanges in Enthalpy

Consider the following expression for a chemical Consider the following expression for a chemical process:process:

H = HH = Hproductsproducts - H - Hreactantsreactants

If If H >0, then qH >0, then qpp >0. The reaction is endothermic >0. The reaction is endothermic

If If H <0, then qH <0, then qpp <0. The reaction is exothermic <0. The reaction is exothermic

Hess’ Law: An ExampleHess’ Law: An Example

Using Hess’ LawUsing Hess’ Law

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

2NO2 (g)

N2 (g) + 2O2 (g)

q

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

Example (cont.)Example (cont.)

Our reaction of interest is:Our reaction of interest is:

NN22(g) + 2O(g) + 2O22(g) 2NO(g) 2NO22(g) (g) H = 68 kJH = 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


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

NN22 (g) + O (g) + O22 (g) 2NO(g) (g) 2NO(g) H = 180 kJH = 180 kJ 2NO (g) + O2NO (g) + O22 (g) 2NO (g) 2NO22(g) (g) H = -112 kJH = -112 kJ

NN22 (g) + 2O (g) + 2O22 (g) 2NO (g) 2NO22(g) (g) H = 68 H = 68 kJkJ


Note the important things about this Note the important things about this example, the sum of example, the sum of H for the two H for the two reaction steps is equal to the reaction steps is equal to the H for the H for the reaction of interest.reaction of interest.

Big point: We can combine reactions of Big point: We can combine reactions of known known H to determine the H to determine the H for the H for the “combined” reaction.“combined” reaction.

Hess’ Law: DetailsHess’ Law: Details

Once can always reverse the Once can always reverse the direction of a reaction when making direction of a reaction when making a combined reaction. When you do a combined reaction. When you do this, the sign of this, the sign of H changes.H changes.

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

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

Details (cont.)Details (cont.) The magnitude of The magnitude of H is directly H is directly

proportional to the quantities involved proportional to the quantities involved (it is an “extensive” quantity).(it is an “extensive” quantity).

As such, if the coefficients of a reaction As such, if the coefficients of a reaction are multiplied by a constant, the value are multiplied by a constant, the value of of H is also multiplied by the same H is also multiplied by the same integer.integer.

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

N2(g) + 4O2(g) 4NO2(g) H = 136 kJ

Using Hess’ LawUsing Hess’ Law

When trying to combine reactions to When trying to combine reactions to form a reaction of interest, one form a reaction of interest, one usually works usually works backwardsbackwards from the from the reaction of interest.reaction of interest.

Example:Example:

What is What is H for the following reaction?H for the following reaction?

3C (g) + 4H3C (g) + 4H22 (g) C (g) C33HH88 (g) (g)


3C (g) + 4H3C (g) + 4H22 (g) C (g) C33HH88 (g) (g) H = H = ??

• You’re given the following reactions:

C (g) + 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


Step 1. Only reaction 1 has C (g). Step 1. Only reaction 1 has C (g). Therefore, we will multiply by 3 to Therefore, we will multiply by 3 to get the correct amount of C (g) with get the correct amount of C (g) with respect to our final equation.respect to our final equation.

C (g) + O2 (g) CO2 (g) H = -394 kJ

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

Initial:

Final:


Step 2. To get CStep 2. To get C33HH88 on the product on the product side of the reaction, we need to side of the reaction, we need to reverse reaction 2.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:


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


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

3C (g) + 4H3C (g) + 4H22O (l) CO (l) C33HH88 (g) + 2O (g) + 2O2 2 H = H = +1038 kJ +1038 kJ

HH22 (g) + 1/2O (g) + 1/2O22 (g) H (g) H22O (l) O (l) H = -286 kJH = -286 kJ

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

Need to multiply second reaction by 4


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

3C (g) + 4H3C (g) + 4H22O (l) CO (l) C33HH88 (g) + 2O (g) + 2O2 2 H = +1038kJ H = +1038kJ

44HH22 (g) + (g) + 22OO22 (g) (g) 44HH22O (l) O (l) H = H = -1144-1144 kJkJ

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


Step 4 (cont.): Step 4 (cont.):

3C (g) + 4H3C (g) + 4H22O (l) CO (l) C33HH88 (g) + 2O (g) + 2O2 2 H = +1038 kJ H = +1038 kJ

44HH22 (g) + (g) + 22OO22 (g) (g) 44HH22O (l) O (l) H = H = -1144-1144 kJ kJ

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

Another ExampleAnother Example

Calculate Calculate H for the following H for the following reaction:reaction:

HH22(g) + Cl(g) + Cl22(g) 2HCl(g)(g) 2HCl(g)

Given the following:Given the following:

NHNH33 (g) + HCl (g) NH (g) + HCl (g) NH44Cl(s) Cl(s) H = -176 kJH = -176 kJ

NN22 (g) + 3H (g) + 3H22 (g) 2NH (g) 2NH33 (g) (g) H = -92 kJH = -92 kJ

NN22 (g) + 4H (g) + 4H22 (g) + Cl (g) + Cl22 (g) 2NH (g) 2NH44Cl(s) Cl(s) H = -629 kJH = -629 kJ

Another Example (cont.)Another Example (cont.)

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

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

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


Step 2. Need ClStep 2. Need Cl22 as a reactant, as a reactant, therefore, add reaction 3 to result therefore, add reaction 3 to result from step 1 and see what is left.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


Step 3. Use remaining known Step 3. Use remaining known reaction in combination with the reaction in combination with the result from Step 2 to get final result from Step 2 to get final reaction.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


Step 3. Use remaining known Step 3. Use remaining known reaction in combination with the reaction in combination with the result from Step 2 to get final result from Step 2 to get final reaction.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

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hess’ law

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