Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 23: Problem 19

Calcium will reduce \(\mathrm{MgO}(\mathrm{s})\) to \(\mathrm{Mg}(\mathrm{s})\) at all temperatures from 0 to \(2000^{\circ} \mathrm{C}\). Use this fact, together with the melting point ( \(839^{\circ} \mathrm{C}\) ) and boiling point \(\left(1484^{\circ} \mathrm{C}\right)\) of calcium, to sketch a plausible graph of \(\Delta G^{\circ}\) as a function of temperature for the reaction \(2 \mathrm{Ca}(\mathrm{s})+\) \(\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CaO}(\mathrm{s})\).

Short Answer

Expert verified
The graph begins with \(\Delta G^{\circ}\) at a negative value at 0°C. There are two points where \(\Delta G^{\circ}\) is zero, which are at the melting point (839°C) and boiling point (1484°C) of Calcium. Otherwise, the \(\Delta G^{\circ}\) is negative from 0 to 2000°C, indicating that the reaction is spontaneous over this entire temperature range.

Step by step solution

01

Identify Reaction Spontaneity

Given that Calcium reduces Magnesium Oxide to Magnesium at all temperatures from 0 to 2000°C, it means the reaction is spontaneous at this temperature range. Hence, \(\Delta G^{\circ}\) is negative throughout this range.
02

Consider the Phase Changes of Calcium

Calcium has a melting point of \(839^{\circ} \mathrm{C}\) and a boiling point of \(1484^{\circ} \mathrm{C}\). At these temperatures, \(\Delta G^{\circ}\) is zero as phase changes are equilibrium processes where Gibbs free energy change is zero.
03

Sketch the \(\Delta G^{\circ}\) vs Temperature Graph

The graph starts with negative \(\Delta G^{\circ}\) at 0°C indicating spontaneous reaction, it then moves to zero at \(839^{\circ} \mathrm{C}\) due to melting point, goes negative again as temperature increases, then to zero at boiling point \(1484^{\circ} \mathrm{C}\), and then again goes negative showing the reaction is still spontaneous at temperature above boiling point of Calcium.

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

Balance the following oxidation-reduction equations. $$\text { (a) } \mathrm{Fe}_{2} \mathrm{S}_{3}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{S}(\mathrm{s})$$ $$\begin{aligned} &\text { (b) } \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\\\ &&\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \end{aligned}$$ $$\begin{aligned} &\text { (c) } \mathrm{Ag}(\mathrm{s})+\mathrm{CN}^{-}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O} \longrightarrow\\\ &&\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \end{aligned}$$

By means of a chemical equation, give an example to represent the reaction of (a) a transition metal with a nonoxidizing acid; (b) a transition metal oxide with \(\mathrm{NaOH}(\mathrm{aq}) ;(\mathrm{c})\) an inner transition metal with \(\mathrm{HCl}(\mathrm{aq})\).

Attempts to make \(\mathrm{CuI}_{2}\) by the reaction of \(\mathrm{Cu}^{2+}(\mathrm{aq})\) and \(\mathrm{I}^{-}(\text {aq })\) produce \(\mathrm{CuI}(\mathrm{s})\) and \(\mathrm{I}_{3}^{-}(\mathrm{aq})\) instead. Without performing detailed calculations, show why this reaction should occur. $$2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{I}^{-}(\mathrm{aq}) \longrightarrow 2 \mathrm{CuI}(\mathrm{s})+\mathrm{I}_{3}^{-}(\mathrm{aq})$$

Equation \((23.18),\) which represents the chromatedichromate equilibrium, is actually the sum of two equilibrium expressions. The first is an acid-base reaction, \(\mathrm{H}^{+}+\mathrm{CrO}_{4}^{2-} \rightleftharpoons \mathrm{HCrO}_{4}^{-}\). The second reaction involves elimination of a water molecule between two \(\mathrm{HCrO}_{4}^{-}\) ions (a dehydration reaction), \(2 \mathrm{HCrO}_{4}^{-} \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}_{2} \mathrm{O} .\) If the ionization constant, \(K_{\mathrm{a}},\) for \(\mathrm{HCrO}_{4}^{-}\) is \(3.2 \times 10^{-7},\) what is the value of \(K\) for the dehydration reaction?

Covalent bonding is involved in many transition metal compounds. Draw Lewis structures, showing any nonzero formal charges, for the following molecules or ions: (a) \(\mathrm{Hg}_{2}^{2+} ;\) (b) \(\mathrm{Mn}_{2} \mathrm{O}_{7} ;\) (c) \(\mathrm{OsO}_{4}\). [Hint: In (b), there is one \(\mathrm{Mn}-\mathrm{O}-\text { Mn linkage in the molecule. }\rfloor\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

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