Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The entropy change when \(36 \mathrm{~g}\) of water evaporates at \(373 \mathrm{~K}\) is \(\left(\Delta \mathrm{H}=40.63 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)\) (a) \(218 \mathrm{~J} \mathrm{~K}^{-1}\) (b) \(150 \mathrm{~J} \mathrm{~K}^{-1}\) (c) \(118 \mathrm{~J} \mathrm{~K}^{-1}\) (d) \(200 \mathrm{~J} \mathrm{~K}^{-1}\)

Short Answer

Expert verified
The entropy change is approximately 218 J K^{-1}.

Step by step solution

01

Identify the Formula for Entropy Change

The formula for the change in entropy during a phase change is given by \( \Delta S = \frac{\Delta H}{T} \), where \( \Delta S \) is the entropy change, \( \Delta H \) is the enthalpy change per mole, and \( T \) is the temperature in Kelvin at which the change occurs.
02

Determine the Number of Moles of Water

First, convert the mass of water from grams to moles using the molar mass of water (\(18 \mathrm{~g/mol}\)).\[\text{Moles of water} = \frac{36 \mathrm{~g}}{18 \mathrm{~g/mol}} = 2 \text{ moles}\]
03

Calculate Total Enthalpy Change for Given Mass

The total enthalpy change is calculated by multiplying the number of moles by the enthalpy change per mole:\[\Delta H_{\text{total}} = 2 \text{ moles} \times 40.63 \mathrm{~kJ/mol} = 81.26 \mathrm{~kJ}\]. Convert \(\Delta H_{\text{total}}\) to \( \mathrm{J} \) (\(1 \mathrm{~kJ} = 1000 \mathrm{~J}\)):\[81.26 \mathrm{~kJ} = 81260 \mathrm{~J}\]
04

Calculate Entropy Change

Use the formula from Step 1 to find \( \Delta S \):\[ \Delta S = \frac{\Delta H_{\text{total}}}{T} = \frac{81260 \mathrm{~J}}{373 \mathrm{~K}} \approx 217.85 \mathrm{~J} \mathrm{~K}^{-1} \]
05

Select the Most Appropriate Answer

The calculated entropy change is approximately \(217.85 \mathrm{~J} \mathrm{~K}^{-1}\), which rounds to \(218 \mathrm{~J} \mathrm{~K}^{-1}\), so the correct answer is (a) \(218 \mathrm{~J} \mathrm{~K}^{-1}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Phase Change
Phase change, also known as a phase transition, describes the process where a substance changes from one state of matter to another, such as from liquid to gas. In the problem, we are dealing with the phase change of water evaporating at 373 K. Here, water changes from liquid to vapor. This process requires an absorption of energy since the molecules move from a densely packed liquid state into a more separated gaseous state. This energy absorbed during the phase change significantly affects the entropy, or the measure of disorder, of the system. Evaporation increases the number of possible states for molecules, thereby increasing the system's entropy.
  • Evaporation: This is the specific phase change from liquid to gas as seen in the exercise.
  • Energy Requirement: The energy needed to overcome intermolecular forces in the liquid.
  • Entropy Increase: Represents the disorder increase resulting from more spread-out gaseous molecules.
Enthalpy Change
Enthalpy change, denoted as \( \Delta H \, \), is a measure of the heat absorbed or released during a chemical reaction under constant pressure. In this exercise, the enthalpy change describes the amount of energy needed to convert water from liquid to vapor at a specific temperature. This value is given as 40.63 kJ/mol. For water undergoing evaporation, the enthalpy change is positive, indicating that the process absorbs heat from the surroundings.To determine the total enthalpy change for the entire mass of water, you need to calculate it based on the number of moles of water under consideration. Multiplying the number of moles by the given enthalpy change per mole gives the total enthalpy change, crucial for finding the entropy change during the phase transition.
  • Energy Input: Required to break the hydrogen bonds present in liquid water.
  • Units: Typically expressed in joules or kilojoules per mole.
  • Relation to Entropy: More energy input generally leads to an increase in entropy in such phase changes.
Moles of Water
To solve the problem effectively, determining the number of moles of water is a vital step. In chemical processes, we often convert mass to moles to use thermodynamic equations. The molar mass of water, \( 18 \text{ g/mol} \, \), serves as the conversion factor between grams and moles.For instance, when given 36 g of water, dividing by the molar mass results in 2 moles of water, as shown in the exercise. Using moles instead of grams provides a basis for calculating other properties, like enthalpy or entropy changes, that depend on the amount of substance on a molecular scale.
  • Conversion: Mass to moles conversion helps standardize calculations.
  • Molar Mass of Water: Essential for calculating moles (18 g/mol).
  • Role in Thermodynamics: Foundational for accurately applying entropic and enthalpy calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The change in entropy, in the conversion of one mole of water at \(373 \mathrm{~K}\) to vapour at the same temperature is (Latent heat of vaporization of water = \(\left.2.257 \mathrm{~kJ} \mathrm{~g}^{-1}\right)\) (a) \(99 \mathrm{JK}^{-1}\) (b) \(129 \mathrm{JK}^{-1}\) (c) \(89 \mathrm{JK}^{-1}\) (d) \(109 \mathrm{JK}^{-1}\)

Which of the following equations represent standard heat of formation of \(\mathrm{C}_{2} \mathrm{H}_{4} ?\) (a) \(2 \mathrm{C}\) (diamond) \(+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (b) \(2 \mathrm{C}\) (graphite) \(+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (c) \(2 \mathrm{C}\) (diamond) \(+4 \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\) (d) \(2 \mathrm{C}\) (graphite) \(+4 \mathrm{H}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})\)

The enthalpy change involved in the oxidation of glucose is \(-2880 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Twenty five per cent of this energy is available for muscular work. If \(100 \mathrm{~kJ}\) of muscular work is needed to walk one kilometre, what is the maximum distance that a person will be able to walk after consuming \(120 \mathrm{~g}\) of glucose? (a) \(7.9 \mathrm{~km}\) (b) \(9.7 \mathrm{~km}\) (c) \(4.8 \mathrm{~km}\) (d) \(8.4 \mathrm{~km}\)

Determine \(\Delta \mathrm{H}\) and \(\Delta \mathrm{E}\) for reversible isothermal evaporation of \(90 \mathrm{~g}\) of water at \(100^{\circ} \mathrm{C}\). Assume that water vapour behaves as an ideal gas and heat of evaporation of water is \(540 \mathrm{cal} \mathrm{g}^{-1}\left(\mathrm{R}=2.0 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) \(48600 \mathrm{cal}, 44870 \mathrm{cal}\) (b) \(43670 \mathrm{cal}, 47700 \mathrm{cal}\) (c) 47700 cal, \(43670 \mathrm{cal}\) (d) \(44870 \mathrm{cal}, 48670 \mathrm{cal}\)

The enthalpy change involved in oxidation of glucose is \(-2880 \mathrm{~kJ} \mathrm{~mol}^{-1}, 25 \%\) of this energy is available for muscular work. If \(100 \mathrm{~kJ}\) of muscular work is needed to walk one \(\mathrm{km}\), the maximum distance \((\mathrm{km})\) that a person will be able to walk after taking \(150 \mathrm{~g}\) of glucose is

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free