Lecture 25 and 26 Notes: Equilibrium

Equilibrium Concepts

• Chemical reactions reach equilibrium with a balance between reactants and products.
• This is a dynamic process where the rate of the forward reaction equals the rate of the reverse reaction.
• Reactant and product concentrations remain constant at equilibrium, though interchange occurs.

Equilibrium Constant

• The equilibrium concentrations can be characterized using the equilibrium constant.

• It's calculated as the ratio of product to reactant concentrations at equilibrium.

  K = \frac{{\text{{[Products]}}}}{{\text{{[Reactants]}}}} where equilibrium concentrations are represented as K = \frac{{\text{{[Products]eq]}}}}{{\text{{[Reactants]eq]}}}} and the powers correspond to stoichiometric coefficients.

• K >> 1 : Equilibrium favors products.

• K << 1 : Equilibrium favors reactants.

• K ≈ 1 : Even balance between reactants and products.

• Changing stoichiometric coefficients alters the value of K. This can be confusing because even with differing coefficients, if the initial amount of the stuff is the same the concentrations at equilibrium are the same. Doubling the coefficients will change the value of K.

Equilibrium Constant Expressions

• Equilibrium constants can be expressed in terms of concentrations or partial pressures.
• Using partial pressures is common for gas-phase reactions, while concentrations are used in solutions.
• Equilibrium constants will have different values whether they are expressed as partial pressures or concentrations, but the concentrations at equilibrium will be the same. For a gas phase reaction, it is typical to write an equilibrium constant in terms of the partial pressures of all of the species at equilibrium. If the reaction is in solution, it makes more sense to express it in terms of concentration (moles per liter).
• For reactions involving solids and liquids, mole fractions are used.
• Concentrations of pure liquids or solids aren't included, as their mole fractions are considered to be one.
• If water is the solvent and also a product, it's not included in the equilibrium constant.

Types of Equilibrium Constants

• Subscripts denote the units used in the equilibrium constant. Subscripts matter for specifying the units of the Equilibrium Constant
  • K_c: Equilibrium constant in units of concentration.
  • K_p: Equilibrium constant in units of pressure.
  • K_a: Acid dissociation constant.
  • K_b: Base dissociation constant.
  • K_w: Autoionization constant of water (water splitting to give a proton and a hydroxide).
  • K_{sp}: Solubility product (for a salt dissolving in solution).
• For K_w, the reactant concentration isn't included because we're in an aqueous solution, which means the mole fraction of water is very close to one.

Calculations with Equilibrium Constants

• If concentrations at equilibrium are known, K can be calculated.
• If K and initial concentrations are known, equilibrium concentrations can be calculated.

Worksheet Question

• For N2(g) + O2(g) \rightleftharpoons 2NO(g), K_p = 2.5 \times 10^{-3}. It's presumed the units are in pressure if not told otherwise.

• If asked for the units of an equilibrium you must consider the stoichiometry of a reaction

  Kp = \frac{{(P{NO})^2}}{{(P{N2}) \cdot (P{O2})}}

• Units: \frac{{\text{{atm}}^2}}{{\text{{atm}} \cdot \text{{atm}}}} = \text{{unitless}} . In this case, the units cancel out because there are two products and two reactants

• Equilibrium constant expressions are based on the ratio of product to reactant concentrations or pressures.

Equilibrium Partial Pressure Calculation

• Given: P{N2} = 0.78 atm, P{O2} = 0.21 atm

  P{NO} = \sqrt{Kp \cdot P{N2} \cdot P{O2}}
  P_{NO} = \sqrt{(2.5 \times 10^{-3}) \cdot (0.78) \cdot (0.21)} = 0.020 \text{{ atm}}

Heterogeneous Systems

• Deal with equilibrium constants for heterogeneous systems where we don't just have one phase present.
• Just gases feature in the equilibrium constant, and we write them as partial pressures.

Equilibrium Constant Examples

• C(s) + H2O(g) \rightleftharpoons CO(g) + H2(g)
• The equilibrium constant is Kp = \frac{{P{CO} \cdot P{H2}}}{{P{H2O}}}. Carbon is excluded because it is a pure solid whose mole fraction is equal to one.
• CaCO3(s) \rightleftharpoons CaO(s) + CO2(g)
• The equilibrium constant is Kp = P{CO_2}. Pure solids are excluded from the equilibrium constant.
• PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)
• The equilibrium constant is K_{sp} = [Pb^{2+}] \cdot [I^-]^2. Pure solids are excluded, and the equation is written in terms of concentrations.
• CO2(s) \rightleftharpoons CO2(g)
• The equilibrium constant is Kp = P{CO_2}. We don't consider solids. There is only one species in this equilibrium constant.
• NH3(aq) \rightleftharpoons NH3(g)
• The equilibrium constant is Kc = \frac{{[NH3(g)]}}{{[NH_3(aq)]}}. You must label your phases here because it's the same species in different phases. Don't mix units, write in terms of concentrations because all of the species are not in the gas phase.
• If all that is left after eliminating pure solids and liquids are gases, then you write it as Kp. If there are solution species present, then you write it as Kc or K_{sp} in terms of concentrations.

