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.)

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Giangiglia Giangiglia · Answer 1 · 2020-01-15T13:31:08+01:00

Answer:

[tex]K=3.9*10^{-82}[/tex]

Explanation:

Reactions:

[tex]NO_2 (g) \longleftrightarrow NO (g) +O (g)[/tex]

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

[tex]O_3 (g) + NO (g) \longleftrightarrow NO_2 (g) + O_2 (g)[/tex]

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

To achive the reaction we combine those two reactions:

[tex]NO_2 (g) + O_3 (g) + NO (g) \longleftrightarrow NO (g) + O (g) + NO_2 (g)[/tex]

[tex]O_3 (g) \longleftrightarrow O (g) + O_2 (g)[/tex]

The equilibrium constant:

[tex]K= K_1*K_2=(6.8*10^{-49})*(5.8*10^{-34})=3.9*10^{-82}[/tex]