The evaporation of water is endothermic: $$ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H_{\text {nan }}=+44.01 \mathrm{~kJ} $$ What minimum mass of water (in \(g\) ) has to evaporate to absorb \(175 \mathrm{~kJ}\) of heat?

Short Answer

Expert verified
The minimum mass of water that must evaporate is approximately \( m = \frac{175 \text{ kJ}}{44.01 \text{ kJ/mol}} \times 18.015 \text{ g/mol} \approx 71.54 \text{ g} \).

Step by step solution

01

Understand the Heat of Evaporation

The given enthalpy change, \( \Delta H_{vap} \) for water is +44.01 KJ per mole of water. This indicates that 44.01 KJ of heat is required to evaporate one mole of water.
02

Calculate Moles of Water to Absorb Given Heat

Use the heat energy required (175 kJ) and the enthalpy change to calculate the number of moles of water needed to absorb this heat. The formula is \( n = \frac{Q}{\Delta H_{vap}} \), where \( n \) is the number of moles, \( Q \) is the heat absorbed, and \( \Delta H_{vap} \) is the heat of evaporation.
03

Convert Moles to Mass

Knowing that the molar mass of water is approximately 18.015 g/mol, convert the moles calculated to grams using the formula \( m = n \times M \), where \( m \) is the mass in grams, \( n \) is the number of moles, and \( M \) is the molar mass.
04

Perform the Calculations

Calculate the number of moles of water required to absorb 175 kJ: \( n = \frac{175}{44.01} \) moles. Then, calculate the mass of water needed: \( m = n \times 18.015 \) grams.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat of Evaporation
The phrase 'heat of evaporation', also known as the enthalpy of vaporization, is a critical concept in thermochemistry that refers to the amount of heat energy required to convert a liquid into a gas at its boiling point. It is an intrinsic property of substances that dictates how much heat is needed to overcome intermolecular forces that hold a liquid together.
For water, the heat of evaporation is relatively high, which explains why sweating is an efficient cooling mechanism for the human body. The heat taken up during this transformation is substantial, allowing for effective release of body heat into the environment when sweat evaporates.
Students often engage with this concept during exercises that require the calculation of how much a substance can cool an environment by evaporating. To enhance understanding, it's essential to visualize how molecules in a liquid state require a specific amount of energy to break free from their liquid form and become gas. This is not a one-size-fits-all value but rather varies significantly across different substances.
Thermochemistry
Thermochemistry is the study of the heat change or transfer that accompanies chemical reactions and physical changes, such as phase transitions. An understanding of thermochemistry is foundational to solving problems involving heat of evaporation or other thermal processes.
In the context of the textbook problem, thermochemistry principles allow us to calculate the thermal energy involved in the transition of water from liquid to vapor. These calculations hinge on the understanding that energy can neither be created nor destroyed, but rather it is transformed or transferred. The energy required for the evaporation of water is stored as potential energy within the gaseous molecules.
Real-life importance:
  • Designing heating and cooling systems.
  • Developing processes in chemical industry.
  • Understanding environmental energy exchanges.
Incorporating interactive simulations or visual aids can significantly enhance students’ comprehension of these abstract concepts, making the study of thermochemistry more engaging and accessible.
Molar Mass
Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It plays a vital role in converting between the mass of a substance and the number of moles, a key step in stoichiometric calculations and when dealing with reactions and phase changes such as evaporation.
The molar mass of water (H2O) is approximately 18.015 g/mol. To calculate any mass of water based on the number of moles, knowledge of the molar mass is essential. Estimating the mass of a substance required for a chemical process or phase change often starts with moles-to-mass or mass-to-moles conversions.
For deeper understanding, students should conduct experiments that involve measuring a substance before and after a reaction. Through these hands-on experiences, they can directly observe and calculate the molar mass, reinforcing the concept's practical utility in solving scientific problems.
Endothermic Processes
Endothermic processes are chemical reactions or physical changes that absorb heat from their surroundings. This is in contrast to exothermic processes, where energy is released. In endothermic reactions, the system gains heat as the surrounding temperature decreases.
Evaporation is a classic example of an endothermic process. When water evaporates, it requires energy in the form of heat from its surroundings, which is used to break the hydrogen bonds between water molecules, allowing them to escape into the air as vapor.
Visual demonstrations of endothermic processes, such as the dissolution of certain salts in water that results in a temperature drop, can be particularly impactful for students. Emphasizing the practical implications, such as the use of endothermic reactions in cold packs, can help learners connect the concept to real-world applications, fostering a more profound and lasting understanding.

