2.4 Hess’s Law

Many chemical reactions occur in a series of steps rather than a single step.

For example, the following reaction describes the burning (combustion) of carbon:

(1) C (s) + O₂(g) → CO₂(g) ΔH = -393.5 kJ

If not enough oxygen is present, CO rather than CO₂ is produced:

(2) C (s) + ½ O₂(g) → CO (g) ΔH = -110.0 kJ

If more oxygen is now added, CO will undergo further combustion with oxygen:

(3) CO (g) + ½ O₂(g) → CO₂ (g) ΔH = -283.0 kJ

Watch what happens if we add together the second and third reaction:

(2) C (s) + ½ O2(g) → CO (g) (3) CO (g) + ½ O2(g) → CO2 (g) (1) C (s) + O2(g) → CO2(g)

Be sure you see how these equations can be added together Add things that are on the same side of the equation: ½O2+ ½O2= 1 O2 and cross things out on opposite side of the equation (the CO) Sometimes you will be left with a "remainder" - things won't cancel completely

Now compare the total energy released in the second and third reactions with the amount of energy released in the original reaction:

ΔHReaction 2	+	ΔHReaction 3	=	ΔHReaction 1
-110.5 kJ	+	(-283.0 kJ)	=	-393.5 kJ
-393.5 kJ			=	-393.5 kJ

The end result is that it doesn't matter if the reaction proceeds all at once or in series of steps; the net energy change is the same. This illustrates Hess's Law of Constant Heat Summation:

Here is another example. You need to make sure you carefully understand how the equations are added together.

Hess's Law The enthalpy change for any reaction depends only on the energy states of the final products and initial reactants and is independent of the pathway or the number of steps between the reactant and product.

Example: Given the intermediate steps in the production of tetraphosphorus decaoxide, P₄O₁₀, calculate ΔH_f for P₄O₁₀

Given the following reactions:

(1) 4 P + 3 O₂ → P₄O₆ ΔH = -1640 kJ

(2) P₄O₆ + 2 O₂ → P₄O₁₀ ΔH = -1344 kJ

Recall the the heat of formation reaction involves the production of one mole of the compound from it's elements. Thus, we want to calculate ΔH for:

(3) 4 P + 5 O₂ → P₄O₁₀

Answer

Carefully examine the reactions you are given, and see how they compare with the final equation for which you are asked to determine ΔH.

For this example, we see that if we add reactions (1) and (2) we can obtain the desired heat of formation reaction. Thus, we can also add together the ΔH values to obtain ΔH for the desired reaction:

(1) 4 P + 3 O2 → P4O6 ΔH = -1640kJ (2) P4O6 + 2 O2 → P4O10 ΔH = -1344 kJ (3) 4 P + 5 O2→ P4O10 ΔH = -2984 kJ Answer

Some Important Things to Know:

• If you reverse an equation, be sure to move the energy term to the other side of the equation as well. If you write the enthalpy term separately from the equation as ΔH, be sure to reverse the sign of ΔH

For example, if 1640 kJ of energy are released during the formation of 1 mole of P₄O₆ (see equation (1)), then the same amount of energy will be required when 1 mole of P₄O₆ is decomposed to its elements:

P₄O₆ + 1640 kJ → 4 P + 3 O₂ OR

P₄O₆ → 4 P + 3 O₂ ΔH = +1640 kJ

If you change the balancing coefficients within an equation, be sure to adjust the value of the energy term.

For example, if 1640 kJ of energy are released when 1 mole of P₄O₆ is formed from its elements:

4 P + 3 O₂ → P₄O₆ ΔH = -1640kJ

then producing 2 moles of P₄O₆ will release 3280 kJ of energy (2 × 1640 kJ):

8 P + 6O₂ → 2 P₄O₆ ΔH = -3280kJ

Tips for for Using Hess's Law:

• Consider the intermediate steps involved in a reaction and ΔH values for each.
• Reverse intermediate reactions and change the sign of ΔH if necessary.
• Multiply intermediate reactions as necessary to balance the overall equation and multiply ΔH values of the steps as required.

• Determine the final ΔH from the algebraic sum of ΔH values for the intermediate reactions.

Another Example:

You are given the following two reactions (Reactions 1 and 2):

(1)	C2H2(g) + 5/2 O2(g) → 2 CO2(g) + H2O(l)	ΔH = -1299 kJ
(2)	C6H6(l) + 15/2 O2(g) → 6 CO2(g) + 3 H2O(l)	ΔH = -3267 kJ
Find ΔH for the following reaction (Reaction 3) and tell whether it is exothermic or endothermic:
(3)	3 C2H2(g) → C6H6(g)

Solution:

1.	Reverse any of the intermediate steps as necessary. You will have to reverse equation 2. Why? Because C6H6 needs to end up on the product side of the equation, but it is on the reactant side in equation (2), the only equation in which it appears. Remember to change the sign of ΔH.
2.	Check to see how you will have to balance the overall equation. You will have to multiply equation (1) by 3 in order to end up with the correct number of moles for the final equation. Remember to also multiply ΔH by 3.
3.	Add the equations together:
	.

3 C2H2(g) + 15/2 O2(g) → 6 CO2(g) + 3 H2O(l)		ΔH = 3(-1299) = -3897
6 CO2(g) + 3 H2O(l) → C6H6(l) + 15/2 O2(g)		ΔH = + 3267 kJ
3 C2H2(g) → C6H6(g)		ΔH = -630 kJ

or

3 C2H2(g) → C6H6(g) + 630 kJ

Since ΔH is negative, the reaction is exothermic.