Chapter 15 - Priciples of Chemical Equilibrium - Example 15-2 - Relating K to the Balanced Chemical Equation - Page 664: Practice Example B

$$K = 6.9 \times 10^{-5}$$

1. Reverse the equation. (K'' = 1/K) $$NO_2(g) \rightleftharpoons NO(g) + \frac 12 O_2(g)$$ $$K'' = 1/(1.2 \times 10^2) = 0.0083$$ 2. Multiply all coeficients by 2. (K'' = $K^2$) $$2 NO_2(g) \rightleftharpoons 2 NO(g) + O_2(g)$$ $$K = 0.0083^2 = 6.9 \times 10^{-5}$$