CHEM 1412 Concept Review: Chemical Equilibrium

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Many reactions in chemistry are reversible, which means that they can proceed in both the forward and reverse directions. Reactants and products of a reversible reaction will eventually reach dynamic equilibrium, which refer to the condition whereby the rate of the forward reaction is equal to the rate of the reverse reaction. At dynamic equilibrium, the concentrations of products and reactants remain essentially unchanged until the system is disturbed from the outside.

Law of Mass Action

According to the Law of Mass Action, for a general reaction of the form:

$LaTeX: aA+bB\:\longleftrightarrow\:dD+eE$ $aA+bB\:\longleftrightarrow\:dD+eE$

the equilibrium ratio of products to reactants will always be the same at a given temperature and can be described by the Equilibrium Constant (K_c) expression below:

$LaTeX: K_c=\frac{\left[D\right]^d\left[E\right]^e}{\left[A\right]^a\left[B\right]^b}$ $K_c=\frac{\left[D\right]^d\left[E\right]^e}{\left[A\right]^a\left[B\right]^b}$

Note: Because the concentration of liquids and solids do not change significantly during the course of a reaction, they are not included in the equilibrium constant. Concentrations represented in the equation above MUST be concentrations at EQUILIBRIUM.

The equilibrium constant expression in terms of partial pressures (K_p) would be expressed as follows:

$LaTeX: K_p=\frac{\left(P_D\right)^d\left(P_E\right)^e}{\left(P_A\right)^a\left(P_B\right)^b}$ $K_p=\frac{\left(P_D\right)^d\left(P_E\right)^e}{\left(P_A\right)^a\left(P_B\right)^b}$

Note: All pressures MUST be in “atm”.

The following equation relates K_c and K_p:

$LaTeX: K_p=K_c\left(RT\right)^{\Delta n}$ $K_p=K_c\left(RT\right)^{\Delta n}$

Where R is the ideal gas law constant , T is the temperature in Kelvins, and ∆n is the change in the number of moles of gas in the reaction, which is given by the equation: $LaTeX: \Delta n=\left(moles\:of\:gaseous\:products\right)-\left(moles\:of\:gaseous\:reactants\right)$ $\Delta n=\left(moles\:of\:gaseous\:products\right)-\left(moles\:of\:gaseous\:reactants\right)$

What “K” tells us…

The numerical value of the equilibrium constant gives us useful information about the nature of the equilibrium including the following:

If $LaTeX: K\ll1$ $K\ll1$ , then the reverse reaction is favored and reactants predominate.
If $LaTeX: K\approx1$ $K\approx1$ , then neither the forward nor the reverse reaction are favored over one another. Reaction proceeds about halfway. Reactants and products are balanced about equally.
If $LaTeX: K\gg1$ $K\gg1$ , then the forward reaction is favored and products predominate.

The Effect of Altering or Adding Equations on the Value of K_eq

The relationship and numerical value of the equilibrium constant for a given reaction are based on how that reaction is expressed in a chemical equation. If a chemical equation is altered, the K_c or K_p will also be altered. The ways in which manipulating an equation can affect the value of the equilibrium constant are as follows:

1. If you reverse the equation, invert the equilibrium constant.

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2. If you multiply the coefficients in an equation by a number, raise the equilibrium constant to the power of that number. (Coefficients in the chemical reaction become exponents in the equilibrium constant.)

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3. If you add two or more individual chemical equations to obtain an overall equation, multiply the corresponding equilibrium constants by each other to obtain the overall equilibrium constant.

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Reaction Quotient (Q)

The reaction quotient is represented below. The concentrations in the reaction quotient can be representative of any time during a reaction, often when the reaction is not at equilibrium. If the concentrations are equilibrium concentrations, then LaTeX: Q_c=K_c $Q_c=K_c$ .

$LaTeX: Q_c=\frac{\left[D\right]^d\left[E\right]^e}{\left[A\right]^a\left[B\right]^b}$ $Q_c=\frac{\left[D\right]^d\left[E\right]^e}{\left[A\right]^a\left[B\right]^b}$

In terms of molar concentrations

$LaTeX: Q_p=\frac{\left(P_D\right)^d\left(P_E\right)^e}{\left(P_A\right)^a\left(P_B\right)^b}$ $Q_p=\frac{\left(P_D\right)^d\left(P_E\right)^e}{\left(P_A\right)^a\left(P_B\right)^b}$

In terms of pressures

Q vs. K Comparison

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Q is less than K, which means there is not enough of the products.

Therefore, equilibrium shifts to the RIGHT $LaTeX: \longrightarrow$ $\longrightarrow$

Q is equal to K, which means the system is already at equilibrium.

Therefore, there is NO SHIFT

Q is greater than K, which means there is too much of the products.

Therefore, equilibrium shifts to the LEFT $LaTeX: \longleftarrow$ $\longleftarrow$

Le Châtelier’s Principle

This fundamental principle states that whenever equilibrium is disturbed, the system will always shift in a direction that minimizes the disturbance. This law can be thought of in light of Newton’s 3^rd law, “for every action, there is an equal and opposite reaction.” For example, if product is added to an equilibrium mixture, the equilibrium will shift towards reactants (to the left) to lower the amount of product.

Please see the supplemental guide “ICE Charts and Equilibrium” linked HERE for more information on how to carry out various calculations related to equilibrium constants and concentrations.