Problem 56 - General Chemistry

Question

With a platinum catalyst, ammonia will burn in oxygen give nitric oxide, NO. $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ \Delta H &=-906 \mathrm{~kJ} \end{aligned} $$ What is the enthalpy change for the following reaction? $$ \mathrm{NO}(g)+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{NH}_{3}(g)+\frac{5}{4} \mathrm{O}_{2}(g) $$

Step-by-Step Solution

Answer

The enthalpy change for the reaction is \(+226.5 \text{ kJ/mol}\).

Step 1: Understanding the problem

We need to find the enthalpy change for the given reverse reaction of ammonia burning to form nitric oxide. The reaction provided is a reverse of a known standard reaction with given enthalpy change.

Step 2: Write the reverse reaction

The original reaction is \(4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2O(g)\) with \(\Delta H = -906 \text{ kJ}\). Reverse this to obtain \(4 \text{NO}(g) + 6 \text{H}_2O(g) \rightarrow 4 \text{NH}_3(g) + 5 \text{O}_2(g)\). The enthalpy change for the reverse reaction is \(+906 \text{ kJ}\).

Step 3: Calculate enthalpy for 1 mol NO

The reversed reaction produces 4 moles of \(\text{NO}\), so to find the enthalpy change for \(1\) mole: divide the total \(\Delta H = +906 \text{ kJ}\) by 4. This gives \(\Delta H = +226.5 \text{ kJ/mol}\) for one mole of \(\text{NO}\) reacting as specified.

Step 4: Adjust the equation stoichiometry

The provided reaction \(\text{NO}(g) + \frac{3}{2} \text{H}_2O(g) \rightarrow \text{NH}_3(g) + \frac{5}{4} \text{O}_2(g)\) aligns perfectly with obtaining per mole basis calculated in Step 3. No further stoichiometry adjustments are needed.

Key Concepts

Chemical Reactions

Chemical reactions are processes that result in the transformation of one set of chemical substances to another. In a chemical reaction, reactants are converted to products. This occurs via the breaking and forming of bonds, which involves changes in energy.
When we talk about chemical reactions, one key concept is enthalpy change, denoted as \( \Delta H \). The enthalpy change represents the heat released or absorbed during a reaction at constant pressure. A negative \( \Delta H \) indicates an exothermic reaction, where heat is released. Conversely, a positive \( \Delta H \) signifies an endothermic reaction, where heat is absorbed.

• Reactants: Starting materials in a chemical reaction
• Products: Substances formed from the reaction
• Enthalpy (\( \Delta H \)): Heat change at constant pressure

This fundamental understanding of chemical reactions and enthalpy change is crucial for interpreting the energetics of reactions, such as the combustion of ammonia.

Ammonia Combustion

Ammonia combustion is a chemical process where ammonia \( (\text{NH}_3) \) reacts with oxygen \( (\text{O}_2) \) to produce nitric oxide \( (\text{NO}) \) and water \( (\text{H}_2O) \). This reaction is a type of exothermic reaction, where significant heat is released, as denoted by the enthalpy change \( \Delta H = -906 \text{ kJ} \).
The balanced chemical equation for the ammonia combustion reaction is:
\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2O(g) \]
This reaction is significant because it highlights the transition of reactants to products through a complex chemical transformation. The process releases a lot of energy, which can be harnessed for various industrial applications, including the production of nitrates.

• Balanced Equation: Ensures the conservation of mass
• Exothermic Reaction: Releases energy in the form of heat
• Industrial Application: Used in the production of fertilizers and explosives

Ammonia combustion is an excellent example of a reaction that is both important for basic chemical education and industry applications.

Catalysis

Catalysis involves the use of a catalyst to speed up a chemical reaction without undergoing permanent change itself. In the combustion of ammonia, platinum is used as a catalyst. Catalysts provide an alternative reaction pathway, usually with lower activation energy, thus allowing reactions to proceed more quickly and efficiently.
In the context of ammonia combustion:

• Platinum Catalyst: Accelerates the reaction without being consumed
• Significance: Reduces the energy barrier for the reaction, saving time and energy
• Reusable: Can be recovered and used for subsequent reaction cycles

Catalysts are vital in many industrial processes as they help in increasing the efficiency of chemical transformations without being depleted in the process.
In essence, catalysts like platinum facilitate the ammonia combustion reaction, making it more feasible for industrial applications.

Reversing Reactions

Reversing reactions involves taking a chemical reaction and altering the direction, so the products become reactants and vice versa. In chemistry, it is essential to understand that every forward reaction has a corresponding reverse reaction. The enthalpy change for the reverse reaction is equal in magnitude but opposite in sign.
For example, if the forward reaction has an enthalpy change of \( \Delta H = -906 \text{ kJ} \), the reverse reaction will have \( \Delta H = +906 \text{ kJ} \). This is because reversing the reaction means undoing the energy change that occurred in the forward process.

• Reverse Reaction: Products become reactants and vice versa
• Enthalpy Change: Equal in magnitude but opposite in sign
• Importance: Understanding equilibrium and reaction feasibility

In chemistry, the concept of reversing reactions helps explain phenomena such as chemical equilibrium, where both forward and reverse reactions occur simultaneously.