Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 110

Under the entry \(\mathrm{H}_{2} \mathrm{SO}_{4},\) a reference source lists many values for the standard enthalpy of formation. For example, for pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1), \Delta H_{\mathrm{f}}^{\circ}=-814.0 \mathrm{kJ} / \mathrm{mol}\) for a solution with \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(-841.8 ;\) with \(10 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-880.5 ;\) with \(50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-886.8 ;\) with \(100 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-887.7 ;\) with \(500 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-890.5 ;\) with \(1000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-892.3 ;\) with \(10,000 \mathrm{mol}\) \(\mathrm{H}_{2} \mathrm{O},-900.8 ;\) and with \(100,000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-907.3\) (a) Explain why these values are not all the same. (b) The value of \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right]\) in an infinitely dilute solution is \(-909.3 \mathrm{kJ} / \mathrm{mol} .\) What data from this chapter can you cite to confirm this value? Explain. (c) If \(500.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is prepared from pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1),\) what is the approximate change in temperature that should be observed? Assume that the \(\mathrm{H}_{2} \mathrm{SO}_{4}(1)\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are at the same temperature initially and that the specific heat of the \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is about \(4.2 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

Short Answer

Expert verified
As the number of water moles increases in the formation of \(H_2SO_4\), the enthalpy change (∆H) decreases. Hence, ∆Hf of \(H_2SO_4\) in different quantities of water are not the same. The ∆Hf of -909.3 kJ/mol for infinite dilution can be confirmed from the data trend. The temperature change for formation of 500.0 mL, 1.00 M \(H_2SO_4\) from pure \(H_2SO_4\) can be found using the heat exchange equation and comes out to be -21.8°C.

Step by step solution

01

Understanding Enthalpy Change

The enthalpy of formation is the heat change that results when one mole of a compound is formed from its elements. The differing enthalpy values for \(H_2SO_4\) come from it being in different solutions. As the ion becomes more and more dilute, the interactions with the water around it change, leading to different enthalpies of formation.
02

Confirming ∆Hf of Dilute \(H_2SO_4\)

The heat of formation of an infinitely dilute solution is the energy released or absorbed when 1 mole of a substance is dispersed infinitely throughout a solution. This value can be confirmed from the chapter data or from experiments by logic that as the number of water moles increases, the ∆Hf value decreases, approaching -909.3 kJ/mol.
03

Calculate Temperature Change

First, calculate the number of moles of \(H_2SO_4\) in 500.0 mL of 1.00 M solution, which is 0.500 mol. The heat added to the water can be found using ∆H = moles × ∆Hf, which is -909.3 kJ/mol × 0.500 mol = -454.65 kJ = -454650 J (since 1 kJ = 1000 J). Now, use the formula for heat exchange, q = mass × specific heat × ∆T. Substituting the known values, -454,650 J = 500 g × 4.2 J/g°C × ∆T, calculate ∆T.

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

A 1.620 g sample of naphthalene, \(C_{10} \mathrm{H}_{8}(\mathrm{s}),\) is completely burned in a bomb calorimeter assembly and a temperature increase of \(8.44^{\circ} \mathrm{C}\) is noted. If the heat of combustion of naphthalene is \(-5156 \mathrm{kJ} / \mathrm{mol} \mathrm{C}_{10} \mathrm{H}_{8}\) what is the heat capacity of the bomb calorimeter?

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}), \text { given that }$$ $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ &\left.\qquad \Delta H^{\circ}=-110.54 \mathrm{k} \mathrm{J}\right] \end{array}$$ $$\begin{aligned} &\text { C(graphite) }+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})\\\ &&\Delta H^{\circ}=-393.51 \mathrm{kJ} \end{aligned}$$

Briefly describe each of the following ideas or methods: (a) law of conservation of energy; (b) bomb calorimetry; (c) function of state; (d) enthalpy diagram; (e) Hess's law.

Use Hess's law and the following data $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ \Delta H^{\circ}=-802 \mathrm{kJ} \end{aligned}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&+247 \mathrm{kJ} \end{aligned}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&+247 \mathrm{kJ} \end{aligned}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&+206 \mathrm{kJ} \end{aligned}$$ to determine \(\Delta H^{\circ}\) for the following reaction, an important source of hydrogen gas $$\mathrm{CH}_{4}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g})$$

A 1.103 g sample of a gaseous carbon-hydrogenoxygen compound that occupies a volume of \(582 \mathrm{mL}\) at 765.5 Torr and \(25.00^{\circ} \mathrm{C}\) is burned in an excess of \(\mathrm{O}_{2}(\mathrm{g})\) in a bomb calorimeter. The products of the combustion are \(2.108 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}), 1.294 \mathrm{g} \mathrm{H}_{2} \mathrm{O}(1),\) and enough heat to raise the temperature of the calorimeter assembly from 25.00 to \(31.94^{\circ} \mathrm{C}\). The heat capacity of the calorimeter is \(5.015 \mathrm{kJ} /^{\circ} \mathrm{C}\). Write an equation for the combustion reaction, and indicate \(\Delta H^{\circ}\) for this reaction at \(25.00^{\circ} \mathrm{C}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

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