Reaction Quotient

• The reaction quotient (Q) is written the same way as the equilibrium constant but at any point in time whether or not the reaction is at equilibrium. Q is only written only in terms of equilibrium concentrations.
• Q = K: The system is at equilibrium.
• Q < K: More reactants than equilibrium. The system will shift towards the products.
• Q > K: More products than equilibrium. The system will shift towards the reactants.

Reaction Quotient Example

• For an organic reaction with K = 1.7, two solutions are prepared with different initial concentrations. If this ratio is smaller than the equilibrium constant we've got to shift in a direction that converts these to that. So if that is smaller than that, it is going to shift towards the products. If we plug these values into the expression we get the number seven, which is much bigger than the equilibrium constant and we've got to have a shift from products to reactants.

Worksheet Question

• Decomposition Reaction: CaCO3(s) \rightleftharpoons CaO(s) + CO2(g). At 900 degrees, K_p = 1.04.
• The reaction quotient (Q) is P{CO2}. If partial pressure is less than K_p, the reaction shifts towards the products.
• Reaction: If there is only calcium carbonate present nothing will happen because there is no calcium oxide, so the reaction cannot take place. So we can calculate the reaction quotient, and we can predict which way something will shift but that depends on what reactants are present.

Calculating Equilibrium Concentrations

• Given K and initial concentrations, you can determine all species' concentrations at equilibrium.
• Isomerization Reaction: cis-2-butene ⇌ trans-2-butene, K = 1.4. The concentrations are particularly simple because they inter-convert one to one.
• If the initial concentration of cis is 0.1 Molar the initial concentration of trans is 0 Molar, what are the concentrations at equilibrium. If there is no product we know we have to shift that way to get to equilibrium. With the presence of both determine where you have to shift.
• If we think about what's going to happen here, this concentration decreases by some amount, which this concentration has to increase by some amount of an unknown. When this decreases by x, this has to increase by x.

Setting Up the Equation

• K = \frac{{[\text{{trans}}]}}{{[\text{{cis}}]}} = \frac{{x}}{{0.1 - x}}
• Rearranging the equation: 0.14 - 1.4x = x \implies x = \frac{{0.14}}{{2.4}} \approx 0.058 Molar.
• Concentration of trans 2-butene is x, which is approximately 0.058 Molar.
• Concentration of cis 2-butene is: 0.1 - x \approx 0.042 Molar.

Reaction with Multiple Species

• H2(g) + I2(g) \rightleftharpoons 2HI(g)
• Equilibrium constant is Kc = 64 (in terms of concentrations). We write it in terms of concentration, or else one would assume it's Kp.

Systematic Approach (ICE Method)

• ICE method
• I = Initial concentrations
• C = Change in concentration
• E = Equilibrium concentrations

ICE Table Setup

• Initially, 0.2 moles of hydrogen and 0.2 moles of iodine were added to a 2-liter flask.

Calculating Initial Concentrations

• Number of moles divided by the volume to calculate concentrations.

ICE Table Example

• If the shift in this is minus x, then the shift in this has to also be minus x because it's a 1-1 stoichiometry. That relates the shifts right, this is one to one minus one and minus one two and two. We can consume all of our reactants which produce our products which are proportional to the stoichiometry of the reaction.

Setting Up the Equation

• Expression for the equilibrium constant is

  Kc = \frac{{[HI]^2}}{{[H2] \cdot [I_2]}} = 64

Substituting to Solve for x

• K_c = \frac{{(2x)^2}}{{(0.1 - x)^2}} = 64

Solving for x

• \sqrt{64} = \frac{{2x}}{{0.1 - x}} \implies x \approx 0.08

Equilibrium Concentrations

• Calculate the equilibrium concentrations.
• Equilibrium concentrations for hydrogen and iodine are: 0.1 - x \approx 0.02 Molar.
• Equilibrium concentration for HI is: 2x \approx 0.16 Molar.
• ICE method gives some practice, so given the equilibrium constant, we can set up an equation and solve it for x, and then work back to calculate the equilibrium concentrations.