5.5: Hess's Law

Learning Objectives

• Use Hess’s Law to calculate the enthalpy change of a reaction by combining equations with known enthalpy changes.
• Explain how reversing or scaling chemical equations affects the enthalpy change, and account for physical states when applying Hess’s Law.

Hess’s Law

There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other reactions with known enthalpy changes. Some reactions are difficult, if not impossible, to study directly. Even when a reaction is easy to measure, it’s often useful to calculate the enthalpy change from other known reactions.

Hess’s Law states that if a process occurs in multiple steps, the total enthalpy change is the sum of the enthalpy changes for each individual step. This is possible because enthalpy is a state function, it depends only on the starting and ending states, not on the path taken to get from one to the other.

As an example, consider the combustion of carbon to form carbon dioxide. This can happen directly:

\[\ce{C(s) + O2(g) -> CO2(g)} \quad \Delta H=-394 kJ/mol \nonumber \]

The reaction can also occur indirectly, in two steps, where carbon monoxide is formed in the first step:

\[\ce{C(s) + 1/2 O2(g) -> CO (g)} \quad \Delta H=-111 kJ/mol \nonumber \]

Carbon monoxide reacts further with oxygen to form carbon dioxide in the second step:

\[\ce{CO(g) + 1/2 O2(g) -> CO2(g)} \quad \Delta H=-283 kJ/mol \nonumber \]

The equation describing the overall reaction is the sum of these two chemical changes:

\[\begin{align*}
&\text{step 1} & \ce{C (s) + 1/2 O2(g)} & \ce{-> CO(g)} \\[2pt]
&\text{step 2} & \ce{CO(g) + 1/2 O2(g)} & \ce{-> CO2(g)} \\[2pt]
\hline
&\text{sum} & \ce{C (s) + \cancel{CO(g)} + O2(g) } & \ce{-> \cancel{CO(g)} + CO2(g)}
\end{align*} \]

Because the \(\ce{CO(g)}\) produced in Step 1 is consumed in Step 2, the net change is:

\[\ce{C(s) + O2(g) -> CO2(g)} \nonumber \]

According to Hess’s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps.

\[
\begin{align*}
&\text{step 1} & \ce{C (s) + 1/2 O2(g)} & \ce{-> \cancel{CO(g)}} &\Delta H=-111\,\text{kJ/mol} \\[2pt]
&\text{step 2} & \ce{\cancel{CO(g)} + 1/2 O2(g)} & \ce{-> CO2(g)} & \Delta H=-283 \,\text{kJ/mol} \\[2pt]
\hline
&\text{sum} & \ce{C (s) + O2(g)} & \ce{-> CO2(g)} & \Delta H=-394\,\text{kJ/mol}
\end{align*} \]

The principle of Hess's Law and enthalpy as a state function is shown graphically in Figure \(\PageIndex{1}\), the total enthalpy change is the same whether the reaction occurs in one step or two. This is the essence of Hess’s Law: enthalpy changes are additive. It is important to specify the physical states of reactants and products in thermochemical equations, as enthalpy values depend on them.

Figure \(\PageIndex{1}\): The formation of \(\ce{CO2(g)}\) from its elements can be thought of as occurring in two steps, which sum to the overall reaction, as described by Hess’s law. The horizontal blue lines represent enthalpies. For an exothermic process, the products are at lower enthalpy than are the reactants.

Recall two important features of \(\Delta H\):

1. \(\Delta H\) is directly proportional to the stoichiometric coefficients of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of \(\ce{NO2(g)}\) is +33.2 kJ/mol:

   \[\ce{ 1/2 N2(g) + O2(g) -> NO2(g)} \quad \Delta H=+33.2 \,\text{kJ/mol} \nonumber \]

   If the reaction is doubled (stoichiometry is doubled), the enthalpy change doubles:

   \[\ce{N2(g) + 2 O2(g) -> 2 NO2(g)} \quad \Delta H=+66.4\, \text{kJ/mol} \nonumber \]

   When a chemical equation is multiplied or divided by a number, the corresponding \(ΔH\) must be multiplied or divided by the same factor.

