Problem 59 - General Chemistry

Question

Ammonia burns in the presence of a copper catalyst to form nitrogen gas. $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) ; \\ \Delta H=-1267 \mathrm{~kJ} \end{aligned} $$ What is the enthalpy change to burn \(35.8 \mathrm{~g}\) of ammonia?

Step-by-Step Solution

Answer

The enthalpy change is approximately -664.18 kJ.

Step 1: Calculate Molar Mass of Ammonia

The molar mass of ammonia (NH₃) is the sum of the atomic masses of nitrogen and hydrogen. \[ \text{Molar mass of NH}_3 = 14.01 + 3 \times 1.01 = 17.04 \text{ g/mol} \]

Step 2: Convert Grams of Ammonia to Moles

To determine the number of moles of ammonia, use the formula: \[ \text{moles of NH}_3 = \frac{\text{mass of NH}_3}{\text{molar mass of NH}_3} \]. Substitute the given values: \[ \frac{35.8 \text{ g}}{17.04 \text{ g/mol}} \approx 2.10 \text{ moles NH}_3 \]

Step 3: Use Reaction Stoichiometry

The balanced chemical equation shows that 4 moles of NH₃ release -1267 kJ of energy. Therefore, we need to calculate the enthalpy change for 2.10 moles of NH₃ using the proportion: \[ \text{Heat released} = \left( \frac{-1267 \text{ kJ}}{4 \text{ moles}} \right) \times 2.10 \text{ moles} \approx -664.18 \text{ kJ} \]

Key Concepts

Ammonia Combustion

Ammonia combustion is a chemical reaction involving ammonia (\(\text{NH}_3\)) reacting with oxygen (\(\text{O}_2\)) to produce nitrogen gas (\(\text{N}_2\)) and water vapor (\(\text{H}_2\text{O}\)). This process typically requires a catalyst like copper to proceed efficiently. In chemical reactions, catalysts are substances that increase the rate of chemical reactions without being consumed. This means they help speed up the reaction without being part of the final products. The role of the copper catalyst in ammonia combustion is to make the reaction occur more easily and quickly, releasing energy in the process.During ammonia combustion, nitrogen is effectively "freed" from ammonia. The reaction proceeds as:- Four molecules of ammonia react with three molecules of oxygen. - This results in the formation of two molecules of nitrogen gas and six molecules of water. The complete reaction can be summarized by the balanced equation:\[4 \text{NH}_3(g) + 3 \text{O}_2(g) \rightarrow 2 \text{N}_2(g) + 6 \text{H}_2\text{O}(g)\]This is an exothermic reaction, meaning it releases heat.

Chemical Stoichiometry

Chemical stoichiometry is the calculation of reactants and products in chemical reactions. It’s based on the laws of conservation of mass and energy, ensuring that atoms are neither created nor destroyed in any chemical reaction.Stoichiometry allows chemists to predict the amounts of substances consumed and produced in a given reaction. It's vital for efficiently conducting chemical processes and for understanding how substances interact at a molecular level.In the context of ammonia combustion, stoichiometry helps us measure the relationship between the quantities of reactants (ammonia and oxygen) and products (nitrogen and water). This balanced equation guides us:\[4 \text{NH}_3(g) + 3 \text{O}_2(g) \rightarrow 2 \text{N}_2(g) + 6 \text{H}_2\text{O}(g)\]Using this, stoichiometry allows us to calculate how much nitrogen gas would be produced from a certain amount of ammonia. Additionally, it allows determining the energy change associated with the reaction by using proper proportions of the enthalpy change.

Enthalpy Calculations

Enthalpy calculations are crucial for understanding the energy changes occurring during chemical reactions. Enthalpy (\(\Delta H\)) represents the heat absorbed or released under constant pressure.In the ammonia combustion reaction, the enthalpy change is given as \(-1267 \text{kJ}\) for reaction of four moles of ammonia. To determine the enthalpy change when a different amount of ammonia is involved, we use stoichiometric relationships.For instance:- Given \(35.8 \text{ g}\) of ammonia.- First, convert this mass to moles using ammonia's molar mass (\(17.04 \text{ g/mol}\)): \[\text{moles of NH}_3 = \frac{35.8 \text{ g}}{17.04 \text{ g/mol}} \approx 2.10 \text{ moles}\]- Then, apply the stoichiometry of the reaction:\[\text{Heat released} = \left( \frac{-1267 \text{kJ}}{4 \text{ moles}} \right) \times 2.10 \text{ moles} \approx -664.18 \text{kJ}\]This result tells us that burning \(35.8 \text{ g}\) of ammonia will release approximately \(-664.18 \text{kJ}\) of energy. This is essential for energy management in chemical processes.