Chapter 20, Problem 36PS

### Chemistry & Chemical Reactivity

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

Chapter
Section

### Chemistry & Chemical Reactivity

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

# Generating hydrogen gas using the least amount of energy possible is very desirable. One proposed scheme is the catalyzed decomposition of formic acid, HCO2H. [S. Ott, Science, Vol. 333, pp. 1714-1715, 2011.]HCO2H(g) ⇄ CO2(g) + H2(g)A particularly interesting aspect of this reaction is that it can be run in reverse and so serve as a way of storing hydrogen. Calculate the equilibrium constant (at 298 K) for the decomposition reaction using thermodynamic data. Is the decomposition reaction product-favored at equilibrium at 298 K? (For HCO2H(g), ΔfH° = −378.6 kJ/mol and S° = 248.70 J/K·mol.)

Interpretation Introduction

Interpretation: The equilibrium constant for the reaction HCO2H(g)CO2(g)+H2(g) at 298K has to be determined from the given data.

Concept introduction:

Equilibrium constant in terms of concentration[KC]: Equilibrium constant can be expressed in terms of concentration.

aA(g)+bB(g)cC(g)+dD(g)Kc=[C]c×[D]d[A]a×[B]b

Equilibrium constant and Gibbs free energy:

ΔG0=RTlnK

Relation between Gibbs free energy, Enthalpy and Entropy:

ΔG=ΔH-TΔSΔG-GibbsfreeenergyΔH-EnthalpyΔS-Entropy

Explanation

Given:

Î”fH0â€‰â€‰=â€‰âˆ’378.6â€‰kJ/molÎ”Sâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰248.70â€‰J/K.molTâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰298â€‰K

Using the relation Î”Gâ€‰â€‰=â€‰â€‰Î”H-TÎ”S, we can calculate the Gibbs free energy

Î”Gâ€‰â€‰=â€‰â€‰Î”Hâ€‰âˆ’TÎ”Sâ€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰â€‰(âˆ’378.6â€‰â€‰kJ/mol)â€‰âˆ’â€‰298â€‰K(Ã—â€‰248.70â€‰J/K.mol)â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰=â€‰âˆ’74

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