equilibrium+calculations+1

Calculations Involving the Equilibrium Constant, Part 1
__1. When equilibrium concentrations are known:__

If you are provided a chemical equation and the concentrations of each substance at equilibrium, then calculating K is as easy as 'plug-and-chug'.

For example:

Calculate the equilibrium constant for the following reaction at equilibrium, given the following information:

Iron (III) hydroxide is slightly soluble in water: Fe(OH) 3 (s) <--> Fe 3+ (aq) + 3 OH - (aq)

At equilibrium, the following concentrations were determined:

[Fe 3+ ] = 8.8 x 10 -11 M

[OH - ] = 2.6 x 10 -10 M

__Solution:__

Write the equilibrium constant expression: Ksp = [Fe 3+ ][OH - ] 3

Then plug in the concentrations: Ksp = (8.8 x 10 -11 )(2.6 x 10 -10 ) 3

And do the math.

Ksp = 1.55 x 10 -39

__2. When the value of K is known, and the equation is changed:__

There are two general rules for this situation. Suppose we have a chemical equation as follows:

2 A (g) <--> 3 B (g)

With a K = 0.25

First, if the reaction equation was simply reversed:

3 B (g) <--> 2 A (g)

then the __first rule__ is that equilibrium constant expression would have to be inverted. This means the value of K is inverted.

So the rule for reversing an equation is to invert K:


 * K -1 (or 1/K)**

For the example, this would be (0.25) -1 = 4.0

The __second rule__ is when a chemical equation's coefficients are changed by multiplying them all by some number n.

For example, if the original chemical equation from above: 2 A (g) <--> 3 B (g), with a K = 0.25, was modified to:

A (g) <--> 3/2 B (g)

Since coefficients become exponents in the equilibrium expression, then the rule is to take K to the power of whatever number n was used in modifying the equation.


 * K n **

In the example, the equation was multiplied by 1/2, so n = 1/2:

K 1/2 = (0.25 1/2 ) = 0.50