Hess’s Law

Hess’s Law• States that the energy change in an overall chemical

reaction is equal to the sum of the energy changes in the individual reactions comprising it.

• The law is based on the principle of conservation of energy and the path independence of energy changes

• Hess's law can be used to predict energy changes that are not easily measured.

Whaaaa?

• A + B C ΔH = -34 kJ

• C + B D ΔH = -87 kJ

• Overall:

• A + 2B D ΔH = -121 kJ

Now does the definition make more sense?

• Hess’s Law• States that the energy change in an overall chemical

reaction is equal to the sum of the energy changes in the individual reactions comprising it.

Example:N2(g) + 2O2(g) 2NO2(g) ΔH = 68 kJ

N2 (g) + O2 (g) 2NO(g) ΔH = 180 kJ2NO (g) + O2 (g) 2NO2(g) ΔH = -112 kJ

Simple Practice – Example 0• Given:

• X Y ΔH = -40 kJ

• X Z ΔH = -95 kJ

• What is ΔH of the reaction Y Z?

Remember this:

1. First decide how to rearrange equations so reactants and products are on appropriate sides of the arrows.

2. If equations had to be reversed, reverse the sign of ΔH

3. If equations had be multiplied to get a correct coefficient, multiply the ΔH by this coefficient.

4. Check to ensure that everything cancels out to give you the exact equation you want

5. Hint** It is often helpful to begin your work backwards from the answer that you want!

Simple Practice – Example 0• Given:

• X Y ΔH = -40 kJ

• X Z ΔH = -95 kJ

• What is ΔH of the reaction Y Z?

Simple Practice – Example 0• Given:

• X Y ΔH = -40 kJ flip this reaction and its ΔH

• X Z ΔH = -95 kJ

• What is ΔH of the reaction Y Z?

Simple Practice – Example 0• Given:

• X Y ΔH = -40 kJ flip this reaction and its ΔH

• X Z ΔH = -95 kJ

• What is ΔH of the reaction Y Z?

• Y X ΔH = +40 kJ

• X Z ΔH = -95 kJ

• Now add them together

Simple Practice – Example 0• Given:

• X Y ΔH = -40 kJ flip this reaction and its ΔH

• X Z ΔH = -95 kJ

• What is ΔH of the reaction Y Z?

• Y X ΔH = +40 kJ

• X Z ΔH = -95 kJ

• Now add them together

• Y Z ΔH = +40 – 95 = -55 kJ

Real Chemistry Example 1• Given:

P4 + 3O2 P4O6 ΔH = -1640.1 kJ

P4 + 5O2 P4O10 ΔH = -2940.1 kJ

Solve for the ΔH of the reaction:

P4O6 + 2O2 P4O10

Real Chemistry Example 1 • Given:

P4 + 3O2 P4O6 ΔH = -1640.1 kJ

P4 + 5O2 P4O10 ΔH = -2940.1 kJ

Solve for the ΔH of the reaction:

P4O6 + 2O2 P4O10

Look for unique chemicals. Which reaction should be flipped?

Real Chemistry Example 1• Given:

P4 + 3O2 P4O6 ΔH = -1640.1 kJ

P4 + 5O2 P4O10 ΔH = -2940.1 kJ

Solve for the ΔH of the reaction:

P4O6 + 2O2 P4O10

P4O6 P4 + 3O2 ΔH = +1640.1 kJ

P4 + 5O2 P4O10 ΔH = -2940.1 kJ

Real Chemistry Example 1Solve for the ΔH of the reaction:

P4O6 + 2O2 P4O10

P4O6 P4 + 3O2 ΔH = +1640.1 kJ

P4 + 5O2 P4O10 ΔH = -2940.1 kJ

Added together gives final reaction

ΔH = +1640.1 kJ -2940.1 kJ = -1300.0 kJ

Real Chemistry Example 2• Givens:

• H2 + F2 2HF ΔH = -537 kJ

• C + 2F2 CF4 ΔH = -680 kJ

• 2C + 2H2 C2H4 ΔH = +52.3 kJ

• Calcualte ΔH of:• C2H4 + 6F2 2CF4 + 4HF ΔH = ?

Real Chemistry Example 2• Givens:

• H2 + F2 2HF ΔH = -537 kJ

• C + 2F2 CF4 ΔH = -680 kJ

• 2C + 2H2 C2H4 ΔH = +52.3 kJ• Flip this reaction

• Calcualte ΔH of:• C2H4 + 6F2 2CF4 + 4HF ΔH = ?

Real Chemistry Example 2• Givens:

• H2 + F2 2HF ΔH = -537 kJ

• C + 2F2 CF4 ΔH = -680 kJ• Multiply this reaction by 2 all the way across

• 2C + 2H2 C2H4 ΔH = +52.3 kJ• Flip this reaction

• Calcualte ΔH of:• C2H4 + 6F2 2CF4 + 4HF ΔH = ?

