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Chapter 21: Problem 12

The Gibbs energies of formation, \(\Delta G_{f}^{Q}\), for \(\mathrm{KO}_{2}(\mathrm{s})\) and \(\mathrm{K}_{2} \mathrm{O}(\mathrm{s})\) are \(-240.59 \mathrm{kJmol}^{-1}\) and \(-322.09 \mathrm{kJmol}^{-1}\) respectively, at \(298 \mathrm{K}\). Calculate the equilibrium constant for the reaction below at \(298 \mathrm{K}\). Is \(\mathrm{KO}_{2}(\mathrm{s})\) thermodynamically stable with respect to \(\mathrm{K}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g})\) at \(298 \mathrm{K} ?\) $$ 2 \mathrm{KO}_{2}(\mathrm{s}) \longrightarrow \mathrm{K}_{2} \mathrm{O}(\mathrm{s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) $$

Short Answer

Expert verified
Substitute the values into the equations in Steps 1 and 2 to get the value for \(\Delta G\) and the equilibrium constant, K. Use the sign of the \(\Delta G\) to determine the thermodynamic stability in Step 3.

Step by step solution

01

Calculate the change in Gibbs free energy for the reaction

The change in Gibbs free energy for the reaction can be calculated using the Gibbs energies of formation of the products and reactants in the reaction. The formula is \(\Delta G_{reaction} = \Delta G_{f}^{products} - \Delta G_{f}^{reactants}\). Substituting the given values, we have, \(\Delta G = [(-322.09 kJ/mol) + (1.5*0)] - 2*(-240.59 kJ/mol)\) where 0 kJ/mol is the standard Gibbs free energy for \(O_2\).
02

Calculate the equilibrium constant

The relationship between the Gibbs free energy and the equilibrium constant is given by the equation \(\Delta G = -R*T*log(K)\), where 'R' is the gas constant (8.314 J/K.mol) and 'T' is the temperature (in Kelvin). We can rearrange this equation to solve for the equilibrium constant, K: \(K = exp(-\Delta G / R*T)\). Here, T is given as 298 K. Substituting these values, we can find the equilibrium constant for the given reaction.
03

Determine the thermodynamic stability of KO2

The sign of the Gibbs free energy change for the reaction determines whether a reaction is thermodynamically favourable. If \(\Delta G\) is negative, the reaction proceeds spontaneously, and the reverse reaction is not spontaneous. If \(\Delta G\) is positive, the reaction does not occur spontaneously, and the reverse reaction is favourable. Use this concept to determine the thermodynamic stability of KO2 with respect to K2O and \(O_2\).

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