   Chapter 20, Problem 36PS

Chapter
Section
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:

ΔfH0=378.6kJ/molΔS=248.70J/K.molT=298K

Using the relation ΔG=ΔH-TΔS, we can calculate the Gibbs free energy

ΔG=ΔHTΔS=(378.6kJ/mol)298K(×248.70J/K.mol)=74

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