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Most popular questions from this chapter

Consider the reaction between sulfur trioxide and water: $$ \mathrm{SO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) $$ A chemist allows \(61.5 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(11.2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) to react. When the reaction is finished, the chemist collects \(54.9 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\). Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants. $$ 2 \mathrm{~K}(s)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{KCl}(s) $$ (a) \(1 \mathrm{~mol} \mathrm{~K} ; 1 \mathrm{~mol} \mathrm{Cl}_{2}\) (b) \(1.8 \mathrm{~mol} \mathrm{~K} ; 1 \mathrm{~mol} \mathrm{Cl}_{2}\) (c) \(2.2 \mathrm{~mol} \mathrm{~K} ; 1 \mathrm{~mol} \mathrm{Cl}_{2}\) (d) \(14.6 \mathrm{~mol} \mathrm{~K} ; 7.8 \mathrm{~mol} \mathrm{Cl}_{2}\)

The combustion of liquid ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) produces carbon dioxide and water. After \(3.8 \mathrm{~mL}\) of ethanol (density \(=0.789 \mathrm{~g} / \mathrm{mL}\) ) is allowed to burn in the presence of \(12.5 \mathrm{~g}\) of oxygen gas, \(3.10 \mathrm{~mL}\) of water (density \(=1.00 \mathrm{~g} / \mathrm{mL}\) ) is collected. Determine the limiting reactant, theoretical yield of \(\mathrm{H}_{2} \mathrm{O}\), and percent yield for the reaction. (Hint: Write a balanced equation for the combustion of ethanol.)

Consider the reaction between \(\mathrm{N}_{2} \mathrm{H}_{4}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) : $$ 2 \mathrm{~N}_{2} \mathrm{H}_{4}(g)+\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 3 \mathrm{~N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ A reaction vessel initially contains \(27.5 \mathrm{~g} \mathrm{~N}_{2} \mathrm{H}_{4}\) and \(74.9 \mathrm{~g}\) of \(\mathrm{N}_{2} \mathrm{O}_{4}\). Calculate the masses of \(\mathrm{N}_{2} \mathrm{H}_{4}, \mathrm{~N}_{2} \mathrm{O}_{4}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) that will be in the reaction vessel after the reactants have reacted as much as possible. Assume \(100 \%\) yield. Hint: The limiting reactant is completely consumed, but the reactant in excess is not. Use the amount of limiting reactant to determine the amount of products that form and the amount of the reactant in excess that remains after complete reaction.

Determine the theoretical yield of \(\mathrm{C}\) when each of the initial quantities of \(\mathrm{A}\) and \(\mathrm{B}\) is allowed to react in the generic reaction: $$ 2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow 2 \mathrm{C} $$ (a) \(2 \mathrm{~mol} \mathrm{~A} ; 4 \mathrm{~mol} \mathrm{~B}\) (b) \(3 \mathrm{~mol} \mathrm{~A} ; 3 \mathrm{~mol} \mathrm{~B}\) (c) \(5 \mathrm{~mol} \mathrm{~A} ; 6 \mathrm{~mol} \mathrm{~B}\) (d) \(4 \mathrm{~mol} \mathrm{~A} ; 5 \mathrm{~mol} \mathrm{~B}\)

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