# Chapter 15 - Questions and Problems - Page 708: 15.31

$$K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$$ $K_c = 5.6 \times 10^{23}$

#### Work Step by Step

1. Multiply the first reaction by 2, to do so, raise $K_c^{'}$ to the power of 2. $$2S(s) + 2O_2(g) \leftrightharpoons 2SO_2(g)$$ $K_c^{'} = (4.2 \times 10^{52})^2 = 1.764\times 10^{105}$ 2. Invert the first reaction: $$2SO_2(g) \leftrightharpoons 2S(s) + 2O_2(g)$$ $K_c^{'} = 1/(1.764 \times 10^{105}) = 5.7 \times 10^{-106}$ 3. Add both reactions; removing the repeated components, the Kc of the resulting reaction is the multiplication of $K_c^{''}$ and the new $K_c^{'}$ $$2SO_2(g) +O_2(g) \leftrightharpoons 2SO_3(g)$$ $K_c = 5.7 \times 10^{-106} \times 9.8 \times 10^{128} = 5.6 \times 10^{23}$ 4. Write the equilibrium constant expression: $$K_c = \frac{[Products]}{[Reactants]} = \frac{[SO_3]^2}{[SO_2]^2[O_2]}$$

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