Equilibrium Chemistry

Introduction to Equilibrium

• Equilibrium: In chemistry, equilibrium pertains to the extent of a chemical reaction and is fundamentally concerned with the concentrations of reactants and products over time. It represents a dynamic state where the forward and reverse reaction rates are equal, leading to no net change in concentrations of reactants and products. This also means no net change in macroscopic properties like pressure and temperature.

• Kinetics: Refers to the speed or rate of a chemical reaction.

• Equilibrium vs. Kinetics:

  • Kinetics answers the question, "How fast does the reaction proceed?"

  • Equilibrium answers the question, "To what extent does the reaction proceed?"

Learning Objectives

• Calculate equilibrium constants in terms of concentration and pressure.

• Predict the direction a reaction will progress based on equilibrium constants.

• Calculate concentrations of reactants and products at equilibrium.

• Predict how an equilibrated system responds to disturbances.

Dynamics of Chemical Reactions at Equilibrium

• At equilibrium, both reactants and products are present, and the reaction can be denoted as:
  $aA + bB \underset{k_b}{\rightleftharpoons} cC + dD$

• Here, $k<em>f$ is the rate constant for the forward reaction, and $k</em>b$ is the rate constant for the backward reaction. Crucially, at equilibrium, the rate of the forward reaction equals the rate of the backward reaction, meaning $k<em>f[A]^a[B]^b = k</em>b[C]^c[D]^d$. This implies that while reactions are still occurring, there is no net change in the concentrations of species.

• From the equality of forward and reverse rates, the equilibrium constant K can be derived as the ratio of the forward and reverse rate constants: $K = \frac{k<em>f}{k</em>b}$.

Concentration of Reactants and Products Over Time

• During a reaction, concentrations of reactants decrease and products increase until they reach constant values at equilibrium. At equilibrium, these concentrations remain steady, signifying that the system has reached a stable state where the net change is zero.

Concept of Equilibrium Constant (K)

• The equilibrium constant expression for a reaction:
  $K = \frac{[C]^c[D]^d}{[A]^a[B]^b}$

• Where $[X]$ denotes the molarity of species X at equilibrium.

• It is important to note that the value of K is specific to a given reaction at a particular temperature. Changing the temperature will change the value of K.

Meaning of K Values

• If K >> 1: Reaction favors products at equilibrium.

• If K << 1: Reaction favors reactants at equilibrium.

• Example: For the reaction
  $H<em>2 + I</em>2 \underset{k_f}{\rightleftharpoons} 2HI \; ; \; K \approx 50 \; \text{at} \; 500 ^\circ C$

Exercises and Problems

• Problem: Write the equilibrium constant expression for
  $2H<em>2O \underset{k</em>f}{\rightleftharpoons} 2H<em>2 + O</em>2$

• Answer: $K = \frac{[H<em>2]^2[O</em>2]}{[H_2O]^2}$

• K for this reaction is: $K \approx 10^{-20}$, indicating reactants are favored.

Relationships Between Different Equilibrium Constants

• For the reactions written in reverse:

  • Invert K:
    $K<em>2 = \frac{1}{K</em>1}$

• For reaction coefficients multiplied by a constant factor:

  • Raise K to that power:
    $K<em>2 = K</em>1^n$

• For multiple reactions added together:
  $K<em>3 = K</em>1 K_2$

Equilibrium Constants KP and KC

• KC and KP relations:
  $K<em>C = \frac{[C]^c[D]^d}{[A]^a[B]^b}$ $K</em>P = \frac{P<em>C^c P</em>D^d}{P<em>A^a P</em>B^b}$

• The relationship between them: $K<em>P = K</em>C (RT)^{\Delta n},$WHERE $\Delta n = c + d - (a + b)$

  • R is the ideal gas constant ($0.08206 \; \text{L atm/(mol K)}$) and T is the absolute temperature in Kelvin.

• Example:

  • For the reaction:
    $N<em>2(g) + 3H</em>2(g) \underset{k<em>f}{\rightleftharpoons} 2NH</em>3(g), \; K_C = 3.7 \times 10^8$

  • Calculation of $K_P$ can follow from the relation stated above.

