Given the following equilibrium constants at \(427^{\circ} \mathrm{C}\) $$\begin{array}{ll}\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{1}=2 \times 10^{-25} \\\\\mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{2}=2 \times 10^{-5} \\\\\mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{3}=5 \times 10^{-29} \\\\\mathrm{NaO}_{2}(s) \rightleftharpoons \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{4}=3 \times 10^{-14}\end{array}$$,determine the values for the equilibrium constants for the following reactions: a. \(\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) b. \(\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\) c. \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)

Step by step solution

01

Identify the given reactions and their equilibrium constants

We are given the following reactions with their respective equilibrium constants: 1. Na2O(s) <=> 2Na(l) + 0.5O2(g) with K1 = 2 x 10^-25 2. NaO(g) <=> Na(l) + 0.5O2(g) with K2 = 2 x 10^-5 3. Na2O2(s) <=> 2Na(l) + O2(g) with K3 = 5 x 10^-29 4. NaO2(s) <=> Na(l) + O2(g) with K4 = 3 x 10^-14

02

Manipulate the given reactions to obtain the desired reactions

We need to obtain the new reactions: a. Na2O(s) + 0.5O2(g) <=> Na2O2(s) b. NaO(g) + Na2O(s) <=> Na2O2(s) + Na(l) c. 2NaO(g) <=> Na2O2(s) Now, let's manipulate the given reactions: For (a), subtract Reaction 1 from Reaction 3: Na2O2(s) + 2Na(l) + O2(g) - (Na2O(s) + 2Na(l) + 0.5O2(g)) = Na2O(s) + 0.5O2(g) <=> Na2O2(s) For (b) add Reaction 1 and Reaction 2: Na2O(s) + 2Na(l) + 0.5O2(g) + NaO(g) <=> Na(l) + 0.5O2(g) + Na2O2(s) + Na(l) For (c), double Reaction 2 and subtract Reaction 3: (2)(NaO(g) <=> Na(l) + 0.5O2(g)) - (Na2O2(s) <=> 2Na(l) + O2(g)) = 2NaO(g) <=> Na2O2(s)

03

Apply rules for combining equilibrium constants

For (a), when we subtract equilibrium expressions (Reaction 3 - Reaction 1), we divide the equilibrium constants (K3 / K1): Ka = K3 / K1 = (5 x 10^-29) / (2 x 10^-25) = 2.5 x 10^-4 For (b), when we add equilibrium expressions (Reaction 1 + Reaction 2), we multiply the equilibrium constants (K1 * K2): Kb = K1 * K2 = (2 x 10^-25) * (2 x 10^-5) = 4 x 10^-30 For (c), when we double Reaction 2 and subtract Reaction 3, first we square the equilibrium constant of Reaction 2 (K2^2) and then divide it by the equilibrium constant of Reaction 3 (K2^2 / K3): Kc = (K2^2) / K3 = (2 x 10^-5)^2 / (5 x 10^-29) = 1.6 x 10^-9 So, the equilibrium constants for the new reactions are: a. Ka = 2.5 x 10^-4 b. Kb = 4 x 10^-30 c. Kc = 1.6 x 10^-9

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