### Determination of the Heat of Formation for PbS#### IntroductionUsing the provided table and the chemical equation below, determine the heat of formation (in kJ/mol) for PbS:\[ 2 \, \text{PbS} \, (s) + 3 \, \text{O}_2 \, (g) \rightarrow 2 \, \text{SO}_2 \, (g) + 2 \, \text{PbO} \, (s) \]\[\Delta H^\circ = -828.4 \, \text{kJ/mol}\]#### Table of Standard Enthalpies of Formation| Substance | \(\Delta H_f^\circ\) (kJ/mol) ||----------------|-----------------------------|| O\(_2\) (g) | 0 || SO\(_2\) (g) | -296.9 || PbO (s) | -217.3 |#### ExplanationThe reaction involves the oxidation of lead sulfide (PbS) to form lead oxide (PbO) and sulfur dioxide (SO\(_2\)). The standard enthalpies of formation for the products and reactants (except for PbS which needs to be found) are provided in the table. Oxygen (\(\text{O}_2\)) is in its elemental state, so its enthalpy of formation is 0 by definition. The overall reaction enthalpy (\(\Delta H^\circ = -828.4 \, \text{kJ/mol}\)) is given and can be used to calculate the unknown enthalpy of formation for PbS using Hess’s law.

The heat of formation for PbS(s) = ?

Using the provided table and the equation below, determine the heat of formation (in kJ/mol) for PbS. 2 PbS (s) + 3 O2 (g) → 2 SO₂ (g) + 2 PbO (s) AH° = -828.4 kJ/mol Substance O₂ (g) SO: (g) PbO (s) Hf (kJ/mol) 0 -296.9 -217.3

Using the provided table and the equation below, determine the heat of formation (in kJ/mol) for PbS. 2 PbS (s) + 3 O2 (g) → 2 SO₂ (g) + 2 PbO (s) AH° = -828.4 kJ/mol Substance O₂ (g) SO: (g) PbO (s) Hf (kJ/mol) 0 -296.9 -217.3

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

### Determination of the Heat of Formation for PbS

#### Introduction

Using the provided table and the chemical equation below, determine the heat of formation (in kJ/mol) for PbS:

\[ 2 \, \text{PbS} \, (s) + 3 \, \text{O}_2 \, (g) \rightarrow 2 \, \text{SO}_2 \, (g) + 2 \, \text{PbO} \, (s) \]

\[\Delta H^\circ = -828.4 \, \text{kJ/mol}\]

#### Table of Standard Enthalpies of Formation

| Substance | \(\Delta H_f^\circ\) (kJ/mol) |
|----------------|-----------------------------|
| O\(_2\) (g) | 0 |
| SO\(_2\) (g) | -296.9 |
| PbO (s) | -217.3 |

#### Explanation

The reaction involves the oxidation of lead sulfide (PbS) to form lead oxide (PbO) and sulfur dioxide (SO\(_2\)). The standard enthalpies of formation for the products and reactants (except for PbS which needs to be found) are provided in the table. Oxygen (\(\text{O}_2\)) is in its elemental state, so its enthalpy of formation is 0 by definition.

The overall reaction enthalpy (\(\Delta H^\circ = -828.4 \, \text{kJ/mol}\)) is given and can be used to calculate the unknown enthalpy of formation for PbS using Hess’s law.

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