Equilibrium Equations and Equilibrium Constants

Reactions that go to completion:

Reactions that are nonreversible are said to go to completion. After the reaction occurs the reaction mixture contains only products and any left over excess reactants. The products do not break down to reform reactants. Burning an organic compound like methane, CH4, is an example of a reaction that goes to completion. When you burn natural gas in air the products are CO2 and H2O. The methane is completely converted to these products. The extra oxygen in the atmosphere is still around but it isn't part of a reversible process. The CO2 and H2O do not react to form CH4 and oxygen. If an equilibrium constant were to apply, it would have a value of infinity. The value for K is large because products dominate the mixture. In fact, there are no reactant methane molecules left. The larger the value for the equilibrium constant the more the reaction goes to completion. Irreversible reactions can be thought to have an infinite equilibrium constant so there are no reactants left.

What does an equilibrium constant tell:

The number values for equilibrium constants are tied to the nature of reactants and products in a reaction. The number values for "K" are gotten from experiments measuring equilibrium concentrations. The number value tells the equilibrium ratio of products to reactants. In an equilibrium mixture both reactants and products coexist. The value for K is large when products dominate the mixture. The value for K is small when the reactants dominate the mixture. The expression for an equilibrium reaction is determined by the coefficients in the balanced equation. Large K > 1 products are "favored" K = 1 neither reactants nor products are favored Small K < 1 reactants are "favored" The term "favored" means that side of the equation has higher numbers of moles and higher concentrations than the other. Large K have values that are huge, 1 x 1034 K = 1 neither reactants nor products are favored Small K have values that are tiny, 4 x 10-41 Reversible reactions have analogous mechanical examples. Image a pair of identical fish tanks connected by a tube with a sliding "gate". One tank has no fish and the other contains 10 identical fish. Predict what will happen when you open the gate that blocks access to the empty tank. Hmm! I hope we agree. The fish will swim around and the 10 fish will eventually distribute themselves between the two tanks. The "equilibrium" condition will have 5 fish in each tank. If you took pictures of the process as the fish swim around they might look like these.   The fish population will reach an equilibrium with five fish in each tank. The ratio of the number of fish in the right hand side to the number in the left hand side will be "1". The value for the equilibrium constant is a "1". Neither side is favored. Now suppose the right hand tank was very attractive. Suppose 9 fish on average stay in the right hand tank leaving only one fish in the reactants or left hand side. The ratio of products to reactants will be ratio = 9/1 = 9. The right hand tank is favored. Products will dominate. Reversible chemical reactions follow a similar pattern. At equilibrium the number of reactant and product molecules stay constant. The identity of individual molecules keeps changing . The reactants are initially the only molecules around. They react to form products. The amount of reactant dwindles and the forward reaction slows down. The product amount increases at the same time the reactants are disappearing. The products "break down" to form reactants. The rate for this reverse reaction increases as the amount of product grows. Ultimately there comes a time when the forward reaction rate and the reverse reaction rate are equal. The mixture is at equilibrium.

How to use balanced equations to determine equilibrium expressions:

How the equilibrium expression is determined Equilibrium expressions are written as a fraction. The activities of products appear in the numerator and the activities of reactants in the denominator. The "activities" are a measure of how much reaction potential a substance has. For gases activity depends on pressure. For dissolved substances activity depends on concentration. The entries are gas pressures for gases and concentrations in moles per liter for dissolved substances. The expression never includes pure solids or pure liquids. These have fixed values and are not a factor in equilibrium changes.   Example: The reversible reaction 2 H2(g) + N2(g) N2H4(g) has an equilibrium expression that looks like this. The brackets are a shorthand code for concentration in units of moles per liter. Notice the exponents are "1" on the nitrogen, N2, and hydrazine, N2H4. The exponent is a "2" on the hydrogen, H2. The exponents are the same as the coefficients in the balanced equation. Summary: The reactants appear in the denominator, bottom, The products appear in the numerator, top. Pure solids and liquids do not change concentration so do not appear in the expression.

Exercise

What is the equilibrium expression for ? N2 (g) + 3 H2 (g) 2 NH3(g) + heat

Answer:

This is the accepted equilibrium expression. Normally heat is not included, but literally it can appear in the expression with the appropriate part of the equation. Here for an exothermic reaction a HEAT term can be placed with products, in the numerator.

Exercise

What is the equilibrium expression for the following reaction? N2(g) + 2 O2(g) 2 NO2(g) + heat

Answer:

This is the accepted equilibrium expression. Normally heat is not included, but literally it can appear in the expression with the appropriate part of the equation. Here for an exothermic reaction a HEAT term can be placed with products, in the numerator.

Exercise

What is the equilibrium expression for the following reactions? CaCl2(s) Ca2+(aq) + 2 Cl-(aq)

Answer:

This is the accepted equilibrium expression. The reactant CaCl2(s) does not appear in the expression because it is a "pure" solid. All solids are assumed to be pure. Normally the "1" in the denominator is omitted. The expression is simplified.

How to use equilibrium concentrations to determine equilibrium constants:

Exercise

a. What is the equilibrium constant for the following reaction? 2 NO2(g) 2 N2O4(g) The concentrations at equilibrium are [NO2] = 0.025 moles / liter ; [N2O4] = 0.0869 moles / liter b. What is the equilibrium concentration for NO2 if the concentration of N2O4 is 0.12 moles/ liter? Remember the equilibrium constant is just that, a constant. It doesn't change.

Answer:

Step 1: Write out the equilibrium expression: This is the accepted equilibrium expression. Step 2: Substitute the equilibrium concentrations in the equilibrium expression. Normally the equilibrium constant is bigger than "1" The products are favored. The reaction tends to favor products. Step 3: Now that the value for K is known, set up the expression for the equilibrium. the equilibrium concentrations in the equilibrium expression. Step 4: Rearrange the equation and solve for x2.