8.5 Hess’s Law
Learning Objective
 Learn how to combine chemical equations and their enthalpy changes.
Now that we understand that chemical reactions occur with a simultaneous change in energy, we can apply the concept more broadly. To start, remember that some chemical reactions are rather difficult to perform. For example, consider the combustion of carbon to make carbon monoxide:
2 C(s) + O_{2}(g) → 2 CO(g) ΔH = ?
In reality, this is extremely difficult to do; given the opportunity, carbon will react to make another compound, carbon dioxide:
C(s) + O_{2}(g) → CO_{2}(g) ΔH = −393.5 kJ
Is there a way around this? Yes. It comes from the understanding that chemical equations can be treated like algebraic equations, with the arrow acting like the equals sign. Like algebraic equations, chemical equations can be combined, and if the same substance appears on both sides of the arrow, it can be canceled out (much like a spectator ion in ionic equations). For example, consider these two reactions:
2 C(s) + 2 O_{2}(g) → 2 CO_{2}(g)
2 CO_{2}(g) → 2 CO(g) + O_{2}(g)
If we added these two equations by combining all the reactants together and all the products together, we would get
2 C(s) + 2 O_{2}(g) + 2 CO_{2}(g) → 2 CO_{2}(g) + 2 CO(g) + O_{2}(g)
We note that 2 CO_{2}(g) appears on both sides of the arrow, so they cancel:
We also note that there are 2 mol of O_{2} on the reactant side, and 1 mol of O_{2} on the product side. We can cancel 1 mol of O_{2} from both sides:
What do we have left?
2 C(s) + O_{2}(g) → 2 CO(g)
This is the reaction we are looking for! So by algebraically combining chemical equations, we can generate new chemical equations that may not be feasible to perform.
What about the enthalpy changes? Hess’s law states that when chemical equations are combined algebraically, their enthalpies can be combined in exactly the same way. Two corollaries immediately present themselves:
 If a chemical reaction is reversed, the sign on ΔH is changed.
 If a multiple of a chemical reaction is taken, the same multiple of the ΔH is taken as well.
What are the equations being combined? The first chemical equation is the combustion of C, which produces CO_{2}:
2 C(s) + 2 O_{2}(g) → 2 CO_{2}(g)
This reaction is two times the reaction to make CO_{2} from C(s) and O_{2}(g), whose enthalpy change is known:
C(s) + O_{2}(g) → CO_{2}(g) ΔH = −393.5 kJ
According to the first corollary, the first reaction has an energy change of two times −393.5 kJ, or −787.0 kJ:
2 C(s) + 2 O_{2}(g) → 2 CO_{2}(g) ΔH = −787.0 kJ
The second reaction in the combination is related to the combustion of CO(g):
2 CO(g) + O_{2}(g) → 2 CO_{2}(g) ΔH = −566.0 kJ
The second reaction in our combination is the reverse of the combustion of CO. When we reverse the reaction, we change the sign on the ΔH:
2 CO_{2}(g) → 2 CO(g) + O_{2}(g) ΔH = +566.0 kJ
Now that we have identified the enthalpy changes of the two component chemical equations, we can combine the ΔH values and add them:
Hess’s law is very powerful. It allows us to combine equations to generate new chemical reactions whose enthalpy changes can be calculated, rather than directly measured.
Example 10
Determine the enthalpy change of
C_{2}H_{4} + 3 O_{2} → 2 CO_{2} + 2 H_{2}O ΔH = ?
from these reactions:
C_{2}H_{2} + H_{2} → C_{2}H_{4} ΔH = −174.5 kJ
2 C_{2}H_{2} + 5 O_{2} → 4 CO_{2} + 2 H_{2}O ΔH = −1,692.2 kJ
2 CO_{2} + H_{2} → 2 O_{2} + C_{2}H_{2} ΔH = −167.5 kJ
Solution
We will start by writing chemical reactions that put the correct number of moles of the correct substance on the proper side. For example, our desired reaction has C_{2}H_{4} as a reactant, and only one reaction from our data has C_{2}H_{4}. However, it has C_{2}H_{4} as a product. To make it a reactant, we need to reverse the reaction, changing the sign on the ΔH:
C_{2}H_{4} → C_{2}H_{2} + H_{2} ΔH = +174.5 kJ
We need CO_{2} and H_{2}O as products. The second reaction has them on the proper side, so let us include one of these reactions (with the hope that the coefficients will work out when all our reactions are added):
2 C_{2}H_{2} + 5 O_{2} → 4 CO_{2} + 2 H_{2}O ΔH = −1,692.2 kJ
We note that we now have 4 mol of CO_{2} as products; we need to get rid of 2 mol of CO_{2}. The last reaction has 2CO_{2} as a reactant. Let us use it as written:
2 CO_{2} + H_{2} → 2 O_{2} + C_{2}H_{2} ΔH = −167.5 kJ
We combine these three reactions, modified as stated:
What cancels? 2 C_{2}H_{2}, H_{2}, 2 O_{2}, and 2 CO_{2}. What is left is
C_{2}H_{4} + 3 O_{2} → 2 CO_{2} + 2H_{2}O
which is the reaction we are looking for. The ΔH of this reaction is the sum of the three ΔH values:
ΔH = +174.5 − 1,692.2 − 167.5 = −1,685.2 kJ
Test Yourself
Given the thermochemical equations
Pb + Cl_{2} → PbCl_{2} ΔH = −223 kJ
PbCl_{2} + Cl_{2} → PbCl_{4} ΔH = −87 kJ
determine ΔH for
2 PbCl_{2} → Pb + PbCl_{4}
Answer
+136 kJ
Key Takeaway
 Hess’s law allows us to combine reactions algebraically and then combine their enthalpy changes the same way.
Exercises

