Problem 76 - Chemistry

Question

Determine the amount of heat (in kJ) given off when \(1.26 \times 10^{4} \mathrm{~g}\) of ammonia is produced according to the equation \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-92.6 \mathrm{~kJ} / \mathrm{mol}\) Assume that the reaction takes place under standardstate conditions at \(25^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The amount of heat given off is approximately -68474 kJ.

Step 1: Find Molar Mass of Ammonia (NH₃)

The first step is to find the molar mass of ammonia (NH₃). Ammonia consists of 1 nitrogen (N) and 3 hydrogen (H) atoms. The molar mass is calculated as follows: the atomic mass of N is approximately 14.01 g/mol and that of H is approximately 1.01 g/mol. Therefore, the molar mass of NH₃ is \( 14.01 + 3 \times 1.01 = 17.04 \) g/mol.

Step 2: Calculate Moles of Ammonia

Next, we convert the given mass of ammonia to moles by using its molar mass. The formula to find moles is given by \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Substituting the values gives \( \frac{1.26 \times 10^{4} \, \text{g}}{17.04 \, \text{g/mol}} \approx 739.2 \, \text{mol} \) of NH₃.

Step 3: Calculate Heat Released

Now that we have the moles of ammonia, we can calculate the heat released using the given enthalpy change \( \Delta H_{\text{rxn}}^{\circ} = -92.6 \, \text{kJ/mol} \). The formula is \( \text{heat} = \text{moles} \times \Delta H_{\text{rxn}}^{\circ} \). Substituting in the values, \( 739.2 \, \text{mol} \times -92.6 \, \text{kJ/mol} = -68473.92 \, \text{kJ} \). The negative sign indicates heat is released.

Key Concepts

Molar Mass Calculation

To perform a molar mass calculation, one must sum up the atomic masses of each element in a molecule, based on the periodic table. Let's explore ammonia (NH₃), which has one nitrogen atom and three hydrogen atoms. The atomic mass of nitrogen (N) is approximately 14.01 g/mol, while hydrogen (H) is about 1.01 g/mol. To find the molar mass of ammonia, one would calculate: - 1 nitrogen: 14.01 g/mol - 3 hydrogens: 3 × 1.01 g/mol = 3.03 g/mol
Adding these together, the molar mass of NH₃ = 14.01 + 3.03 = 17.04 g/mol. This value is crucial in converting grams of a substance to moles, as seen in the given exercise. Understanding molar mass aids in tackling many chemistry problems, especially those involving reactions and enthalpy calculations.

Moles Conversion

Moles conversion is an essential step in chemistry, especially when dealing with reactions. Given a specific mass of a substance, converting it to moles involves dividing by its molar mass. This step answers "how many moles of the substance do I have?"

• Use the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \)
• For the exercise, the mass of ammonia is \( 1.26 \times 10^{4} \text{g} \)
• The calculated molar mass of NH₃ is 17.04 g/mol

Applying the formula helps us convert mass to moles:
\[ \frac{1.26 \times 10^{4} \text{g}}{17.04 \text{g/mol}} = 739.2 \text{mol NH₃} \]
This conversion allows us to relate the physical amount of a substance to chemical quantities, key in calculating energy changes in chemical reactions.

Heat Calculation

In chemistry, heat calculation often involves using enthalpy changes, especially in reaction energetics. The concept of heat calculation relates the energy change during the formation or breakdown of substances.
Given that the problem provides an enthalpy change of \( \Delta H_{\text{rxn}}^{\circ} = -92.6 \, \text{kJ/mol} \), where the negative sign denotes the release of heat. The formula to find the total heat released or absorbed in a reaction is:
\[ \text{heat} = \text{moles} \times \Delta H_{\text{rxn}}^{\circ} \]
Using the 739.2 moles of ammonia calculated earlier:
\[ 739.2 \, \text{mol} \times -92.6 \, \text{kJ/mol} = -68473.92 \, \text{kJ} \]
This calculation indicates that 68,473.92 kJ of heat is released in the reaction. Understanding and performing such calculations help to predict whether a reaction will release or absorb energy, which is critical in industries like chemical manufacturing and environmental science.