Problem 73 - General Chemistry

Question

Ammonia will burn in the presence of a platinum catalyst to produce nitric oxide, \(\mathrm{NO}\). $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ What is the heat of reaction at constant pressure? Use the following thermochemical equations: $$ \begin{gathered} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) ; \Delta H=180.6 \mathrm{~kJ} \\ \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) ; \Delta H=-91.8 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) ; \Delta H=-483.7 \mathrm{~kJ} \end{gathered} $$

Step-by-Step Solution

Answer

The heat of reaction is \(-906.3\) kJ.

Step 1: Understanding the Given Reaction

We are given the reaction \(4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2 \text{O}(g)\) and need to find the heat of reaction, \(\Delta H\), using Hess's Law and the given thermochemical equations.

Step 2: Analyzing Thermochemical Equations

The given thermochemical equations are: \( \text{N}_2(g) + \text{O}_2(g) \rightarrow 2 \text{NO}(g); \Delta H = 180.6 \text{ kJ} \), \( \text{N}_2(g) + 3 \text{H}_2(g) \rightarrow 2 \text{NH}_3(g); \Delta H = -91.8 \text{ kJ} \), and \( 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(g); \Delta H = -483.7 \text{ kJ} \). We will manipulate these to match the desired reaction.

Step 3: Manipulate Equation for \(\text{NH}_3\) to \(\text{N}_2\) and \(\text{H}_2\)

We need 4 moles of \(\text{NH}_3\). The equation \( \text{N}_2(g) + 3 \text{H}_2(g) \rightarrow 2 \text{NH}_3(g) \) must be reversed: \( 2 \text{NH}_3(g) \rightarrow \text{N}_2(g) + 3 \text{H}_2(g) \). Reversing the equation changes \( \Delta H = 91.8 \text{ kJ} \). We multiply by 2: \( 4 \text{NH}_3(g) \rightarrow 2 \text{N}_2(g) + 6 \text{H}_2(g); \Delta H = 183.6 \text{ kJ} \).

Step 4: Produce \(\text{NO}\) from \(\text{N}_2\) and \(\text{O}_2\)

The equation \( \text{N}_2(g) + \text{O}_2(g) \rightarrow 2 \text{NO}(g) \) can be used directly. It's needed twice: \( 2 \times (\text{N}_2(g) + \text{O}_2(g) \rightarrow 2 \text{NO}(g)) \) gives \( \Delta H = 2 \times 180.6 \text{ kJ} = 361.2 \text{ kJ} \).

Step 5: Form \(\text{Water}\) from \(\text{H}_2\) and \(\text{O}_2\)

We need 6 moles of \(\text{H}_2\text{O}\), using \( 2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(g) \). Multiply by 3: \( 6 \text{H}_2(g) + 3 \text{O}_2(g) \rightarrow 6 \text{H}_2\text{O}(g); \Delta H = 3 \times (-483.7) \text{ kJ} = -1451.1 \text{ kJ} \).

Step 6: Add All Adjusted Equations

Combine all changes: \(4 \text{NH}_3(g)\rightarrow 2 \text{N}_2(g) + 6 \text{H}_2(g), 2 \text{N}_2(g) + 2 \text{O}_2(g) \rightarrow 4 \text{NO}(g), 6 \text{H}_2(g) + 3 \text{O}_2(g) \rightarrow 6 \text{H}_2\text{O}(g)\). This results in the target reaction: \(4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(g)\).

Step 7: Calculate Overall \(\Delta H\)

Sum the adjusted \(\Delta H\) values: \(183.6 \text{ kJ} + 361.2 \text{ kJ} - 1451.1 \text{ kJ} = -906.3 \text{ kJ} \). This is the overall heat of reaction at constant pressure.

Key Concepts

Thermochemical Equations

Thermochemical equations are an invaluable tool in understanding chemical reactions. They not only show the reactants and products but also the enthalpy change, \(\Delta H\), associated with the reaction. This change in enthalpy tells us if a reaction is endothermic (absorbing heat) or exothermic (releasing heat).

When using thermochemical equations, it's crucial to pay attention to the stoichiometry of the reaction—essentially, the balance of atoms in the reactants and products. The coefficients in a thermochemical equation not only indicate the number of molecules or moles involved but implicitly relate to the heat exchanged. For example:

• \(1\,\text{N}_2(g) + 3\,\text{H}_2(g) \rightarrow 2\,\text{NH}_3(g)\); \(\Delta H = -91.8\,\text{kJ}\)
• \(2\,\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\,\text{H}_2\text{O}(g)\); \(\Delta H = -483.7\,\text{kJ}\)

In these equations, the enthalpy change corresponds to the formation or decomposition of specific molecules. It’s important to reverse the sign of \(\Delta H\) if you reverse the reaction, as demonstrated when deriving equations in Hess's Law applications.

Enthalpy Change

Enthalpy change, \(\Delta H\), is a key part of understanding energy exchanges in chemical reactions. It measures the total heat content in a system and reflects the energy needed to break bonds in reactants compared to the energy released when bonds form in the products.

Enthalpy changes are classified broadly into two categories:

• Endothermic reactions: These reactions absorb heat, leading to a positive \(\Delta H\). An example is photosynthesis, where plants absorb sunlight to convert carbon dioxide and water into glucose.
• Exothermic reactions: These release heat, resulting in a negative \(\Delta H\). Combustion and many of the spontaneous chemical reactions fall into this category.

Hess's Law provides a method to calculate \(\Delta H\) for reactions by simplifying complex processes into a series of stepwise reactions for which enthalpy changes are known. These enthalpy changes are then summed to find the overall reaction's heat change. This flexibility is due to the fact that \(\Delta H\) is a state function and does not depend on the path taken but only on the initial and final states of the system.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into new substances, called products. These reactions are fundamental to everything from industrial manufacturing to the basic metabolic processes in our bodies.

In the process of a chemical reaction, bonds in the reactants are broken, and new bonds form in the products. This bond transformation requires energy changes, either absorbing or releasing heat, depending on the nature of the reaction. For example:

• Combustion, such as burning wood, is a highly exothermic reaction where organic matter reacts with oxygen releasing heat and light.
• The neutralization of an acid and a base to form water and a salt, particularly exothermic and often used in calorimetry to calculate energy change.

Every chemical reaction can be described using a chemical equation, balancing the quantity of each type of atom and illustrating the transformation and energy transfer involved. The balanced equation allows chemists to understand the proportions of reactants and products and lays the groundwork for calculating necessary parameters like enthalpy change using thermochemical equations.