Question

Consider the following data. NO2(g) equilibrium reaction arrow identifying the requirement for light NO(g) + O(g) K = 6.8 ✕ 10-49 O3(g) + NO(g) equilibrium reaction arrow NO2(g) + O2(g) K = 5.8 ✕ 10-34 Calculate a value for the equilibrium constant for the reaction below. (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

Answer 1

Answer:

$K=3.9*10^{-82}$

Explanation:

Reactions:

$NO_2 (g) \longleftrightarrow NO (g) +O (g)$

With: $K_1=\frac{[NO][O]}{[NO_2]}=6.8*10^{-49}$

$O_3 (g) + NO (g) \longleftrightarrow NO_2 (g) + O_2 (g)$

With: $K_2=\frac{[NO_2}{[O_3][NO]}=5.8*10^{-34}$

To achive the reaction we combine those two reactions:

$NO_2 (g) + O_3 (g) + NO (g) \longleftrightarrow NO (g) + O (g) + NO_2 (g)$

$O_3 (g) \longleftrightarrow O (g) + O_2 (g)$

The equilibrium constant:

$K= K_1*K_2=(6.8*10^{-49})*(5.8*10^{-34})=3.9*10^{-82}$