# Using Hess' Law, how do you calculate the standard heat of formation of Copper(I) Oxide given the following data?

## $C u O \left(s\right) \to C u \left(s\right) + \frac{1}{2} {O}_{2}$ $\Delta H = 157.3 k J$$/$$m o l$ $4 C u O \left(s\right) \to 2 C {u}_{2} O \left(s\right) + {O}_{2} \left(g\right)$ $\Delta H 292.0 k J$$/$$m o l$

Apr 17, 2016

You can do it like this:

#### Explanation:

Hess' Law states that the overall enthalpy change of a process is independent of the route taken.

In thermodynamics we are interested in initial and final states.

You need to construct a Hess Cycle using the information given:

You can see that, in energy terms, the $\textcolor{b l u e}{\text{blue}}$ route is equal to the $\textcolor{red}{\text{red}}$ route as the arrows start and finish in the same place. This is in accordance with Hess' Law.

So we can write:

$\left(4 \times 157.3\right) + \Delta H = 292$

$\therefore \Delta H = - 337.2 \text{kJ}$

Enthalpy of formation refers to the formation of 1 mole of a substance from its elements in their standard states under standard conditions.

We have found $\Delta H$ for:

$4 C u + 2 {O}_{2} \rightarrow 2 C {u}_{2} O + {O}_{2}$

Which is the same as:

$4 C u + {O}_{2} \rightarrow 2 C {u}_{2} O$

This refers to the formation of 2 moles of copper(I) oxide. We need the enthalpy change for the formation of 1 mole of copper(I) oxide.

$\therefore \Delta {H}_{f} \left[C {u}_{2} O\right] = \frac{\Delta H}{2} = - \frac{337.2}{2} = - 168.6 \text{kJ/mol}$