The equilibrium position of a reaction may be changed by:•Adding or removing a reactant or product•Changing the pressure by changing the volume (equilibria involving gases)•Dilution (for equilibria in solution) •Changing the temperature. •Le Chatelier’s Principle is very useful in determining how the position of equilibrium can be changed to ensure more product is formed.•Le Chatelier’s Principle states that if a system is at equilibrium and the temperature, pressure or concentrations of the species are changed, the reaction will proceed in such a direction as to oppose this change. Changes in concentration and changes in pressure for gaseous systems will alter the equilibrium position but not the value of the equilibrium constant, K. The addition of a reactant moves the position of equilibrium forward, producing more product. The value of the equilibrium constant, K, is unchanged. An increase in pressure for a gaseous system moves the position of equilibrium to the side of the equation that has fewer moles of substance, and hence lower pressure. The value of the equilibrium constant, K, is unchanged. When solution is diluted the reaction will go in the direction that will form the greater amount of particles. Eg. Fe^3+(aq) + SCN^-(aq) <-----> Fe(SCN)^2+ (aq) 2 particles 1 particle There will be a net back reaction. The direction of the equilibrium will change for an increase in temperature depending on whether the reaction is endothermic or exothermic. The addition of a catalyst does not change the position of equilibrium or the value of the equilibrium constant ,K. Equilibrium is simply reached more quickly. The addition of an inert gas at constant volume does not change the position of equilibrium or the value of the equilibrium constant, K. The total pressure is increased. → indicates a net forward reaction← indicates a net back reaction•Add reactants: →•Add Products: ←•Increase Pressure: direction of the smaller amount of particles•Dilution: direction of larger amount of particles•Increase Temperature:•Exothermic Reaction ←•Endothermic Reaction →•Catalyst: Increased rate but no change in equilibrium. •There are two things we would like to do in industry to make processes more viable: increase the RATE and YIELD of the reaction.
Image courtesy of http://en.wikipedia.org/wiki/Image:Lechatelier.jpg, Public Domain, https://commons.wikimedia.org/w/index.php?curid=428795
In the previous section, you learned about reactions that can reach a state of equilibrium, in which the concentration of reactants and products aren't changing. If these amounts are changing, we should be able to make a relationship between the amount of product and reactant when a reaction reaches equilibrium.
Equilibrium reactions are those that do not go to completion, but are in a state where the reactants are reacting to yield products and the products are reacting to produce reactants. In a reaction at equilibrium, the equilibrium concentrations of all reactants and products can be measured. The equilibrium constant (\(K\)) is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants. Sometimes, subscripts are added to the equilibrium constant symbol \(K\), such as \(K_\text{eq}\), \(K_\text{c}\), \(K_\text{p}\), \(K_\text{a}\), \(K_\text{b}\), and \(K_\text{sp}\). These are all equilibrium constants and are subscripted to indicate special types of equilibrium reactions. There are some rules about writing equilibrium constant expressions that need to be learned:
Write the equilibrium constant expression for: \[\ce{CO} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons \ce{CH_4} \left( g \right) + \ce{H_2O} \left( g \right) \nonumber \]
\[K = \dfrac{\left[ \ce{CH_4} \right] \left[ \ce{H_2O} \right]}{\left[ \ce{CO} \right] \left[ \ce{H_2} \right]^3} \nonumber \] *Note that the coefficients become exponents. Also, note that the concentrations of products in the numerator are multiplied. The same is true of the reactants in the denominator.
Write the equilibrium constant expression for: \[2 \ce{TiCl_3} \left( s \right) + 2 \ce{HCl} \left( g \right) \rightleftharpoons 2 \ce{TiCl_4} \left( s \right) + \ce{H_2} \left( g \right) \nonumber \]
\[K = \dfrac{\left[ \ce{H_2} \right]}{\left[ \ce{HCl} \right]^2} \nonumber \] *Note that the solids have a value of 1, and multiplying or dividing by 1 does not change the value of K.