Define Hess’s law.

What does Hess’s law require us to do to the ΔH of a thermochemical equation if we reverse the equation?

If the ΔH for
C_{2}H_{4} + H_{2} → C_{2}H_{6}
is −65.6 kJ, what is the ΔH for this reaction?
C_{2}H_{6} → C_{2}H_{4} + H_{2}

If the ΔH for
2 Na + Cl_{2} → 2NaCl
is −772 kJ, what is the ΔH for this reaction:
2 NaCl → 2 Na + Cl_{2}

If the ΔH for
C_{2}H_{4} + H_{2} → C_{2}H_{6}
is −65.6 kJ, what is the ΔH for this reaction?
2 C_{2}H_{4} + 2 H_{2} → 2 C_{2}H_{6}

If the ΔH for
2 C_{2}H_{6} + 7 O_{2} → 4 CO_{2} + 6 H_{2}O
is −2,650 kJ, what is the ΔH for this reaction?
6 C_{2}H_{6} + 21 O_{2} → 12 CO_{2} + 18 H_{2}O

The ΔH for
C_{2}H_{4} + H_{2}O → C_{2}H_{5}OH
is −44 kJ. What is the ΔH for this reaction?
2 C_{2}H_{5}OH → 2 C_{2}H_{4} + 2 H_{2}O

The ΔH for
N_{2} + O_{2} → 2 NO
is 181 kJ. What is the ΔH for this reaction?
NO → 1/2 N_{2} + 1/2 O_{2}

Determine the ΔH for the reaction
Cu + Cl_{2} → CuCl_{2}
given these data:
2 Cu + Cl_{2} → 2 CuCl ΔH = −274 kJ
2 CuCl + Cl_{2} → 2 CuCl_{2} ΔH = −166 kJ 
Determine ΔH for the reaction
2 CH_{4} → 2 H_{2} + C_{2}H_{4}
given these data:
CH_{4} + 2 O_{2} → CO_{2} + 2 H_{2}O ΔH = −891 kJ
C_{2}H_{4} + 3 O_{2} → 2 CO_{2} + 2 H_{2}O ΔH = −1,411 kJ
2 H_{2} + O_{2} → 2 H_{2}O ΔH = −571 kJ 
Determine ΔH for the reaction
Fe_{2}(SO_{4})_{3} → Fe_{2}O_{3} + 3SO_{3}
given these data:
4 Fe + 3O_{2} → 2 Fe_{2}O_{3} ΔH = −1,650 kJ
2 S + 3 O_{2} → 2 SO_{3} ΔH = −792 kJ
2 Fe + 3 S + 6 O_{2} → Fe_{2}(SO_{4})_{3} ΔH = −2,583 kJ 
Determine ΔH for the reaction
CaCO_{3} → CaO + CO_{2}
given these data:
2 Ca + 2 C + 3 O_{2} → 2 CaCO_{3} ΔH = −2,414 kJ
C + O_{2} → CO_{2} ΔH = −393.5 kJ
2 Ca + O_{2} → 2 CaO ΔH = −1,270 kJ
Answers
1.
If chemical equations are combined, their energy changes are also combined.
3.
ΔH = 65.6 kJ
5.
ΔH = −131.2 kJ
7.
ΔH = 88 kJ
9.
ΔH = −220 kJ
11.
ΔH = 570 kJ