   • Note: The unit kJ/mol refers to the enthalpy change per mole of reaction as written, including all reactants and products in their specified stoichiometric coefficients. It does not refer to the enthalpy per mole of an individual substance.

2. Reversing a reaction changes the sign of the enthalpy change. The decomposition of NO₂(g) is the reverse of the formation reaction above:
   \[\ce{NO2(g) -> 1/2 N2(g) + O2(g)} \quad \Delta H=-33.2 \,\text{kJ/mol} \nonumber \]
   Since the direction of heat flow is reversed, the enthalpy change has the same magnitude but the opposite sign.

Example \(\PageIndex{1}\)

When carbon is burned with limited amounts of oxygen gas (O₂), carbon monoxide (CO) is the main product:

\[ \left ( 1 \right ) \; \ce{2C (s) + O2 (g) -> 2 CO (g)} \quad \Delta H=-221.0 \; \text{kJ/mol} \nonumber \]

When carbon is burned in excess O₂, carbon dioxide (CO₂) is produced:

\[ \left ( 2 \right ) \; \ce{C (s) + O2 (g) -> CO2 (g)} \quad \Delta H=-393.5 \; \text{kJ/mol} \nonumber \]

What is \(ΔH\) for the following reaction?

\[ \ce{2CO (g) + O2 (g) -> 2CO2 (g)} \nonumber \]

Given: two balanced chemical equations and their \(ΔH\) values

Solution:

To apply Hess’s Law, we need to combine the following equations to arrive at the desired reaction.

Step 1: Reverse the first reaction so that CO is a reactant and change the sign of \(ΔH\):

\[ \left ( 1 \right ) \; \ce{2 CO (g) -> 2C (s) + O2 (g)} \quad \Delta H=+221.0 \; \text{kJ/mol} \nonumber \]

Step 2: Multiply the second reaction by 2 so that the stoichiometric coefficient of CO₂ is 2, and adjust \(ΔH\) accordingly:

\[ \left ( 2 \right ) \; \ce{2C (s) + 2O2 (g) -> 2CO2 (g)} \quad \Delta H=2(-393.5 \; \text{kJ/mol})= -787.0 \; \text{kJ/mol}\nonumber \]

Step 3: Add the two reactions together:

\[ \begin{align*}&\text{step 1}& \ce{2 CO (g) &-> \cancel{2 C(s)} + \cancel{O_2 (g)} } & \Delta H & = +221.0 \; \text{kJ/mol} \\ &\text{step 2}&\ce{\cancel{2 C (s)} + \cancel{2} O2 (g) &-> 2 CO2 (g)} & \Delta H & =-787.0 \; \text{kJ/mol} \\[2ex] \hline &\text{sum}&\ce{2 CO (g) + O2 (g) &-> 2 CO2 (g)} & \Delta H &=-566.0 \; \text{kJ/mol} \end{align*} \nonumber \]

Exercise \(\PageIndex{1}\)

The reaction of acetylene (C₂H₂) with hydrogen (H₂) can produce either ethylene (C₂H₄) or ethane (C₂H₆):

\[ \begin{matrix}
C_{2}H_{2}\left ( g \right ) + H_{2}\left ( g \right ) \rightarrow C_{2}H_{4}\left ( g \right ) & \Delta H = -175.7 \; kJ/mol \; C_{2}H_{2} \\
C_{2}H_{2}\left ( g \right ) + 2H_{2}\left ( g \right ) \rightarrow C_{2}H_{6}\left ( g \right ) & \Delta H = -312.0 \; kJ/mol \; C_{2}H_{2}
\end{matrix} \nonumber \]

What is ΔH for the following reaction?

\[\ce{C2H4 (g) + H2 (g) \rightarrow C2H6 (g) }\nonumber\]

Answer

Reverse the first reaction so that C₂H₄ appears as a reactant, matching the target equation. Change the sign of ΔH accordingly:

\[\ce{C2H4 (g) \rightarrow C2H2 (g) + H2 (g)} \quad \Delta H = +175.7 \; \text{kJ/mol} \nonumber\]

\[ \begin{align*} \ce{\cancel{C2H2(g)} +\cancel{2}H2 (g) &-> C2H6(g)  } & \Delta H & = -312.0 \; \text{kJ/mol} \\ \ce{C2H4 (g) &-> \cancel{C2H2 (g)}+\cancel{H2 (g)}} & \Delta H & =+175.7 \; \text{kJ/mol} \\[2ex] \hline \ce{C2H4 (g) + H2 (g) &-> C2H6 (g)} & \Delta H &=-136.3 \; \text{kJ/mol} \end{align*} \nonumber \]

