Question

Consider the reaction: \(4 \mathrm{CO}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+\) \(\mathrm{N}_{2}(g)\). Using the following information, determine \(\Delta H^{\circ}\) for the reaction at \(25^{\circ} \mathrm{C}\). $$\begin{array}{ll}\mathrm{NO}(g) & \Delta H_{\mathrm{f}}^{\circ}=+91.3 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{CO}_{2}(g) & \Delta H_{\mathrm{f}}^{\mathrm{o}}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \Delta H^{\circ}=-116.2 \mathrm{~kJ} \\ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) & \Delta H^{\circ}=-566.0 \mathrm{~kJ} \end{array}$$

Short answer

\( \Delta H^{\circ} = -1640.4 \, \text{kJ} \).

Step-by-step solution

Identify Known Data

We have the formation enthalpies: \( \Delta H_{\mathrm{f}}^{\circ}(\mathrm{CO}_{2}) = -393.5 \, \text{kJ/mol} \). Additionally, from the given reactions: \( 2 \mathrm{NO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) \) with \( \Delta H^{\circ} = -116.2 \, \text{kJ} \) and \( 2 \mathrm{CO}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{g}) \) with \( \Delta H^{\circ} = -566.0 \, \text{kJ} \). Formation enthalpy for NO is \( \Delta H_{\mathrm{f}}^{\circ}(\mathrm{NO}) = +91.3 \, \text{kJ/mol} \).

Write Formation Reactions

The formation reaction for \( \mathrm{NO}_{2}(g) \) is: \( \mathrm{N}_{2}(g) + 2\mathrm{O}_{2}(g) \rightarrow 2\mathrm{NO}_{2}(g) \). This implies \( \Delta H_{\mathrm{f}}^{\circ}(\mathrm{NO}_{2}(g)) = \frac{-116.2}{2} + 2 \times 91.3 \, \text{kJ/mol} \).

Calculate Enthalpy for NO2 Formation

Calculate the enthalpy for \( \mathrm{NO}_{2}(g): \Delta H^{\circ}(\mathrm{NO}_{2}(g)) = \frac{-116.2}{2} + 2 \times 91.3 \approx 33.2 \, \text{kJ/mol} \).

Express Reaction in Terms of Formation

Express the target reaction: \( 4 \mathrm{CO}(g) + 2 \mathrm{NO}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g) + \mathrm{N}_{2}(g) \) in terms of formation using \( \Delta H_{\mathrm{f}}^{\circ}(\mathrm{CO}_{2}(g)) \) and \( \Delta H_{\mathrm{f}}^{\circ}(\mathrm{NO}_{2}(g)) \).

Apply Hess's Law

Using Hess's Law: \( \Delta H^{\circ} = \left[4 \Delta H_{\mathrm{f}}^{\circ}(\mathrm{CO}_{2}) + 0 \right] - \left[4 \times 0 + 2 \times 33.2 \right] \,.\) Hence, \( \Delta H^{\circ} = 4(-393.5) - 2(33.2) \).

Calculate Total Enthalpy Change

Calculate \( \Delta H^{\circ} = 4(-393.5) - 2(33.2) = -1574 - 66.4 = -1640.4 \, \text{kJ} \).

Final Result

Therefore, the enthalpy change \( \Delta H^{\circ} \) for the reaction is \(-1640.4 \, \text{kJ} \).

Key concepts

Enthalpy Change

Enthalpy change, often denoted as \( \Delta H \), is a key concept in thermodynamics that refers to the heat absorbed or released during a chemical reaction at constant pressure. It's an important parameter that helps in understanding whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).
For instance, in our exercise, the goal is to find the overall enthalpy change for the given chemical reaction:

• \(4 \mathrm{CO}(g)+2 \mathrm{NO}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+ \mathrm{N}_{2}(g)\)

Using Hess's Law, we calculate the enthalpy change by summing the enthalpy changes for each step involved in the reaction pathway. This ultimately helps us to realize the net enthalpy change, which was determined to be \(-1640.4 \text{ kJ}\). This value tells us the reaction is highly exothermic, releasing a lot of energy as heat.

Formation Enthalpy

Formation enthalpy, expressed as \( \Delta H_{\text{f}}^{\circ} \), is the enthalpy change that occurs when one mole of a compound is formed from its elements in their standard states. This concept is crucial because it allows us to compute the enthalpy change of complex reactions using Hess's Law. In the exercise, we use formation enthalpy values to break down the reaction into simpler steps.
Our step involves working with known formation enthalpy values:

• \( \Delta H_{\text{f}}^{\circ}(\mathrm{CO}_2) = -393.5 \text{ kJ/mol} \)
• \( \Delta H_{\text{f}}^{\circ}(\mathrm{NO}) = +91.3 \text{ kJ/mol} \)
• Calculate formation enthalpy for \( \mathrm{NO}_2 \) using known reactions and formation enthalpies.

The formation enthalpy for \( \mathrm{NO}_2 \) is computed as approximately \( 33.2 \text{ kJ/mol} \). By substituting these values into Hess's Law formula, we find the overall enthalpy change for the given reaction.

Chemical Reaction

A chemical reaction is a process where reactants are transformed into products through the breaking and forming of chemical bonds. This process is represented by a chemical equation that specifies the reactants, products, and their respective quantities. In the exercise, the given chemical reaction is:

• \(4 \mathrm{CO}(g) + 2 \mathrm{NO}_{2}(g) \rightarrow 4 \mathrm{CO}_{2}(g) + \mathrm{N}_{2}(g)\)

In this reaction, carbon monoxide (\( \mathrm{CO} \)) and nitrogen dioxide (\( \mathrm{NO}_{2} \)) react to form carbon dioxide (\( \mathrm{CO}_{2} \)) and nitrogen gas (\( \mathrm{N}_{2} \)). A chemical reaction is accompanied by energy changes, typically in the form of heat, thus involving the concept of enthalpy change. Understanding the enthalpy involved in chemical reactions helps predict whether a reaction will be spontaneous or require additional energy.