Question

Given: $$ \begin{array}{ll}{2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CuO}(s)} & {\Delta H^{\circ}=-288 \mathrm{kJ}} \\\ {\mathrm{Cu}_{2} \mathrm{O}(s) \longrightarrow \mathrm{CuO}(s)+\mathrm{Cu}(s)} & {\Delta H^{\circ}=11 \mathrm{kJ}}\end{array} $$ Calculate the standard enthalpy of formation \(\left(\Delta H_{f}^{\circ}\right)\) for \(\mathrm{CuO}(s) .\)

Short answer

The standard enthalpy of formation of solid copper oxide (CuO) is \(-299 \text{ kJ/mol}\).

Step-by-step solution

Rearrange and manipulate the given equations

First, let's reverse the second given equation and adjust the coefficients to get CuO on the product side: 2. $$\mathrm{CuO}_{(s)} + \mathrm{Cu}_{(s)} \longrightarrow \mathrm{Cu}_{2}\mathrm{O}_{(s)} \qquad\qquad\Delta H^{\circ}= -11 \text{ kJ}$$ We have reversed the second equation and multiplied by 2 because if we add equations 1 and 2, \(\mathrm{Cu}_{2}\mathrm{O}_{(s)}\) in the reactants will be canceled, and we would get the formation reaction of \(\mathrm{CuO}(s)\) from the reactants \(\mathrm{O_{2}}(g)\) and \(\mathrm{Cu}(s)\), which we want.

Apply Hess's Law to calculate the enthalpy change

Now, let's add the first and second equations and their enthalpy changes to create a new combined equation: 1. $$2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CuO}(s) \qquad\qquad\Delta H^{\circ}=-288 \text{ kJ}$$ + 2. $$\mathrm{CuO}_{(s)} + \mathrm{Cu}_{(s)} \longrightarrow \mathrm{Cu}_{2}\mathrm{O}_{(s)} \qquad\qquad\Delta H^{\circ}= -11 \text{ kJ}$$ \(4 \mathrm{CuO}(s) + \mathrm{Cu}(s) \longrightarrow 2 \mathrm{Cu}_{2}\mathrm{O}(s)+ 2 \mathrm{Cu}_{2}\mathrm{O}(s)+\mathrm{O}_{2}(g)\) Simplifying the above equation gives: $$\mathrm{Cu}(s) + \dfrac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{CuO}(s)$$ Adding the enthalpy changes of the two equations according to Hess's Law: $$\Delta H^{\circ}_{\text{combined}}= -288 \mathrm{kJ} - 11 \mathrm{kJ} = -299 \mathrm{kJ}$$

Calculate the standard enthalpy of formation of CuO

We have obtained the new combined equation with the enthalpy change \(\Delta H^{\circ}_{\text{combined}}\). Since the enthalpy change of this combined equation represents the formation of one mole of solid copper oxide (CuO) from its elements, it is also the standard enthalpy of formation of CuO: $$\Delta H_{f}^{\circ}(\text{CuO}) = \Delta H^{\circ}_{\text{combined}} = -299 \text{ kJ/mol}$$ So, the standard enthalpy of formation of solid copper oxide (CuO) is -299 kJ/mol.

Key concepts

Hess's Law

Understanding Hess's Law can significantly simplify thermodynamic calculations. It states that the total enthalpy change during the course of a chemical reaction is the same, regardless of the multiple steps taken to achieve the reaction. Hess's Law relies on the fact that enthalpy is a state function, which means its value is independent of the path taken.
This principle allows chemists to use known enthalpy changes of simpler reactions to calculate unknown enthalpies, even if they do not occur directly in one step. To use Hess’s Law effectively, reactions are arranged and manipulated so that when you add them together, the unwanted intermediates cancel out, leaving a desired net reaction.
Bulleting key points for clarity:

• The enthalpy change of a reaction is the same regardless of the number of steps or the path taken.
• Hess's Law helps in calculating enthalpy changes indirectly.
• This approach combines known reactions to find enthalpy changes of unknown or complex reactions.

CuO

Copper(II) oxide, or CuO, is a black solid used in various applications such as ceramics, pigment production, and even as a catalyst. In chemical terms, it is composed of copper in the +2 oxidation state and oxygen. This compound is the subject of many studies due to its role in redox reactions and its electrical properties.
When discussing its enthalpy of formation, we're referring to the energy change when one mole of copper(II) oxide is formed from its elements (Cu and O) in their standard states. This is a critical measure as it helps understand the energy involved in breaking and forming bonds during the formation of CuO.
Listed below are important aspects of CuO:

• CuO is used in multiple industrial processes due to its stability and oxidative properties.
• The formation of CuO involves the reaction of elemental copper and oxygen.
• The enthalpy of formation provides insights into the energy profile of this compound.

Enthalpy Change

Enthalpy change is a crucial concept in thermodynamics, representing the heat absorbed or released during a chemical reaction at constant pressure. Positive enthalpy change indicates endothermic reactions where the system absorbs heat, while negative values describe exothermic reactions releasing heat to the surroundings.
In the context of calculating the enthalpy of formation, this term specifically refers to the change when a compound is formed directly from its elements. By combining equations to eliminate intermediates, as we did for CuO, Hess’s law allows us to deduce the enthalpy change effectively.
Let's break down the essential points regarding enthalpy change:

• It measures the heat exchange at constant pressure during a reaction.
• Negative enthalpy change implies an exothermic process, while positive is endothermic.
• Enthalpy changes help predict reaction feasibility and energy requirements.