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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

Tin(IV) oxide, cassiterite, is the main ore of tin. It crystallizes in a rutile-like unit cell with tin(IV) ions taking the place of Ti4+ ions (Study Question 12.6). (The O2− ions marked x are wholly within the unit cell.)

(a) How many tin(IV) ions and oxide ions are there per unit cell of this oxide?

(b) Is it thermodynamically feasible to transform solid SnO2 into liquid SnCl4 by reaction of the oxide with gaseous HCl? What is the equilibrium constant for this reaction at 25 °C?

(a)

Interpretation Introduction

Interpretation:

To predict the number of tin and oxide ions present in per unit cell of tin oxide.

Concept introduction:

The formula for tin oxide is SnO2. Tin has an oxidation number of +4 in this compound. SnO2 also known as, cassiterite is the main ore of tin.

Tin oxides crystallizes in a rutile like unit cell in which Sn4+ and O2 ions are present. The compound should be electrically neutral.

Explanation

    

                    Figure 1

For tin ions, eight atoms are present at the eight corner of the cell and one atom is present at the centre...

(b)

Interpretation Introduction

Interpretation:

Check whether it is thermodynamically feasible to transform solid SnO2 into liquid SnCl4 by reactionof the oxide with gaseous HCl or not. The equailibrium constant for this raction at 250C is to be determined.

Concept introduction:

The free energy change is expressed as,

ΔrG°=nΔfG°(products)nΔfG°(reactants)

The positive value of ΔrG° indicates the thermodynamic unfeasibility of a reaction.

If the free energy change value is known then the equilibrium constant can be determined using the following equation.

Kp=eΔrGRT

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Chapter 21 Solutions

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Sect-21.11 P-2.4ACPSect-21.11 P-2.5ACPCh-21 P-1PSCh-21 P-2PSCh-21 P-3PSCh-21 P-4PSCh-21 P-5PSCh-21 P-6PSCh-21 P-7PSCh-21 P-8PSCh-21 P-9PSCh-21 P-10PSCh-21 P-11PSCh-21 P-12PSCh-21 P-13PSCh-21 P-14PSCh-21 P-15PSCh-21 P-16PSCh-21 P-17PSCh-21 P-18PSCh-21 P-19PSCh-21 P-20PSCh-21 P-21PSCh-21 P-22PSCh-21 P-23PSCh-21 P-24PSCh-21 P-25PSCh-21 P-26PSCh-21 P-27PSCh-21 P-28PSCh-21 P-29PSCh-21 P-30PSCh-21 P-31PSCh-21 P-32PSCh-21 P-33PSCh-21 P-34PSCh-21 P-35PSCh-21 P-36PSCh-21 P-37PSCh-21 P-38PSCh-21 P-39PSCh-21 P-40PSCh-21 P-41PSCh-21 P-42PSCh-21 P-43PSCh-21 P-44PSCh-21 P-45PSCh-21 P-46PSCh-21 P-47PSCh-21 P-48PSCh-21 P-49PSCh-21 P-50PSCh-21 P-51PSCh-21 P-52PSCh-21 P-53PSCh-21 P-54PSCh-21 P-55PSCh-21 P-56PSCh-21 P-57PSCh-21 P-58PSCh-21 P-59PSCh-21 P-60PSCh-21 P-61PSCh-21 P-62PSCh-21 P-63PSCh-21 P-64PSCh-21 P-65PSCh-21 P-66PSCh-21 P-67PSCh-21 P-68PSCh-21 P-69PSCh-21 P-70PSCh-21 P-71PSCh-21 P-72PSCh-21 P-73PSCh-21 P-74PSCh-21 P-75PSCh-21 P-76PSCh-21 P-77PSCh-21 P-78PSCh-21 P-79PSCh-21 P-80PSCh-21 P-81PSCh-21 P-82PSCh-21 P-83PSCh-21 P-84PSCh-21 P-85PSCh-21 P-86PSCh-21 P-87PSCh-21 P-88PSCh-21 P-89GQCh-21 P-90GQCh-21 P-91GQCh-21 P-92GQCh-21 P-93GQCh-21 P-94GQCh-21 P-95GQCh-21 P-96GQCh-21 P-97GQCh-21 P-98GQCh-21 P-99GQCh-21 P-100GQCh-21 P-101GQCh-21 P-102GQCh-21 P-103GQCh-21 P-105GQCh-21 P-106GQCh-21 P-107GQCh-21 P-108GQCh-21 P-110GQCh-21 P-111GQCh-21 P-112GQCh-21 P-113GQCh-21 P-114GQCh-21 P-115ILCh-21 P-116ILCh-21 P-117ILCh-21 P-118ILCh-21 P-119ILCh-21 P-120ILCh-21 P-121SCQCh-21 P-122SCQCh-21 P-123SCQCh-21 P-124SCQCh-21 P-125SCQCh-21 P-126SCQCh-21 P-127SCQCh-21 P-128SCQCh-21 P-129SCQCh-21 P-130SCQCh-21 P-131SCQ

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