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Generating hydrogen gas using the least amount of energy possible is very desirable. One suggested scheme is the catalyzed decomposition of formic acid, HCO2H{HCO}_2 {H}.

HCO2H(g)CO2( g)+H2( g){HCO}_2 {H}({g}) \rightleftarrows {CO}_2({~g})+{H}_2({~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. Determine 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{HCO}_2 {H}({g}), \Delta_f {H}^{\circ}=-378.6 {~kJ} / {mol} and S=248.70 J/Kmol.)\left.S^{\circ}=248.70 {~J} / {K} \cdot {mol}.\right)

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In this task, we calculate the equilibrium constant (at 298 K) for the decomposition reaction of HCO2H\text{HCO}_2\text{H} using thermodynamic data. In addition, we determine whether the decomposition reaction is product-favored at equilibrium at 298 K.

ΔfH°(HCO2H(g))=378.6kJ/molΔ_fH° (\text{HCO}_2\text{H(g)}) = −\,378.6\, \text{kJ/mol} S°(HCO2H(g))=248.70J/KmolS° (\text{HCO}_2\text{H(g)}) = 248.70\, \text{J/Kmol}

Catalyzed decomposition reaction:

HCO2H(g)CO2(g)+H2(g)\text{HCO}_2\text{H(g)}\, \rightleftarrows\, \text{CO}_2\text{(g)}\, +\, \text{H}_2\text{(g)}

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