Question

Write the expression for the equilibrium constant for the following reactions ? 2SO2(g) + O2(g) -------> 2SO3(g)

What is equilibrium constant?

Answer 1

`K_c = (["SO"_3]^2)/(["SO"_2]^2 * ["O"_2])`

Explanation:

For a chemical equilibrium, the equilibrium constant is defined as the ratio between the product of the equilibrium concentrations of the products and the product of the equilibrium concentrations of the reactants, all raised to the power of their respective stoichiometric coefficients.

For a general form equilibrium reaction

`color(blue)(aA + bB rightleftharpoons cC + dD)" "`, where

`A`, `B`, `C`, and `D` are either gases or liquids

the equilibrium constant, `K_c`, will take the form

`color(blue)(K_c = ([C]^c * [D]^d)/([A]^a * [B]^b))`

Here

`[A]`, `[B]`, `[C]`, `[D]` - the equilibrium concentrations of the chemical species that take part in the reaction.

Now take a look at the equilibrium given to you

`color(red)(2)"SO"_text(2(g]) + "O"_text(2(g]) rightleftharpoons color(blue)(2)"SO"_text(3(g])`

Notice that your reaction has two reactants, `"SO"_2` and `"O"_2`, and one product, `"SO"_3`. This tells you that the equilibrium constant will feature three terms, each corresponding to the equilibrium concentration of these three chemical species.

More specifically, the equilibrium constant will be

`K_c = ( ["SO"_3]^color(blue)(2))/( ["SO"_2]^color(red)(2) * ["O"_2])`

SIDE NOTE Since you're dealing with an equilibrium reaction, you can write the equilibrium constant for the reverse reaction, `K_c^'`, by using the same approach

`color(blue)(2)"SO"_text(3(g]) rightleftharpoons color(red)(2)"SO"_text(2(g]) + "O"_text(2(g])`

This time, you have one product and two reactants, so the equilibrium constant will take the form

`K_c^' = (["SO"_2]^color(red)(2) * ["O"_2])/(["SO"_3]^color(blue)(2))`

Notice that `K_c^' = 1/K_c`.