Answer :
To determine how much heat is released when 50.0 g of ammonia is condensed at its boiling point of [tex]\(-33^{\circ} C\)[/tex], we need to follow these steps:
1. Understand the enthalpy of vaporization: The enthalpy of vaporization is the amount of energy required to convert a mole of a liquid into vapor without a temperature change. In this case, the standard enthalpy of vaporization for ammonia ([tex]\(NH_3\)[/tex]) is provided as [tex]\(23.3 \text{ kJ/mol}\)[/tex]. When condensation occurs, the same amount of energy is released, hence it will be negative because energy is being released.
2. Determine the molar mass of ammonia: The chemical formula for ammonia is [tex]\(NH_3\)[/tex]. The molar mass can be calculated by adding the atomic masses of nitrogen (approximately 14.01 g/mol) and three hydrogen atoms (approximately 1.01 g/mol each). Therefore, the molar mass of ammonia is approximately:
[tex]\[
14.01 + 3 \times 1.01 = 17.03 \text{ g/mol}
\][/tex]
3. Calculate the moles of ammonia in 50.0 g: To find the number of moles, use the formula:
[tex]\[
\text{moles of } NH_3 = \frac{\text{mass in grams}}{\text{molar mass in g/mol}}
\][/tex]
[tex]\[
\text{moles of } NH_3 = \frac{50.0 \text{ g}}{17.03 \text{ g/mol}}
\][/tex]
From this calculation, you get approximately 2.936 moles of ammonia.
4. Calculate the heat released during condensation: The quantity of heat released when ammonia condenses is the product of the moles of ammonia and the enthalpy of vaporization (note that the enthalpy will be negative since heat is being released):
[tex]\[
\text{heat released} = -(\text{moles of } NH_3) \times (\Delta H_{\text{vap}})
\][/tex]
[tex]\[
\text{heat released} = -(2.936) \times (23.3 \text{ kJ/mol}) = -68.4 \text{ kJ}
\][/tex]
Therefore, when 50.0 g of ammonia is condensed at [tex]\(-33^{\circ} C\)[/tex], approximately [tex]\(68.4 \text{ kJ}\)[/tex] of heat is released. The correct answer is option A: -68.4 kJ.
1. Understand the enthalpy of vaporization: The enthalpy of vaporization is the amount of energy required to convert a mole of a liquid into vapor without a temperature change. In this case, the standard enthalpy of vaporization for ammonia ([tex]\(NH_3\)[/tex]) is provided as [tex]\(23.3 \text{ kJ/mol}\)[/tex]. When condensation occurs, the same amount of energy is released, hence it will be negative because energy is being released.
2. Determine the molar mass of ammonia: The chemical formula for ammonia is [tex]\(NH_3\)[/tex]. The molar mass can be calculated by adding the atomic masses of nitrogen (approximately 14.01 g/mol) and three hydrogen atoms (approximately 1.01 g/mol each). Therefore, the molar mass of ammonia is approximately:
[tex]\[
14.01 + 3 \times 1.01 = 17.03 \text{ g/mol}
\][/tex]
3. Calculate the moles of ammonia in 50.0 g: To find the number of moles, use the formula:
[tex]\[
\text{moles of } NH_3 = \frac{\text{mass in grams}}{\text{molar mass in g/mol}}
\][/tex]
[tex]\[
\text{moles of } NH_3 = \frac{50.0 \text{ g}}{17.03 \text{ g/mol}}
\][/tex]
From this calculation, you get approximately 2.936 moles of ammonia.
4. Calculate the heat released during condensation: The quantity of heat released when ammonia condenses is the product of the moles of ammonia and the enthalpy of vaporization (note that the enthalpy will be negative since heat is being released):
[tex]\[
\text{heat released} = -(\text{moles of } NH_3) \times (\Delta H_{\text{vap}})
\][/tex]
[tex]\[
\text{heat released} = -(2.936) \times (23.3 \text{ kJ/mol}) = -68.4 \text{ kJ}
\][/tex]
Therefore, when 50.0 g of ammonia is condensed at [tex]\(-33^{\circ} C\)[/tex], approximately [tex]\(68.4 \text{ kJ}\)[/tex] of heat is released. The correct answer is option A: -68.4 kJ.