Chemistry 12

Topics: Chemical equilibrium, Equilibrium constant, Thermodynamics Pages: 10 (1915 words) Published: May 6, 2013
Chemistry 12

Unit 2 Notes - Equilibrium

Chemistry 12 Tutorial 5 – The Equilibrium Constant (Keq)
What is Keq ?
The "K" in Keq stands for "Constant". The "eq" means that the reaction is at equilibrium.

Very roughly, Keq tells you the ratio of Products/Reactants for a given reaction at equilibrium at a certain temperature.
[Products] [Reactants]

K eq =

It's not quite this simple when we deal with real substances. Let's take an example. It has been found for the reaction: 2HI(g ) H2(g) + I2(g)

that if you take the [H2], the [I2] and the [HI] in an equilibrium mixture of these at 423 °C, the expression: [H 2 ] [I 2 ] = 0.0183 [HI] 2 The value of K eq The K eq expression

The value of this ratio stays at 0.0183 regardless of what we might try to do with the concentrations. The only thing that changes the value of Keq for a given reaction is the temperature!

Writing Keq Expressions
In the example just above this, the equation was: 2HI(g) and the Keq expression was: Keq =

H2(g) + I2(g)
[H 2 ] [I 2 ] [HI] 2

Unit 2 Notes – Equilibrium

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Chemistry 12

Unit 2 Notes - Equilibrium

Notice a couple of things here. The concentrations of the products are on the top (numerator) and the concentration of the reactant is on the bottom. (denominator). Also, notice that the coefficient "2" in the "2HI" in the equation ends up as an exponent for [HI] 2 in the Keq expression. Thus we have [HI] in the denominator.

Notice that in the Equilibrium Constant Expression (Keq ), whatever is written on the right of the arrow in the equation (products) is on top and whatever is written on the left of the arrow in the equation (reactants) is on the bottom. This is always the case in a Keq expression, regardless of which reaction (forward or reverse) predominates at a certain time. *********************************************************

The Keq Expressions for Solids and Liquids
Consider the following reaction: CaCO3(s) CaO(s) + CO2(g)

You might expect the Keq expression to be something like:  Keq = [CaO(s)] [CO2(g)] __________________

[CaCO3(s)] But when you consider a solid, the number of moles per litre or molecules in a certain volume is constant. The molecules everywhere in the solid are about the same distance apart and are the same size:

In a Solid, equal volumes anywhere within the solid have an equal number of molecules.. Therefore we say that the concentration of a solid is constant.

Unit 2 Notes – Equilibrium

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Chemistry 12

Unit 2 Notes - Equilibrium

Going back to our example: Consider the following reaction: CaCO3(s)

CaO(s)  + CO2(g)

You might expect the Keq expression to be something like: Keq = [CaO(s)] [CO2(g)] __________________ [CaCO3(s)] Since CaO and CaCO3 are solids, we can assume that their concentrations are constant. We can therefore rewrite the Keq expression as follows:  Keq = (a constant) [CO2(g)] ____________________

(a constant) Now, if we rearrange:  Keq (a constant) (a constant) = [CO2(g)]

You'll notice that now, on the left side, we have an expression which consists of only constants. Chemists simply combine all these constants on the left and call it the equilibrium constant (Keq ) In other words, the concentrations of the solids are incorporated into the value for Keq. Therefore, the Keq expression for the equation: CaCO3(s) is simply: CaO(s)  + CO2(g)

Keq = [CO2]

The bottom line is: When we write the Keq expression for a reaction with solids, we simply leave out the solids. Liquids also have a fairly constant concentration. They don't expand or contract that much even with changes in temperature.

Unit 2 Notes – Equilibrium

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Chemistry 12

Unit 2 Notes - Equilibrium

The same argument that was used for solids can also be used for liquids. Thus, we can expand the last statement: When we write the Keq expression for a reaction with solids or liquids, we simply leave out the solids and the liquids....

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