6.4 Hess's Law

Hess's Law is a fundamental principle in thermochemistry based on the fact that enthalpy is a state function. This means the total enthalpy change for a chemical reaction is independent of the pathway taken from the initial state (reactants) to the final state (products). The change in enthalpy is the same whether the reaction occurs in a single step or in multiple sequential steps.

The law, stated by G.H. Hess in 1840, can be summarized as:

Mathematically, for any cyclic process, the sum of all enthalpy changes is zero: $\Sigma \Delta H(\text{Cycle})=0$

Consider a reaction where reactant A transforms into product B.

• Direct Path: $A\to B$ with an enthalpy change of $\Delta H$.
• Indirect Path: $A\to C\to D\to B$ with enthalpy changes $\Delta H_1$, $\Delta H_2$, and $\Delta H_3$ for each step.

According to Hess's Law, the enthalpy change for the direct path is equal to the sum of the enthalpy changes for the indirect path: $\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

Principle and Application

Hess's Law is particularly useful for determining the enthalpy changes of reactions that are difficult or impossible to measure directly in a calorimeter. This includes:

• Reactions that are very slow.
• Reactions that do not go to completion.
• Reactions that produce side products.
• Formation of compounds that cannot be synthesized directly from their elements (e.g., $\mathrm{CC}{\mathrm{l}}_4$).
• Combustion of elements that form a protective oxide layer (e.g., $\mathrm{Al}$, $\mathrm{B}$).

By using Hess's Law, we can calculate the desired enthalpy change by combining the known enthalpy changes of other related reactions.

Calculating Enthalpies of Reaction from Enthalpies of Formation ($\Delta H_f^{\circ }$)

A common application of Hess's Law is to calculate the standard enthalpy change of a reaction ($\Delta H_{\text{reaction}}^{\circ }$) using the standard enthalpies of formation ($\Delta H_f^{\circ }$) of the reactants and products.

The general formula is: $\Delta H_{\text{reaction}}^{\circ }=\sum n_p\Delta H_f^{\circ }(\text{products})-\sum n_r\Delta H_f^{\circ }(\text{reactants})$

Where:

• $n_p$ and $n_r$ are the stoichiometric coefficients of the products and reactants, respectively.
• Important Note: The standard enthalpy of formation ($\Delta H_f^{\circ }$) of any element in its most stable standard state is defined as zero.

Calculation of Enthalpy of Combustion ($\Delta H_c^{\circ }$)

The enthalpy of combustion, $\Delta H_c^{\circ }$, is the heat evolved when one mole of a substance burns completely in oxygen under standard conditions. It can be calculated using the same formula as above, with the products typically being $\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$ and ${\mathrm{H}}_2\mathrm{O}(\mathrm{l})$.

$\Delta H_c^{\circ }=\sum n_p\Delta H_f^{\circ }(\text{products})-\sum n_r\Delta H_f^{\circ }(\text{reactants})$

Worked Examples

Example 6.6: Combustion of Carbon

The complete combustion of carbon to carbon dioxide can occur in one step or two steps.

Single Step: ${\mathrm{C}}_{(\mathrm{s})}+{\mathrm{O}}_{2(\mathrm{g})}\to \mathrm{C}{\mathrm{O}}_{2(\mathrm{g})}\Delta H^{\circ }=-393.5\text{kJ}$

Two Steps:

1. Formation of carbon monoxide: ${\mathrm{C}}_{(\mathrm{s})}+\frac{1}{2}{\mathrm{O}}_{2(\mathrm{g})}\to \mathrm{C}{\mathrm{O}}_{(\mathrm{g})}\Delta H_1^{\circ }=-110.52\text{kJ}$
2. Combustion of carbon monoxide: $\mathrm{C}{\mathrm{O}}_{(\mathrm{g})}+\frac{1}{2}{\mathrm{O}}_{2(\mathrm{g})}\to \mathrm{C}{\mathrm{O}}_{2(\mathrm{g})}\Delta H_2^{\circ }=-282.98\text{kJ}$

Applying Hess's Law:

The total enthalpy change for the two-step process is the sum of the individual steps: $\Delta H^{\circ }=\Delta H_1^{\circ }+\Delta H_2^{\circ }=(-110.52)+(-282.98)=-393.50\text{kJ}$

This matches the enthalpy change for the single-step reaction, demonstrating the path independence.

