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

Explanation:

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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:

Whereas solid carbon is not inserted in the equilibrium expression.

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adding or remove a gaseous reactant or product changes the concentrations. if the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. if more products are added, the system will shift to form more reactants). conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. if reactants are removed, the system will shift to form more reactants. if the concentration of products is decreased, the equilibrium will shift toward the products).

explanation: