# 7.6: Hess's Law

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##### 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:

$\ce{2C(s) + O2(g) → 2CO(g)} \quad ΔH = ? \nonumber$

In reality, this is extremely difficult to do. Given the opportunity, carbon will react to make another compound, carbon dioxide:

$\ce{2C(s) + O2(g) → 2CO2(g)} \quad ΔH = −393.5 kJ\nonumber$

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:

\begin{align*} \ce{2C(s) + 2O2(g) &→ 2CO2(g)} \\[4pt] \ce{2CO2(g) &→ 2CO(g) + O2(g)}\end{align*} \nonumber

If we added these two equations by combining all the reactants together and all the products together, we would get

$\ce{2C(s) + 2O2(g) + 2CO2(g) → 2CO2(g) + 2CO(g) + O2(g)}\nonumber$

We note that $$\ce{2CO2(g)}$$ appears on both sides of the arrow, so they cancel:

$\ce{2C(s) + 2O_{2}(g) + \cancel{2CO_{2}(g)}\rightarrow \cancel{2CO_{2}(g)}+2CO(g)+O_{2}(g)}\nonumber$

We also note that there are 2 mol of O2 on the reactant side, and 1 mol of O2 on the product side. We can cancel 1 mol of O2 from both sides:

$\ce{2C(s) + 2O_{2}(g)\rightarrow 2CO(g)+O_{2}(g)}\nonumber$

What do we have left?

$\ce{2C(s) + O2(g) → 2CO(g)}\nonumber$

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:

1. If a chemical reaction is reversed, the sign on $$ΔH$$ is changed.
2. 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 CO2:

$\ce{2C(s) + 2O2(g) → 2CO2(g)}\nonumber$

This reaction is two times the reaction to make $$\ce{CO2}$$ from $$\ce{C(s)}$$ and $$\ce{O2(g)}$$, whose enthalpy change is known:

$\ce{C(s) + O2(g) → CO2(g)} \quad ΔH = −393.5 kJ\nonumber$

According to the first corollary, the first reaction has an energy change of two times −393.5 kJ, or −787.0 kJ:

$\ce{2C(s) + 2O2(g) → 2CO2(g)} \quad ΔH = −787.0 kJ\nonumber$

The second reaction in the combination is related to the combustion of CO(g):

$\ce{2CO(g) + O2(g) → 2CO2(g)} \quad ΔH = −566.0 kJ\nonumber$

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:

$\ce{2CO2(g) → 2CO(g) + O2(g)} \quad ΔH = +566.0 kJ\nonumber$

Now that we have identified the enthalpy changes of the two component chemical equations, we can combine the $$ΔH$$ values and add them:

$\begin{cases} 2C(s)+2O_{2}(g)\rightarrow 2CO_{2}(g)\; \; \; \; \; \; \Delta H=-787.0kJ\\ \cancel{2CO_{2}(g)}\rightarrow 2CO(g))+\cancel{O_{2}(g)}\; \; \, \, \Delta H=+566.0kJ\\ -------------------------\\ 2C(s)+O_{2}(g)\rightarrow 2CO(g)\; \; \; \; \; \; \; \; \; \; \, \Delta H=-787.0+566.0=-221.0kJ\\ \end{cases}\nonumber$

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

$\ce{C2H4 + 3O2 → 2CO2 + 2H2O} \quad ΔH = ?\nonumber$

from these reactions:

$\ce{C2H2 + H2 → C2H4} \quad ΔH = −174.5 kJ\nonumber$

$\ce{2C2H2 + 5O2 → 4CO2 + 2H2O} \quad ΔH = −1,692.2 kJ\nonumber$

$\ce{2CO2 + H2 → 2O2 + C2H2} \quad ΔH = −167.5 kJ\nonumber$

###### 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 C2H4 as a reactant, and only one reaction from our data has C2H4. However, it has C2H4 as a product. To make it a reactant, we need to reverse the reaction, changing the sign on the ΔH:

$\ce{C2H4 → C2H2 + H2} \quad ΔH = +174.5 kJ\nonumber$

We need CO2 and H2O 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 of our reactions are added):

$\ce{2C2H2 + 5O2 → 4CO2 + 2H2O} \quad ΔH = −1,692.2 kJ\nonumber$

We note that we now have 4 mol of CO2 as products; we need to get rid of 2 mol of CO2. The last reaction has 2CO2 as a reactant. Let us use it as written:

$\ce{2CO2 + H2 → 2O2 + C2H2} \quad ΔH = −167.5 kJ\nonumber$

We combine these three reactions, modified as stated:

What cancels? 2C2H2, H2, 2O2, and 2CO2. What is left is

$\ce{C2H4 + 3O2 → 2CO2 + 2H2O}\nonumber$

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

##### Exercise $$\PageIndex{1}$$

Given the thermochemical equations

$\ce{Pb + Cl2 → PbCl2} \quad ΔH = −223 kJ\nonumber$

$\ce{PbCl2 + Cl2 → PbCl4} \quad ΔH = −87 kJ\nonumber$

determine $$ΔH$$ for

$\ce{2PbCl2 → Pb + PbCl4}\nonumber$