13.4 Equilibrium Calculations

Equilibrium Concepts

Chemical Equilibrium and Reversible Reactions

Chemical equilibrium describes what happens when a reversible reaction reaches a steady state. The forward and reverse reactions don't stop; they just proceed at the same rate, so the concentrations of reactants and products hold constant over time.

Reversible reactions can proceed in both directions and are represented by a double arrow (⇌). This "dynamic equilibrium" is a common source of confusion: the reaction is still happening in both directions, but because the rates are equal, you see no net change in concentrations.

Changes in Equilibrium Concentrations

Le Chatelier's principle states that a system at equilibrium will respond to a stress by shifting to counteract that stress and re-establish equilibrium. The main types of stress are changes in concentration, pressure/volume, and temperature.

Concentration changes shift equilibrium toward the opposite side:

• Adding reactants or removing products shifts equilibrium to the right (toward products). For example, adding $NaOH$ to a reaction where it's a reactant pushes the system to make more products.
• Adding products or removing reactants shifts equilibrium to the left (toward reactants). For example, injecting extra $NH_3$ into a system where it's a product drives the reverse reaction.

Pressure and volume changes affect gaseous equilibria:

• Increasing pressure (decreasing volume) shifts equilibrium toward the side with fewer moles of gas. In the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$, that's the product side (2 moles vs. 4 moles).
• Decreasing pressure (increasing volume) shifts equilibrium toward the side with more moles of gas.

Temperature changes depend on whether the reaction is exothermic or endothermic:

• Increasing temperature favors the endothermic direction ($\Delta H > 0$), because the system absorbs the added heat.
• Decreasing temperature favors the exothermic direction ($\Delta H < 0$), because the system releases heat.

A catalyst speeds up both the forward and reverse reactions equally. It helps the system reach equilibrium faster but does not change the equilibrium position or the value of $K$.

Calculation of Equilibrium Constants

The equilibrium constant $K$ expresses the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to the power of its stoichiometric coefficient.

For a general reaction $aA + bB \rightleftharpoons cC + dD$:

$K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

A few rules to remember:

• Pure solids and pure liquids are not included in the $K$ expression. For example, in the decomposition of $CaCO_3(s)$, the solid calcium carbonate doesn't appear in the expression. Similarly, $H_2O(l)$ is excluded when it's the solvent.
• A large $K$ (much greater than 1) means products are heavily favored at equilibrium. A small $K$ (much less than 1) means reactants are favored.

The reaction quotient $Q$ uses the same formula as $K$ but with current concentrations rather than equilibrium concentrations. Comparing $Q$ to $K$ tells you which direction the reaction needs to shift:

• $Q < K$: Too few products relative to equilibrium. The reaction shifts right (toward products).
• $Q > K$: Too many products relative to equilibrium. The reaction shifts left (toward reactants).
• $Q = K$: The system is already at equilibrium.

The common ion effect is a specific application of Le Chatelier's principle. If you add an ion that's already present in the equilibrium system (say, adding $Na_2SO_4$ to a solution already containing $SO_4^{2-}$), the equilibrium shifts to reduce that ion's concentration.

ICE Table for Equilibrium Problems

An ICE table organizes the concentrations at three stages of a reaction: Initial (before any shift), Change (how much each species gains or loses), and Equilibrium (final concentrations). Here's how to use one, with the reaction $N_2O_4 \rightleftharpoons 2NO_2$ as an example.

Step 1: Write the balanced equation.

$N_2O_4 \rightleftharpoons 2NO_2$

Step 2: Set up the table with known initial concentrations.

Step 3: Substitute the equilibrium expressions into the $K$ expression.

$K = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(2x)^2}{0.500 - x}$

Step 4: Solve for $x$. If $K$ is given, you'll either rearrange into a quadratic or, when $x$ is very small compared to the initial concentration, use the small-x approximation (drop $x$ from the denominator to simplify the algebra). Always check that the approximation is valid: $x$ should be less than about 5% of the initial concentration.

Step 5: Calculate the equilibrium concentrations using your solved $x$ value. For this example, if $x = 0.0457$:

• $[N_2O_4]_e = 0.500 - 0.0457 = 0.454 \text{ M}$
• $[NO_2]_e = 2(0.0457) = 0.0914 \text{ M}$

Step 6: Verify your answer by plugging the equilibrium concentrations back into the $K$ expression. If the result matches the given $K$, you've solved it correctly.