How can I represent an exothermic reaction in a potential energy diagram?

Dec 19, 2014

An exothermic reaction is a reaction in which energy is given off, or, in other words, a reaction that has a $\Delta H < 0$ (see: enthalpy).

Here's a potential energy diagram for an exothermic reaction, the combustion of glucose

Notice that the potential energy of the reactants (${C}_{6} {H}_{12} {O}_{6} + 6 {O}_{2}$) is smaller than the potential energy of the products ($6 C {O}_{2} + 6 {H}_{2} O$); this difference in potential energy (on this diagram labeled as the energy released by the forward raection or required for the reverse reaction ) represents $\Delta H$.

We know that $\Delta H = {H}_{f} - {H}_{i}$, where

${H}_{f}$ represents the enthalpy (potential energy) of the products;
${H}_{i}$ represents the enthalpy (potential energy) of the reactans;

Since ${H}_{f}$ is smaller than ${H}_{i}$. $\Delta H$ will be smaller than zero (negative), a value that characterizes exothermic reactions.

So, let's say you would have to draw the diagram for the combustion of glucose. You'd start with the balanced chemical equation

${C}_{6} {H}_{12} {O}_{6} + 6 {O}_{2} \to 6 C {O}_{2} + 6 {H}_{2} O$

Then you'd calculate ${H}_{i}$ as the sum of the standard state enthalpies ($\Delta {H}_{f}^{\circ}$ - usually given to you) of the reactans - each multiplied by their stoichiometric coefficients, and ${H}_{f}$ as the sum of the standard state enthalpies of the products - again, each multiplied by their stoichiometric coefficients.

After this point, you'd just graph these values, along with $\Delta H = {H}_{f} - {H}_{i} < 0$ (in this case).