### Preparation of Oxygen by Joseph PriestleyThe element oxygen was first prepared by Joseph Priestley in 1774. Priestley achieved this by heating mercury(II) oxide. The chemical reaction involved is:\[ \text{HgO(s)} \rightarrow \text{Hg(l)} + \frac{1}{2}\text{O}_2(\text{g}), \Delta H^\circ = 90.84 \, \text{kJ/mol} \]### Estimation of Temperature for Reaction SpontaneityTo determine the temperature at which this reaction becomes spontaneous under standard state conditions, we consider the entropy (\(S^\circ\)) values for each species:- \(S^\circ(\text{Hg}) = 76.02 \, \text{J/K} \cdot \text{mol}\)- \(S^\circ(\text{O}_2) = 205.0 \, \text{J/K} \cdot \text{mol}\)- \(S^\circ(\text{HgO}) = 70.29 \, \text{J/K} \cdot \text{mol}\)### Analysis:Using these thermodynamic values and the Gibbs free energy equation, one can estimate the temperature at which the reaction becomes spontaneous.

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The element oxygen was prepared by Joseph Priestley in 1774 by heating mercury(II) oxide: HgO(s) → Hg() + 1/2O2(g), AH = 90.84 kJ/mol. - Estimate the temperature at which this reaction will become spontaneous under standard state conditions. S (Hg) = 76.02 J/K • mol S(O2) = 205.0 J/K • mol S(HgO)= 70.29 J/K • mol

The element oxygen was prepared by Joseph Priestley in 1774 by heating mercury(II) oxide: HgO(s) → Hg() + 1/2O2(g), AH = 90.84 kJ/mol. - Estimate the temperature at which this reaction will become spontaneous under standard state conditions. S (Hg) = 76.02 J/K • mol S(O2) = 205.0 J/K • mol S(HgO)= 70.29 J/K • mol

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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