Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 21

The molar enthalpy of fusion of water is 6.009 \(\mathrm{kJ} / \mathrm{mol}\) at \(0^{\circ} \mathrm{C}\) . Explain what this statement means.

Short Answer

Expert verified
The molar enthalpy of fusion of water refers to the amount of heat energy required to convert one mole of ice at \(0^{\circ} \mathrm{C}\) into water at the same temperature. The given value of 6.009 \( \mathrm{kJ} / \mathrm{mol}\) means that exactly 6.009 kilojoules of heat energy are needed for this transformation.

Step by step solution

01

Definition of Molar Enthalpy

Molar enthalpy, denoted as \(\Delta H_m\), is basically the heat absorbed or released by one mole of a substance during a process carried out at constant pressure. It is measured in \( \mathrm{kJ} / \mathrm{mol}\) (kilojoules per mole).
02

Understanding Fusion

Fusion, in this case, is the process in which a substance changes from a solid state to a liquid state. It is often called 'melting'. It is an endothermic process, which means it requires heat energy to occur.
03

Applying the Concepts to Water

The molar enthalpy of fusion of water is 6.009 \( \mathrm{kJ} / \mathrm{mol}\) at \(0^{\circ} \mathrm{C}\). This statement means that exactly 6.009 kilojoules of heat energy are required to convert one mole (about 18 grams) of ice at \(0^{\circ} \mathrm{C}\) into water at the same temperature.

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.

Start your free trial

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

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 critical point for carbon tetrachloride is at \(283^{\circ} \mathrm{C}\) and \(4.5 \times 10^{3}\) kPa. The triple point is at \(-87.0^{\circ} \mathrm{C}\) and 28.9 \(\mathrm{kPa}\) . The normal boiling point is \(76.7^{\circ} \mathrm{C}\) . Sketch the phase diagram.

How does a pressure cooker work?

Why does ice float in water even though most solids sink in the pure liquid?

Carbon tetrachloride, \(\mathrm{CCl}_{4},\) is nonpolar. What forces hold the molecules together?

Calculating Vapor Pressure by Using a Table The graphing calculator can run a program that calculates a table for the vapor pressure in atmospheres at different temperatures (K) given the number of moles of a gas and the volume of the gas (V). Given a 0.50 mol gas sample with a volume of \(10 \mathrm{L},\) you can calculate the pressure at 290 \(\mathrm{K}\) by using a table. Use this program to make the table. Next, use the table to perform the calculations. Go to Appendix C. If you are using a TI- -83 Plus, you can download the program VAPOR and data and run the application as directed. If you are using another calculator, your teacher will provide you with key- strokes and data sets to use. After you have run the program, answer the questions. a. What is the pressure for 1.3 mol of a gas with a volume of 8.0 \(\mathrm{L}\) and a temperature of 320 \(\mathrm{K} ?\) b. What is the pressure for 1.5 mol of a gas with a volume of 10.0 \(\mathrm{L}\) and a temperature of 340 \(\mathrm{K} ?\) Two gases are measured at 300 \(\mathrm{K}\) . One has an amount of 1.3 \(\mathrm{mol}\) and a volume of \(7.5 \mathrm{L},\) and the other has an amount of 0.5 \(\mathrm{mol}\) and a volume of 10.0 \(\mathrm{L} .\) Which gas has the lesser pressure?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation
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