Write the equilibrium constant expression for: \[\ce{P_4} \left( s \right) + 6 \ce{Cl_2} \left( g \right) \rightleftharpoons 4 \ce{PCl_3} \left( s \right) \nonumber \]
\[K = \dfrac{1}{\left[ \ce{Cl_2} \right]^6} \nonumber \] *Note that the only product is a solid, which is defined to have a value of 1. That leaves just 1 on top in the numerator.
Write the equilibrium constant expression for: \[\ce{H_2O} \left( l \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{OH^-} \left( aq \right) \nonumber \]
\[K = \left[ \ce{H^+} \right] \left[ \ce{OH^-} \right] \nonumber \] *Note that the water is the solvent, and thus has a value of 1. Dividing by 1 does not change the value of K.
The equilibrium constant value is the ratio of the concentrations of the products over the reactants. This means that we can use the value of \(K\) to predict whether there are more products or reactants at equilibrium for a given reaction. What can the value of Keq tell us about a reaction?
Here are some examples to consider:
If the equilibrium constant is 1 or nearly 1, it indicates that the molarities of the reactants and products are about the same. If the equilibrium constant value is a large number, like 100, or a very large number, like \(1 \times 10^{15}\), it indicates that the products (numerator) are a great deal larger than the reactants. This means that at equilibrium, the great majority of the material is in the form of products and it is said that the "products are strongly favored". If the equilibrium constant is small, like 0.10, or very small, like \(1 \times 10^{-12}\), it indicates that the reactants are much larger than the products and the reactants are strongly favored. With large \(K\) values, most of the material at equilibrium is in the form of products and with small \(K\) values, most of the material at equilibrium is in the form of the reactants. The equilibrium constant expression is an equation that we can use to solve for \(K\) or for the concentration of a reactant or product.
Determine the value of \(K\) for the reaction \[\ce{SO_2} \left( g \right) + \ce{NO_2} \left( g \right) \rightleftharpoons \ce{SO_3} \left( g \right) + \ce{NO} \left( g \right) \nonumber \] when the equilibrium concentrations are: \(\left[ \ce{SO_2} \right] = 1.20 \: \text{M}\), \(\left[ \ce{NO_2} \right] = 0.60 \: \text{M}\), \(\left[ \ce{NO} \right] = 1.6 \: \text{M}\), and \(\left[ \ce{SO_3} \right] = 2.2 \: \text{M}\).
Step 1: Write the equilibrium constant expression: \[K = \dfrac{\left[ \ce{SO_3} \right] \left[ \ce{NO} \right]}{\left[ \ce{SO_2} \right] \left[ \ce{NO_2} \right]} \nonumber \] Step 2: Substitute in given values and solve: \[K = \dfrac{\left( 2.2 \right) \left( 1.6 \right)}{\left( 1.20 \right) \left( 0.60 \right)} = 4.9 \nonumber \]
Consider the following reaction: \[\ce{CO} \left( g \right) + \ce{H_2O} \left( g \right) \rightleftharpoons \ce{H_2} \left( g \right) + \ce{CO_2} \left( g \right) \nonumber \] with \(K = 1.34\). If the \(\left[ \ce{H_2O} \right] = 0.100 \: \text{M}\), \(\left[ \ce{H_2} \right] = 0.100 \: \text{M}\), and \(\left[ \ce{CO_2} \right] = 0.100 \: \text{M}\) at equilibrium, what is the equilibrium concentration of \(\ce{CO}\)?
Step 1: Write the equilibrium constant expression: \[K = \dfrac{\left[ \ce{H_2} \right] \left[ \ce{CO_2} \right]}{\left[ \ce{CO} \right] \left[ \ce{H_2O} \right]} \nonumber \] Step 2: Substitute in given values and solve: \[1.34 = \dfrac{\left( 0.100 \right) \left( 0.100 \right)}{\left[ \ce{CO} \right] \left( 0.100 \right)} \nonumber \] Solving for \(\left[ \ce{CO} \right]\), we get: \(\left[ \ce{CO} \right] = 0.0746 \: \text{M}\)
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