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Consider the reaction: S(s) O2(g)SO2(g) Write the equilibrium constant for this reaction in terms of the equilibrium constants, Ka and Kb, for reactions a and b below: a.) 2 S(s) 3 O2(g) 2 SO3(g) Ka b.) SO2(g) 1/2 O2(g) SO3(g) Kb

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sebassandin

Answer:

[tex]Ka=\frac{[SO_3]^2}{[O_2]^3} \\\\Kb=\frac{[SO_3]^3}{[SO_2][O_2]^{1/2}}[/tex]

Explanation:

Hello!

In this case, according to the reactions:

a.) 2 S(s) 3 O2(g) ⇔ 2 SO3(g) Ka

b.) SO2(g) + 1/2 O2(g) ⇔ SO3(g) Kb

Thus, according to the law of mass action, we can write Ka and Kb as follows:

[tex]Ka=\frac{[SO_3]^2}{[O_2]^3} \\\\Kb=\frac{[SO_3]^3}{[SO_2][O_2]^{1/2}}[/tex]

Whereas solid carbon is not inserted in the equilibrium expression.

Best regards!

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