7.6: Hess’s Law
 Page ID
 64049
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:
2C(s) + O_{2}(g) → 2CO(g) ΔH = ?
In reality, this is extremely difficult to do; given the opportunity, carbon will react to make another compound, carbon dioxide:
2C(s) + O_{2}(g) → 2CO_{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:
2C(s) + 2O_{2}(g) → 2CO_{2}(g)2CO_{2}(g) → 2CO(g) + O_{2}(g)
If we added these two equations by combining all the reactants together and all the products together, we would get
2C(s) + 2O_{2}(g) + 2CO_{2}(g) → 2CO_{2}(g) + 2CO(g) + O_{2}(g)
We note that 2CO_{2}(g) appears on both sides of the arrow, so they cancel:
\[2C(s)+2O_{2}(g)+\not{2CO_{2}(g)}\rightarrow \not{2CO_{2}(g)}+2CO(g)+O_{2}(g)\]
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:
\[2C(s)+2O_{2}(g)\rightarrow 2CO(g)+O_{2}(g)\]
What do we have left?
2C(s) + O_{2}(g) → 2CO(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}:
2C(s) + 2O_{2}(g) → 2CO_{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:
2C(s) + 2O_{2}(g) → 2CO_{2}(g) ΔH = −787.0 kJ
The second reaction in the combination is related to the combustion of CO(g):
2CO(g) + O_{2}(g) → 2CO_{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:
2CO_{2}(g) → 2CO(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:
\[\begin{cases} 2C(s)+2O_{2}(g)\rightarrow 2CO_{2}(g)\; \; \; \; \; \; \Delta H=787.0kJ\\ \not{2CO_{2}(g)}\rightarrow 2CO(g))+\not{O_{2}(g)}\; \; \, \, \Delta H=+566.0kJ\\ \\ 2C(s)+O_{2}(g)\rightarrow 2CO(g)\; \; \; \; \; \; \; \; \; \; \, \Delta H=787.0+566.0=221.0kJ\\ \end{cases}\]
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 \(\PageIndex{1}\):
Determine the enthalpy change of
C_{2}H_{4} + 3O_{2} → 2CO_{2} + 2H_{2}O ΔH = ?from these reactions:
C_{2}H_{2} + H_{2} → C_{2}H_{4} ΔH = −174.5 kJ2C_{2}H_{2} + 5O_{2} → 4CO_{2} + 2H_{2}O ΔH = −1,692.2 kJ2CO_{2} + H_{2} → 2O_{2} + C_{2}H_{2} ΔH = −167.5 kJSolution
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 kJWe 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):
2C_{2}H_{2} + 5O_{2} → 4CO_{2} + 2H_{2}O ΔH = −1,692.2 kJWe 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:
2CO_{2} + H_{2} → 2O_{2} + C_{2}H_{2} ΔH = −167.5 kJWe combine these three reactions, modified as stated:
What cancels? 2C_{2}H_{2}, H_{2}, 2O_{2}, and 2CO_{2}. What is left is
C_{2}H_{4} + 3O_{2} → 2CO_{2} + 2H_{2}Owhich 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
Exercise \(\PageIndex{1}\)
Test Yourself
Given the thermochemical equations
Pb + Cl_{2} → PbCl_{2} ΔH = −223 kJPbCl_{2} + Cl_{2} → PbCl_{4} ΔH = −87 kJdetermine ΔH for
2PbCl_{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.
Exercise \(\PageIndex{1}\)
 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
2Na + Cl_{2} → 2NaClis −772 kJ, what is the ΔH for this reaction:
2NaCl → 2Na + 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?
2C_{2}H_{4} + 2H_{2} → 2C_{2}H_{6} 
If the ΔH for
2C_{2}H_{6} + 7O_{2} → 4CO_{2} + 6H_{2}Ois −2,650 kJ, what is the ΔH for this reaction?
6C_{2}H_{6} + 21O_{2} → 12CO_{2} + 18H_{2}O 
The ΔH for
C_{2}H_{4} + H_{2}O → C_{2}H_{5}OHis −44 kJ. What is the ΔH for this reaction?
2C_{2}H_{5}OH → 2C_{2}H_{4} + 2H_{2}O 
The ΔH for
N_{2} + O_{2} → 2NOis 181 kJ. What is the ΔH for this reaction?
NO → 1/2N_{2} + 1/2O_{2} 
Determine the ΔH for the reaction
Cu + Cl_{2} → CuCl_{2}given these data:
2Cu + Cl_{2} → 2CuCl ΔH = −274 kJ2CuCl + Cl_{2} → 2CuCl_{2} ΔH = −166 kJ 
Determine ΔH for the reaction
2CH_{4} → 2H_{2} + C_{2}H_{4}given these data:
CH_{4} + 2O_{2} → CO_{2} + 2H_{2}O ΔH = −891 kJC_{2}H_{4} + 3O_{2} → 2CO_{2} + 2H_{2}O ΔH = −1,411 kJ2H_{2} + O_{2} → 2H_{2}O ΔH = −571 kJ 
Determine ΔH for the reaction
Fe_{2}(SO_{4})_{3} → Fe_{2}O_{3} + 3SO_{3}given these data:
4Fe + 3O_{2} → 2Fe_{2}O_{3} ΔH = −1,650 kJ2S + 3O_{2} → 2SO_{3} ΔH = −792 kJ2Fe + 3S + 6O_{2} → Fe_{2}(SO_{4})_{3} ΔH = −2,583 kJ
 Determine ΔH for the reaction
given these data:
2Ca + 2C + 3O_{2} → 2CaCO_{3} ΔH = −2,414 kJC + O_{2} → CO_{2} ΔH = −393.5 kJ2Ca + O_{2} → 2CaO ΔH = −1,270 kJ
Answers
 If chemical equations are combined, their energy changes are also combined.


ΔH = 65.6 kJ


ΔH = −131.2 kJ


ΔH = 88 kJ


ΔH = −220 kJ


ΔH = 570 kJ