Problem 123 - Chemistry: An Atoms-Focused Approach

Question

Use the information in thermochemical equations (1) through ( 3 ) to calculate the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the reaction in equation (4). (1) \(\mathrm{Pb}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{PbO}(s) \quad \quad \Delta H_{\mathrm{rxn}}^{\circ}=-219 \mathrm{kJ}\) (2) \(\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) \Delta H_{\text {rxn }}^{\circ}=-394 \mathrm{kJ}\) (3) \(\mathrm{PbCO}_{3}(s) \rightarrow \mathrm{PbO}(s)+\mathrm{CO}_{2}(g) \quad \Delta H_{\text {rxn }}^{\circ}=86 \mathrm{kJ}\) (4) \(2 \mathrm{Pb}(s)+2 \mathrm{C}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{PbCO}_{3}(s) \quad \Delta H_{\mathrm{rxn}}^{\circ}=?\)

Step-by-Step Solution

Answer

The value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the reaction is -1312 kJ.

Step 1: Analyze the target equation

We have the following target equation (4): $$2 \mathrm{Pb}(s)+2 \mathrm{C}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{PbCO}_{3}(s)$$ Our task is to manipulate the given thermochemical equations (1), (2), and (3) to match the target equation and find the \(\Delta H_{\mathrm{rxn}}^{\circ}\) for it.

Step 2: Adjust the stoichiometry of the given equations

First, let's adjust the stoichiometry of the given equations by multiplying so that they match the target equation. For equation (1), we need \(2 \mathrm{Pb}(s)\) on the left side, so we will multiply the whole equation by 2: $$2 \mathrm{Pb}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{PbO}(s)$$ $$2 \Delta H_{\mathrm{rxn}}^{\circ}=-438 \mathrm{kJ}$$ For equation (2), we need \(2 \mathrm{C}(s)\) on the left side, so we will also multiply the whole equation by 2: $$2 \mathrm{C}(s)+2 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)$$ $$2 \Delta H_{\mathrm{rxn}}^{\circ}=-788 \mathrm{kJ}$$ Equation (3) already has \(2 \mathrm{PbCO}_{3}(s)\) on the right side, so we don't need to change it.

Step 3: Combine the adjusted equations

Now let's combine the adjusted equations (1), (2), and (3) to obtain equation (4). $$2 \mathrm{Pb}(s)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{PbO}(s)$$ $$2 \mathrm{C}(s)+2 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g)$$ $$\mathrm{PbCO}_{3}(s) \rightarrow \mathrm{PbO}(s)+\mathrm{CO}_{2}(g)$$ $$---------------------------------------------------$$ $$2 \mathrm{Pb}(s)+2 \mathrm{C}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{PbCO}_{3}(s)$$ Now, we have built equation (4).

Step 4: Find the \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the target equation

Based on the Hess's Law, the enthalpy change of a reaction is the sum of enthalpy changes of its sub-reactions. Therefore, the \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the target equation would be: $$\Delta H_{\mathrm{rxn}}^{\circ} = 2 \Delta H_{\text {1}}^{\circ} + 2 \Delta H_{\text {2}}^{\circ} - \Delta H_{\text {3}}^{\circ}$$ $$\Delta H_{\mathrm{rxn}}^{\circ} = -438 + (-788) - 86$$ $$\Delta H_{\mathrm{rxn}}^{\circ} = -1312 \mathrm{kJ}$$ Thus, the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the reaction in equation (4) is -1312 kJ.

Key Concepts

Enthalpy

Enthalpy, represented by the symbol \(H\), refers to the total heat content of a system at constant pressure. It is a measure used in thermodynamics to understand how energy is transferred in reactions. When discussing reactions, particularly chemical reactions, the change in enthalpy \(\Delta H\) is of paramount interest.\\Key aspects of enthalpy include:

• Positive \(\Delta H\) indicates an endothermic reaction where heat is absorbed.
• Negative \(\Delta H\) signifies an exothermic reaction where heat is released.

\In a reaction such as \(2 \mathrm{Pb}(s)+2 \mathrm{C}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{PbCO}_{3}(s)\), the enthalpy change (\(\Delta H_{\mathrm{rxn}}^{\circ}\)) is calculated based on how heat flows during the making and breaking of chemical bonds. When analyzing thermochemical equations, such as those given in the exercise, the enthalpy changes are crucial to understanding whether energy is being absorbed or released.

Hess's Law

Hess's Law is a fundamental principle in thermochemistry that states: The total enthalpy change for a reaction is the same, regardless of whether it occurs in one step or multiple steps. This law is incredibly useful for calculating reaction enthalpies that are difficult to measure directly.\\To apply Hess's Law:

• Identify the target equation whose \(\Delta H_{\mathrm{rxn}}^{\circ}\) you need to find.
• Adjust the given thermochemical equations so they add up to the target equation.
• Sum the \(\Delta H\) values of component equations to find the \(\Delta H\) for the target reaction.

\In our example, we rearrange and scale equations (1), (2), and (3) to match the target equation (4), then sum their enthalpy changes. By doing this, we ensure we are in accordance with Hess's Law and properly calculate the \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the reaction.

Stoichiometry

Stoichiometry deals with the relative quantities of reactants and products in chemical reactions. It allows chemists to predict the amount of products formed in a reaction based on the quantities of reactants used, and vice versa.\\Key points to consider in stoichiometry:

• The coefficients in a balanced chemical equation indicate the ratios of molecules or moles involved in the reaction.
• These ratios can be used to calculate amounts of reactants consumed and products formed.

\For equation (4), where \(2 \mathrm{Pb}(s)+2 \mathrm{C}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{PbCO}_{3}(s)\), stoichiometry was used to adjust equations (1), (2), and (3) to ensure the number of atoms for each element were balanced. By multiplying these equations to match the stoichiometry of the target equation, we could apply Hess's Law effectively to find the enthalpy change.