Activity in Equilibrium

• Activity: Describes how effective a concentration is in contributing to the reaction. It is a dimensionless quantity that measures the 'effective' concentration or partial pressure of a species in a mixture, especially relevant in non-ideal solutions or at high concentrations where interparticle interactions are significant. Using activities instead of concentrations makes the equilibrium constant truly constant, regardless of the solution's properties, whereas using concentrations often leads to values that vary with ionic strength.

• It can be represented as:
  $a = \gamma C$

• Where $a$ is the activity, $\gamma$ is the activity coefficient, and $C$ is the concentration with the standard state concentration being 1 for ideal solutions. For ideal solutions, $\gamma = 1$, and activity equals concentration. For gases, the standard state is 1 atm; for solutes, it's 1 M; for pure solids and liquids, it's their pure form.

Role of Pure Liquids and Solids in Reactions

• In reactions involving pure liquids or solids, such as water or a solid reactant like C(s):

  • The concentration of the solvent or the solid is essentially constant and very high. Its "concentration" (or activity) is fundamentally incorporated into the value of the equilibrium constant itself. Thus, it does not appear explicitly in the equilibrium constant expression (K). This is because their activities are considered to be 1 by convention.

• Example:
  $H<em>2CO</em>3 \text{ formation} \Rightarrow CO<em>2(g) + H</em>2O(l) \to H<em>2CO</em>3(aq)$

Example Calculation of Equilibrium Constant

• For a given equilibrium state of a reaction H2 + I2 \rightleftharpoons 2HI $:

  • If the conditions are
    [H2] = 0.2 \; M, \; [I2] = 0.2 \; M, \; [HI] = 1.41 \; M $</p></li><li><p>Calculate K:<br>$ K = \frac{[HI]^2}{[H2][I2]} = \frac{(1.41)^2}{(0.2)(0.2)} \approx 50 $</p></li></ul></li></ul><h6>Changes in Reaction Conditions</h6><ul><li><p><strong>Le Chatelier's Principle</strong>: When a system at equilibrium is disturbed, it will shift to minimize the disturbance. Reaction shifts can be predicated based on:</p><ol><li><p>Changes in concentrations.</p></li><li><p>Changes in pressure (for gas-phase reactions).</p></li><li><p>Changes in temperature, especially concerning exothermic and endothermic reactions.</p></li></ol></li><li><p><strong>Example</strong>: For an endothermic reaction,$ C(s) + H2O(g) \rightleftharpoons CO(g) + H2(g) $(Heat is absorbed, so treat it as a reactant:$ C(s) + H2O(g) + \text{Heat} \rightleftharpoons CO(g) + H2(g) $$)

    • Actions to consider:

    • (a) Adding more C: Since C(s) is a pure solid, adding more of it has no effect on the equilibrium position, as its concentration/activity is constant.

    • (b) Removing H2O: H2O(g) is a reactant. Removing a reactant will cause the equilibrium to shift to the left (towards reactants) to produce more H2O, thereby trying to replenish the removed substance.

    • (c) Removing CO: CO(g) is a product. Removing a product will cause the equilibrium to shift to the right (towards products) to produce more CO, trying to replenish the removed substance.

    • (d) Decreasing the reaction volume: Decreasing volume increases total pressure. The system will shift to the side with fewer moles of gas to relieve this pressure. Reactant side has 1 mole of gas (H2O), product side has 2 moles of gas (CO + H2). So, the equilibrium will shift to the left (towards reactants).

    • (e) Adding H2: H2(g) is a product. Adding a product will cause the equilibrium to shift to the left (towards reactants) to consume the added H2.

    • (f) Adding a catalyst: A catalyst speeds up both the forward and reverse reactions equally. Therefore, adding a catalyst will not shift the equilibrium position but will allow the system to reach equilibrium faster.

    • (g) Increasing the temperature: This is an endothermic reaction, meaning heat is absorbed (it's a reactant). Increasing the temperature (adding heat) will cause the equilibrium to shift to the right (towards products) to absorb the excess heat.