Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 24

Care must be taken in preparing solutions of solutes that liberate heat on dissolving. The heat of solution of \(\mathrm{NaOH}\) is \(-44.5 \mathrm{kJ} / \mathrm{mol} \mathrm{NaOH} .\) To what maximum temperature may a sample of water, originally at \(21^{\circ} \mathrm{C},\) be raised in the preparation of \(500 \mathrm{mL}\) of \(7.0 \mathrm{M}\) NaOH? Assume the solution has a density of \(1.08 \mathrm{g} / \mathrm{mL}\) and specific heat of \(4.00 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

Short Answer

Expert verified
The maximum temperature that can be reached in the preparation of 500 mL of 7.0 M NaOH from a water sample originally at 21 degrees Celsius is \(21 + 72.20 = 93.20\) degrees Celsius, assuming the density and specific heat remain constant. The sample of water can reach a maximum final temperature of approximately 93.20 degrees Celsius.

Step by step solution

01

Calculation of moles of NaOH

Firstly, calculate the moles of NaOH in the solution using the formula Molarity (M) = moles/volume(L). Given, Molarity (M) = 7 M and Volume (V) = 500 mL = 0.5 L. Hence, moles of NaOH = 7 * 0.5 = 3.5 mol
02

Calculation of heat released

Next, calculate the amount of heat released by the NaOH using the formula Heat = moles of solute * heat of solution. Given, heat of solution for NaOH = -44.5 kJ/mol. Hence, Heat released = -44.5 kJ/mol * 3.5 mol = -155.75 kJ = -155750 J. The negative sign indicates that the heat is being released into the solution.
03

Calculation of temperature increase

Now, use this proper value for the heat released in the next calculation. We apply the formula for energy change, \(q = mc\Delta T \) where q is the heat transferred, m is the mass, c is the specific heat, and \(\Delta T \) is the change in temperature. We know that the specific heat of the solution is 4.00 J/g*degree Celsius, the density is 1.08 g/mL and we have 500 mL solution. The mass of the solution is the density times the volume. Hence Mass (m) = 1.08 g/mL * 500 mL = 540 g. We rearrange the formula to solve for \(\Delta T \), so \(\Delta T = q/(mc) \) . Hence \(\Delta T = -155750 J/(540 g * 4.00 J/g*degree Celsius) = -72.20 degree Celsius\)
04

Calculation of final temperature

Finally, calculate the final temperature by adding the temperature change to the initial temperature. Given, the initial temperature is 21 degree Celsius. Hence Final Temperature = Initial Temperature + Temperature Change = 21 degree Celsius - 72.20 degree Celsius = -51.20 degree Celsius which is not physically possible. Since \(\Delta T \) is the absolute temperature difference, the actual temperature rise would be 72.20 degree Celsius.

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

Construct a concept map to show the use of enthalpy for chemical reactions.

Look up the specific heat of several elements, and plot the products of the specific heats and atomic masses as a function of the atomic masses. Based on the plot, develop a hypothesis to explain the data. How could you test your hypothesis?

A 74.8 g sample of copper at \(143.2^{\circ} \mathrm{C}\) is added to an insulated vessel containing \(165 \mathrm{mL}\) of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}(\mathrm{l})(d=1.26 \mathrm{g} / \mathrm{mL}),\) at \(24.8^{\circ} \mathrm{C} .\) The final temperature is \(31.1^{\circ} \mathrm{C}\). The specific heat of copper is \(0.385 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} .\) What is the heat capacity of glycerol in \(\mathrm{Jmol}^{-1}\) \(^{\circ} \mathrm{C}^{-1} ?\)

The internal energy of a fixed quantity of an ideal gas depends only on its temperature. A sample of an ideal gas is allowed to expand at a constant temperature (isothermal expansion). (a) Does the gas do work? (b) Does the gas exchange heat with its surroundings? (c) What happens to the temperature of the gas? (d) What is \(\Delta U\) for the gas?

Construct a concept map to show the interrelationships between path-dependent and pathindependent quantities in thermodynamics.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

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