Exercise \(\PageIndex{2}\)

Calculate \(ΔH\) for the process:

\[\ce{N2(g) + 2 O2(g) -> 2 NO2(g)} \nonumber \]

Using the following information:

\[\ce{N2(g) + O2(g) -> 2 NO(g)} \quad \Delta H=180.5\, \text{kJ/mol} \nonumber \]

\[\ce{NO(g) + 1/2 O2(g) -> NO2(g)} \quad \Delta H=-57.06\, \text{kJ/mol} \nonumber \]

Answer

Multiplying the second equation by 2 will give the desired coefficient of NO₂:

\[\ce{2NO(g) + O2(g) -> 2NO2(g)} \quad \Delta H=2(-57.06\, \text{kJ/mol})=-114.12\, \text{kJ/mol} \nonumber \]

Adding this to the first equation yields the target reaction:

\[\ce{N2(g) + O2(g) -> 2 NO(g)} \quad \Delta H=180.5\, \text{kJ/mol} \nonumber \]

\[\Delta H=-114.12\, \text{kJ/mol} + 180.5\, \text{kJ/mol} =+66.4 \, \text{kJ/mol}\nonumber \]

Example \(\PageIndex{2}\): Hess’s Law

Chlorine monofluoride can react with fluorine to form chlorine trifluoride:

\[\ce{ClF(g) + F2(g) -> ClF3(g)} \quad \Delta H=? \nonumber \]

Use the following reactions to determine the \(ΔH\) for the above reaction:

\[\ce{2 OF2(g) -> O2(g) + 2 F2(g)} \quad \Delta H_{(i i)}=-49.4\,\text{kJ/mol} \nonumber \] \[\ce{2 ClF(g) + O2(g) -> Cl2O(g) + OF2(g)} \quad \Delta H_{(i i i)}=+214.0\,\text{kJ/mol} \nonumber \] \[\ce{ClF3(g) + O2(g) -> 1/2 Cl2O(g) + 3/2 OF2(g)} \quad \Delta H_{(i v)}=+236.2\,\text{kJ/mol} \nonumber \]

Solution

Our goal is to manipulate and combine reactions (ii), (iii), and (iv) such that they add up to reaction (i):

\[\ce{ClF(g) + F2(g) -> ClF3(g)} \quad \Delta H=? \nonumber \]

Going from left to right in (i), we first see that ClF(g) is needed as a reactant. This can be obtained by multiplying reaction (iii) by \(\ce{1/2}\) which means that the \(ΔH\) change is also multiplied by \(\ce{1/2}\):

\[\ce{ ClF(g) + 1/2 O2(g) -> 1/2 Cl2O(g) + 1/2 OF2(g)} \quad \Delta H=\frac{1}{2}(214.0)\,\text{kJ/mol}=+107.0\,\text{kJ/mol} \nonumber \]

Next, we see that F₂ is also needed as a reactant. To get this, reverse and halve reaction (ii), which means that the \(ΔH\) changes sign and is halved:

\[\ce{1/2 O2(g) + F2(g) -> OF2(g)} \quad \Delta H=+24.7\,\text{kJ/mol} \nonumber \]

To get \(\ce{ClF3}\) as a product, reverse (iv), changing the sign of \(ΔH\):

\[\ce{1/2 Cl2O(g) + 3/2 OF2(g) -> ClF3(g) + O2(g)} \quad \Delta H=-236.2\,\text{kJ/mol}  \nonumber \]