Figure 6.4: Energy cycle for the reaction between $\mathrm{C}(\mathrm{s})$ and ${\mathrm{O}}_{2(\mathrm{g})}$ to produce $\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$.

Example 6.7: Enthalpy of Formation of Methane

Calculate the standard enthalpy of formation for methane ($\mathrm{C}{\mathrm{H}}_4$) using the following combustion data.

Target Reaction: ${\mathrm{C}}_{(\mathrm{s})}+2{\mathrm{H}}_{2(\mathrm{g})}\to \mathrm{C}{\mathrm{H}}_{4(\mathrm{g})}\Delta H_f^{\circ }=?$

Given Reactions:

(i) $\mathrm{C}{\mathrm{H}}_{4(\mathrm{g})}+2{\mathrm{O}}_{2(\mathrm{g})}\to \mathrm{C}{\mathrm{O}}_{2(\mathrm{g})}+2{\mathrm{H}}_2{\mathrm{O}}_{(\mathrm{l})}\Delta H^{\circ }=-890.4{\text{kJ mol}}^{-1}$

(ii) ${\mathrm{H}}_{2(\mathrm{g})}+\frac{1}{2}{\mathrm{O}}_{2(\mathrm{g})}\to {\mathrm{H}}_2{\mathrm{O}}_{(\mathrm{l})}\Delta H^{\circ }=-285.8{\text{kJ mol}}^{-1}$

(iii) ${\mathrm{C}}_{(\mathrm{s})}+{\mathrm{O}}_{2(\mathrm{g})}\to \mathrm{C}{\mathrm{O}}_{2(\mathrm{g})}\Delta H^{\circ }=-393.5{\text{kJ mol}}^{-1}$

Solution:

1. Manipulate the given equations to match the target reaction:

   • Equation (iii): Use as is — provides ${\mathrm{C}}_{(\mathrm{s})}$ as a reactant. $\Delta H^{\circ }=-393.5\text{kJ}$
   • Equation (ii): Multiply by 2 — provides $2{\mathrm{H}}_{2(\mathrm{g})}$ as reactants. $\Delta H^{\circ }=2\times (-285.8)=-571.6\text{kJ}$
   • Equation (i): Reverse the reaction — places $\mathrm{C}{\mathrm{H}}_{4(\mathrm{g})}$ as a product. Sign of $\Delta H^{\circ }$ is changed: $\Delta H^{\circ }=+890.4\text{kJ}$

2. Combine the manipulated equations:

\begin{align*} \mathrm{C_{(s)} + O_{2(g)}} &\rightarrow \mathrm{CO_{2(g)}} & \Delta H^{\circ} &= -393.5 \,\text{kJ} \\ \mathrm{2H_{2(g)} + O_{2(g)}} &\rightarrow \mathrm{2H_2O_{(l)}} & \Delta H^{\circ} &= -571.6 \,\text{kJ} \\ \mathrm{CO_{2(g)} + 2H_2O_{(l)}} &\rightarrow \mathrm{CH_{4(g)} + 2O_{2(g)}} & \Delta H^{\circ} &= +890.4 \,\text{kJ} \end{align*}

3. Cancel species appearing on both sides ($\mathrm{C}{\mathrm{O}}_2$, $2{\mathrm{H}}_2\mathrm{O}$, $2{\mathrm{O}}_2$ cancel):

${\mathrm{C}}_{(\mathrm{s})}+2{\mathrm{H}}_{2(\mathrm{g})}\to \mathrm{C}{\mathrm{H}}_{4(\mathrm{g})}$

4. Sum the enthalpy changes:

$\Delta H_f^{\circ }(\mathrm{C}{\mathrm{H}}_4)=(-393.5)+(-571.6)+(+890.4)=-74.7{\text{kJ mol}}^{-1}$

The standard enthalpy of formation of methane is $-74.7{\text{kJ mol}}^{-1}$.