AP Chemistry 6.8-6.9: Enthalpy of Formation and Hess's Law

Standard state

A reference set of conditions used to tabulate thermodynamic properties. The IUPAC standard state uses a pressure of $1\ \mathrm{bar}$ and aqueous solutions at $1\ \mathrm{M}$; temperature is not fixed, although $298.15\ \mathrm{K}$ is commonly assumed.

Meaning of the notation $\Delta H^\circ$

It denotes an enthalpy change measured or defined for substances in their standard states. Unless another temperature is specified, many tables assume $298.15\ \mathrm{K}$.

Thermochemical equation

A balanced chemical equation that includes the physical states of all substances and the enthalpy change for the reaction as written. The stated $\Delta H$ applies to the exact stoichiometric amounts shown.

Why must physical states be included in a thermochemical equation?

The enthalpy change depends on whether each substance is solid, liquid, aqueous, or gaseous. For example, forming liquid water from hydrogen and oxygen releases a different amount of heat than forming water vapor.

How does multiplying a thermochemical equation affect its enthalpy change?

Because enthalpy is an extensive property, multiplying every stoichiometric coefficient by a factor also multiplies $\Delta H$ by that factor. Dividing the coefficients divides $\Delta H$ correspondingly.

What happens to $\Delta H$ when a thermochemical equation is reversed?

Reversing the reaction reverses the direction of heat flow, so the sign of $\Delta H$ changes. For example, $\Delta H=-286\ \mathrm{kJ}$ becomes $+286\ \mathrm{kJ}$.

Standard enthalpy of formation, $\Delta H_f^\circ$

The enthalpy change for forming exactly $1\ \mathrm{mol}$ of a compound from its constituent elements in their most stable forms under standard-state conditions.

What is the standard enthalpy of formation of an element in its standard form?

It is defined as zero. Examples include $\Delta H_f^\circ[\mathrm{O_2(g)}]=0$, $\Delta H_f^\circ[\mathrm{H_2(g)}]=0$, and $\Delta H_f^\circ[\mathrm{C(graphite)}]=0$.

How do you write a standard enthalpy-of-formation equation for a compound?

Write the elements in their standard forms as reactants and form exactly one mole of the compound as the product. Fractional coefficients are allowed; for example, $2\mathrm{C(s,graphite)}+3\mathrm{H_2(g)}+\frac{1}{2}\mathrm{O_2(g)}\rightarrow\mathrm{C_2H_5OH(l)}$.

How can the standard enthalpy of formation of ozone be obtained from $3\mathrm{O_2(g)}\rightarrow2\mathrm{O_3(g)}$ with $\Delta H^\circ=+286\ \mathrm{kJ}$?

Divide the reaction and its enthalpy change by 2 to form one mole of ozone: $\frac{3}{2}\mathrm{O_2(g)}\rightarrow\mathrm{O_3(g)}$, giving $\Delta H_f^\circ[\mathrm{O_3(g)}]=+143\ \mathrm{kJ\ mol^{-1}}$.

What standard enthalpy relationship can be used to calculate a reaction enthalpy from formation data?

Use $\Delta H_{\mathrm{rxn}}^\circ=\sum n\Delta H_f^\circ(\text{products})-\sum n\Delta H_f^\circ(\text{reactants})$, where each formation enthalpy is multiplied by its stoichiometric coefficient.

What is the correct procedure for using $\Delta H_f^\circ$ values to calculate $\Delta H_{\mathrm{rxn}}^\circ$?

First write and balance the reaction, including physical states. Multiply each substance's $\Delta H_f^\circ$ by its stoichiometric coefficient, sum the product terms, sum the reactant terms, and subtract: $\Delta H_{\mathrm{rxn}}^\circ=\sum n\Delta H_f^\circ(\text{products})-\sum n\Delta H_f^\circ(\text{reactants})$.

Hess’s law

If an overall reaction can be represented as the sum of several reactions, its enthalpy change equals the algebraic sum of the enthalpy changes of those reactions. Hess’s law follows from enthalpy being a state function.

How are equations manipulated when applying Hess’s law?

Reverse any equation whose direction is opposite to the desired reaction and change the sign of its $\Delta H$. Multiply an equation by a factor when necessary and multiply its $\Delta H$ by the same factor; then add equations and cancel intermediate species.

Use Hess’s law to determine the enthalpy for $\mathrm{C(s)}+\mathrm{O_2(g)}\rightarrow\mathrm{CO_2(g)}$ given $\mathrm{C(s)}+\frac{1}{2}\mathrm{O_2(g)}\rightarrow\mathrm{CO(g)}$, $\Delta H=-111\ \mathrm{kJ}$, and $\mathrm{CO(g)}+\frac{1}{2}\mathrm{O_2(g)}\rightarrow\mathrm{CO_2(g)}$, $\Delta H=-283\ \mathrm{kJ}$.

Add the two equations; CO cancels, yielding the target reaction. Therefore, $\Delta H=(-111\ \mathrm{kJ})+(-283\ \mathrm{kJ})=-394\ \mathrm{kJ}$.

Why can Hess’s law determine enthalpy changes for reactions that are difficult to measure directly?

Because the enthalpy change depends only on the initial and final states, a difficult reaction can be replaced mathematically by a combination of reactions with known enthalpy changes. The sum of those known changes gives the desired value.