Now check to make sure that these reactions add up to the reaction we want:

\[\begin{align*}
&\text{step 1} & \ce{ClF(g) + \bcancel{1/2O2(g)}} & \ce{-> \cancel{1/2 Cl2O(g)} + \cancel{1/2 OF2(g)}} & \Delta H&=+107.0\,\text{kJ/mol} \\[2pt]
&\text{step 2} & \ce{\bcancel{1/2O2(g)} + F2(g)} & \ce{-> \cancel{OF2(g)}} &\Delta H&=+24.7 \,\text{kJ/mol} \\[2pt]
&\text{step 3} & \ce{\cancel{1/2 Cl2O(g)} + \cancel{3/2 OF2(g)}} & \ce{-> ClF3(g) + \bcancel{O2(g)}} &\Delta H&=-236.2 \,\text{kJ/mol} \\[2pt]
\hline
&\text{sum} & \ce{ClF(g) + F2(g)} & \ce{-> ClF3(g)} & \Delta H&=-104.5\,\text{kJ}
\end{align*} \nonumber \]

Since summing these three modified reactions yields the reaction of interest, summing the three modified \(ΔH\) values will give the desired \(ΔH\):

\[\Delta H=(+107.0)+(24.7) +(-236.2)=-104.5\,\text{kJ/mol} \nonumber \]

Exercise \(\PageIndex{3}\)

Aluminum chloride can be formed from its elements:

\[\ce{2 Al(s) + 3 Cl2(g) -> 2 AlCl3(s)} \quad \Delta H=? \nonumber \]

Use the reactions here to determine the \(ΔH\) for reaction (i):

\[\ce{HCl(g) -> HCl(aq)} \quad \Delta H_{(i i)}=-74.8 \,\text{kJ/mol} \nonumber \] \[\ce{H2(g) + Cl2(g) -> 2 HCl(g)} \quad \Delta H_{(i i i)}=-185 \,\text{kJ/mol} \nonumber \] \[\ce{AlCl3(aq) -> AlCl3(s)} \quad \Delta H_{(i v)}=+323 \,\text{kJ/mol} \nonumber \] \[\ce{2Al(s) + 6 HCl(aq) -> 2AlCl3(aq) +3 H2(g)} \quad \Delta H_{(v)}=-1049 \,\text{kJ/mol} \nonumber \]

Answer

Multiply equation (iv) by 2 to get the desired product:
\[\ce{2AlCl3(aq) -> 2AlCl3(s)} \quad \Delta H_{(i v)}=2(+323)=+646 \,\text{kJ/mol} \nonumber \]

To cancel 2AlCl₃(aq), use equation (v):
\[\ce{2Al(s) + 6 HCl(aq) -> 2AlCl3(aq) +3 H2(g)} \quad \Delta H_{(v)}=-1049 \,\text{kJ/mol} \nonumber \]

Multiply equation (iii) by three to cancel 3H₂ and have 3Cl₂ as a reactant:
\[\ce{3H2(g) + 3Cl2(g) -> 6 HCl(g)} \quad \Delta H_{(i i i)}=3(-185)=-555 \,\text{kJ/mol} \nonumber \]

Finally to cancel HCl(g) and HCl(aq), multiply equation (ii) by 6:
\[\ce{6HCl(g) -> 6HCl(aq)} \quad \Delta H_{(i i)}=6(-74.8)=-448.8 \,\text{kJ/mol} \nonumber \]

The target equation is obtained by summing the four adjusted reactions, so the total enthalpy change is the sum of their \(ΔH\) values:

\[ \Delta H_{(i)}=(+646 \,\text{kJ/mol})+(-1049 \,\text{kJ/mol})+(-555 \,\text{kJ/mol})+(-448.8 \,\text{kJ/mol}) =-1407\,\text{kJ/mol}\nonumber\]

Summary

Hess’s Law states that the enthalpy change of an overall reaction is equal to the sum of the enthalpy changes of its individual steps (real or hypothetical). This principle follows from the fact that enthalpy is a state function—it depends only on the initial and final states, not the path taken.

Because enthalpy is an extensive property, changing the stoichiometry of a reaction will change the magnitude of \(ΔH\). Reversing a chemical equation changes the sign of \(ΔH\), but not its magnitude. The physical states of reactants and products must also be considered, since phase changes like melting or vaporization involve enthalpy changes of their own.

To apply Hess’s Law, choose reactions with known enthalpy changes, adjust them to match the target reaction (by reversing or scaling as needed), and then add the corresponding \(ΔH\) values to find the overall enthalpy change.