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

Determine the relationship between the given reactions and the desired reaction

Subtract the first reaction from the third reaction: $(\mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2\mathrm{Na}(l)+\mathrm{O}_{2}(g)) - (\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g)) \\\ \Rightarrow (\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s))$

02

Multiply the equilibrium constants

To find the equilibrium constant for the desired reaction, we must divide the equilibrium constant of the third reaction by the equilibrium constant of the first reaction: \(K_a = \frac{K_{3}}{K_{1}} = \frac{5 \times 10^{-29}}{2 \times 10^{-25}} = 2.5 \times 10^{-4}\) b. Determine the equilibrium constant for \(\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\).

03

Determine the relationship between the given reactions and the desired reaction

Subtract the first reaction from the second reaction and add the result to the third reaction: $((\mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g)) - (\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g))) + (\mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g)) \\ \Rightarrow (\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l))$

04

Multiply the equilibrium constants

To find the equilibrium constant for the desired reaction, we must multiply the equilibrium constants of the second and third reactions and divide by the equilibrium constant of the first reaction: \(K_b = \frac{K_{2} \times K_{3}}{K_{1}} = \frac{(2 \times 10^{-5}) (5 \times 10^{-29})}{2 \times 10^{-25}} = 5 \times 10^{-9}\) c. Determine the equilibrium constant for \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\).

05

Determine the relationship between the given reactions and the desired reaction

Subtract the first reaction twice from the third reaction: $2((\mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g))) - 2(\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g)) + (\mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g)) \\ \Rightarrow (2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s))$

06

Multiply the equilibrium constants

To find the equilibrium constant for the desired reaction, we must square the equilibrium constant of the second reaction and multiply by the equilibrium constant of the third reaction, then divide by the square of the equilibrium constant of the first reaction: \(K_c = \frac{(K_{2}^2) K_{3}}{K_{1}^2} = \frac{(2 \times 10^{-5})^2 (5 \times 10^{-29})}{(2 \times 10^{-25})^2} = 1 \times 10^{-8}\)

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