Real Chemistry Example 2• Givens:

• H2 + F2 2HF ΔH = -537 kJ• Multiply this reaction by 2 all the way across

• C + 2F2 CF4 ΔH = -680 kJ• Multiply this reaction by 2 all the way across

• 2C + 2H2 C2H4 ΔH = +52.3 kJ• Flip this reaction

• Calcualte ΔH of:• C2H4 + 6F2 2CF4 + 4HF ΔH = ?

Real Chemistry Example 2• Givens:

• H2 + F2 2HF ΔH = -537 kJ• Multiply this reaction by 2 all the way across

• C + 2F2 CF4 ΔH = -680 kJ• Multiply this reaction by 2 all the way across

• 2C + 2H2 C2H4 ΔH = +52.3 kJ• Flip this reaction

• Calcualte ΔH of:• C2H4 + 6F2 2CF4 + 4HF ΔH = -2486 kJ

Another easy one!Given:

S(s) + O2 (g) SO2 (g) ∆H= -297 kJ

2SO2 (g) + O2 (g) 2SO3 (g) ∆H= - 198 kJ

Solve for:

2S(s) + 3O2 (g) 2SO3(g) ∆H = ?

Challenge Example!Given:B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g) ΔH = 2035 kJ

H2O (l) → H2O (g) ΔH = 44 kJ

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

2 B (s) + 3 H2 (g) → B2H6 (g) ΔH = 36 kJ

Find the ΔH of:

2 B (s) + (3/2) O2 (g) → B2O3 (s)

Homework• Worksheet 2 (all 8 problems)

• Prelab “Enthalpy of Reaction” – do in lab book

Example

Given:B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g) (ΔH = 2035 kJ/mol)

H2O (l) → H2O (g) (ΔH = 44 kJ/mol)

H2 (g) + (1/2) O2 (g) → H2O (l) (ΔH = -286 kJ/mol)

2 B (s) + 3 H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol)

leave this reaction alone

Find the ΔHf of:

2 B (s) + (3/2) O2 (g) → B2O3 (s)

Example

Given:B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g) (ΔH = 2035 kJ/mol)

H2O (l) → H2O (g) (ΔH = 44 kJ/mol)

H2 (g) + (1/2) O2 (g) → H2O (l) (ΔH = -286 kJ/mol)

2 B (s) + 3 H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol)

flip this reaction and its sign

Find the ΔHf of:

2 B (s) + (3/2) O2 (g) → B2O3 (s)

Steps…• Step 1: Manipulate Equations

• Fix # of moles (multiply everything)

• Reverse if needed (change sign of ∆H)

• Step 2: Add equations and cancel out the common terms on both sides• Remember net ionic equations??

2 B (s) + (3/2) O2 (g) → B2O3 (s) B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g) (ΔH = 2035 kJ/mol)

Needs to be reversed…

B2H6 (g) + 3 O2 (g) → B2O3 (s) + 3 H2O (g) (ΔH = -2035 kJ/mol)

H2O (l) → H2O (g) (ΔH = 44 kJ/mol)

Needs to be reversed & Fix # moles…

3 H2O (g) → 3 H2O (l) (ΔH = -132 kJ/mol)

H2 (g) + (1/2) O2 (g) → H2O (l) (ΔH = -286 kJ/mol)

Needs to be reversed & Fix # moles…

3 H2O (l) → 3 H2 (g) + (3/2) O2 (g) (ΔH = 858 kJ/mol)

2 B (s) + 3 H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol)

(no change needed!)

Add equations and cancel out the common terms on both sides

B2H6 (g) + 3 O2 (g) → B2O3 (s) + 3 H2O (g) (ΔH = -2035 kJ/mol)

3 H2O (g) → 3 H2O (l) (ΔH = -132 kJ/mol)

3 H2O (l) → 3 H2 (g) + (3/2) O2 (g) (ΔH = 858 kJ/mol)

2 B (s) + 3 H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol)

2 B (s) + (3/2) O2 (g) → B2O3 (s) ∆H = -1273 kJ/mol

Practice Problem:2S(s) + 3O2 (g) 2SO3(g) ∆H = ?

S(s) + O2 (g) SO2 (g) ∆H= -297 kJ

2SO2 (g) + O2 (g) 2SO3 (g) ∆H= - 198 kJ

Manipulate:

2 S(s) + 2O2 (g) 2SO2 (g) ∆H = - 594 kJ

2 SO2(g) + O2 (g) 2 SO3 (g) ∆H = - 198 kJ

Addition:

2S (s) + 3O2 (g) + 2SO2 (g) 2SO2 (g) + 2SO3 (g)

∆H = - 792 kJ

Net:

2S (s) + 3O2 (g) 2 SO3 (g) ∆H = - 792 kJ

Is this reaction endo or exo? EXOTHERMIC

Spontaneous or not? SPONTANEOUS