Question

The \(\Delta H\) for \(\mathrm{N}_{2}+\mathrm{O}_{2} \rightarrow 2 \mathrm{NO}\) is \(181 \mathrm{~kJ}\). What is the \(\Delta H\) for this reaction? \(\mathrm{NO} \rightarrow 1 / 2 \mathrm{~N}_{2}+1 / 2 \mathrm{O}_{2}\)

Short answer

\( \Delta H = -90.5 \mathrm{~kJ} \) for the reaction.

Step-by-step solution

Identify the Given Information

We are given the reaction \( \mathrm{N}_{2} + \mathrm{O}_{2} \rightarrow 2 \mathrm{NO} \) with \( \Delta H = 181 \mathrm{~kJ} \). This means forming 2 moles of \( \mathrm{NO} \) requires 181 kJ of energy.

Analyze the Target Reaction

The target reaction given is \( \mathrm{NO} \rightarrow \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2} \). This is essentially the reverse of the formation of \( \mathrm{NO} \) from \( \mathrm{N}_{2} \) and \( \mathrm{O}_{2} \).

Understand the Relationship Between Reactions

When a reaction is reversed, the sign of \( \Delta H \) is also reversed. The original reaction has \( \Delta H = 181 \mathrm{~kJ} \) for the forward direction. For the reverse, \( \Delta H = -181 \mathrm{~kJ} \).

Adjust for Stoichiometry

The original \( \Delta H \) value is for the formation of 2 moles of \( \mathrm{NO} \). The given reaction \( \mathrm{NO} \rightarrow \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2} \) deals with 1 mole of NO. Therefore, we use half of the reverse \( \Delta H \), which is \(-\frac{181}{2} \mathrm{~kJ}\).

Calculate the Final \( \Delta H \)

Calculate \( \Delta H = -\frac{181}{2} = -90.5 \mathrm{~kJ} \). This is the energy change for the decomposition of one mole of \( \mathrm{NO} \).

Key concepts

Reversible Reactions

In the realm of chemistry, reversible reactions play a fundamental role. A reversible reaction is one that can proceed in both directions — from reactants to products, and from products back to reactants. In chemical notation, this is often represented by a double-headed arrow. In the context of enthalpy change, when a chemical reaction proceeds in the reverse direction, the sign of the enthalpy change (\(\Delta H\)) is also reversed, which signifies the opposite energy effect.
For example, given the reaction:

• \(\mathrm{N}_{2} + \mathrm{O}_{2} \rightarrow 2 \mathrm{NO}\)
• The forward reaction has an enthalpy change of \(181 \mathrm{~kJ}\).
• In the reverse reaction: \(2 \mathrm{NO} \rightarrow \mathrm{N}_{2} + \mathrm{O}_{2}\), the enthalpy change is \(-181 \mathrm{~kJ}\).

Understanding reversible reactions enables us to predict how energy changes when a reaction proceeds in different directions, providing insight into conservation of energy and equilibrium states.

Thermochemistry

Thermochemistry is the branch of chemistry that focuses on the study of energy and heat related to chemical reactions. It deals with the measurement of heat changes (enthalpy changes) that occur during chemical processes. A central term in thermochemistry is enthalpy (\(H\)), which is used to quantify the heat change at constant pressure.
In a chemical reaction, the enthalpy change (\(\Delta H\)) represents the heat absorbed or released:

• \( \Delta H > 0 \): Endothermic reaction (heat absorbed).
• \( \Delta H < 0 \): Exothermic reaction (heat released).

In the exercise, the enthalpy change for forming \(2 \mathrm{NO}\) is \(181 \mathrm{~kJ}\), indicating an endothermic process. When this reaction is reversed, the sign of \(\Delta H\) changes as it becomes an exothermic process.

Stoichiometry

Stoichiometry is the area of chemistry that deals with the quantitative relationships of reactants and products in a chemical reaction. It involves understanding and calculating the proportions of elements in compounds, as determined by chemical equations.
Consider the stoichiometric relationships in the reaction:

• Reactants: \(\mathrm{N}_{2} + \mathrm{O}_{2}\)
• Products: \(2 \mathrm{NO}\)

In the exercise, the given \(\Delta H = 181 \mathrm{~kJ}\) applies to the formation of \(2 \mathrm{NO}\). However, when dealing with the reverse reaction \(\mathrm{NO} \rightarrow \frac{1}{2} \mathrm{N}_{2} + \frac{1}{2} \mathrm{O}_{2}\), only one mole of \(\mathrm{NO}\) is involved. By dividing the enthalpy change for the complete reversal by 2, the adjusted \(\Delta H\) becomes \(-90.5 \mathrm{~kJ}\) for the decomposition of one mole of \(\mathrm{NO}\). This showcases the applicability of stoichiometry in calculating accurate enthalpy values for reactions when the quantities of reactants/products are altered.