Chem 1412 - Chapter 13

Chapter 1: Introduction

• Calculating Equilibrium Constant:

  • In the previous scenario, equilibrium atmospheres of reaction species were provided, facilitating the calculation of the equilibrium constant.
  • Often, initial concentrations are given instead.

• Example:

  • Initial concentration of $H_2$: 0.2 atmospheres.
  • Initial concentration of $I_2$: 0.2 atmospheres.
  • To calculate the equilibrium constant, equilibrium concentrations of reactants are required.

• ICE Table:

  • ICE (Initial, Change, Equilibrium) table is used to organize and use the given information.
  • Initial Concentrations:
    • $H_2$: 0.20 atmospheres
    • $I_2$: 0.20 atmospheres
    • $HI$: 0 (initial concentration of products is typically zero)
  • Equilibrium Concentration of Product:
    • $HI$: 0.10 (given)

• Changes in Concentration:

  • $H<em>2$ and $I</em>2$ decrease by x amount.
  • $HI$ increases by 2x amount.
  • Given: $x = 0.05$

• Equilibrium Constant Calculation:

  • Equilibrium constant expression: $K = \frac{[HI]^2}{[H<em>2][I</em>2]}$

• Variations of the Problem:

  • Previous example: equilibrium concentrations were provided.
  • Current example: initials for reactants and equilibrium for the product are provided.
  • Other variations may include having different known equilibrium concentrations.

Chapter 2: Know The Equilibrium

• Determining Equilibrium Concentrations:

  • Equilibrium concentration of $H_2$ is known (0.1).
  • If x can be determined, equilibrium concentrations for $I_2$ and $HI$ can be calculated.
  • x equals the change in concentration for $H_2$.

• Using Only Initial Concentrations:

  • If only initial concentrations are given, the quadratic formula may be required to solve for x.
  • Equilibrium expression for only initials $K = \frac{[2x]^2}{[initial - x][initial - x]}$, this simplifies to $K = \frac{4x^2}{(initial - x)^2}$

• Determining Shift to Reach Equilibrium:

  • Equilibrium constant (K) is positive: $1.5 \times 10^{-2}$.
  • Current concentrations are given, and the objective is to determine if the system is at equilibrium or needs to shift (right or left) to reach equilibrium.

• Reaction Quotient (Q):

  • Since the system is not at equilibrium, the term "Q" (reaction quotient) is used instead of "K".
  • Calculate Q based on current concentrations and compare it to K to determine which way the reaction needs to shift.

• Mathematical Interpretation:

  • A greater numerator in the Q expression indicates a shift to the right (towards products).
  • If Q is greater than K, the numerator needs to decrease, indicating a shift to the left (towards reactants).
  • If the denominator is larger than the numerator, Q is less than one, favoring reactants.

• Example:

  • K is $1.5 \times 10^{-2}$, which equals 0.015.
  • Q is calculated based on current concentrations and found to be greater than K.
  • The reaction needs to shift to the left (towards reactants) to reach equilibrium.

Chapter 3: Additional Equilibrium Concentrations

• Molarity vs. Atmosphere:

  • Molarity is moles per liter.
  • If the volume is not one liter, divide the concentration by the volume before using it in the equilibrium constant expression.

• Volume Effects on Equilibrium Calculations:

  • If the volume is other than one, divide the initial concentrations by the volume before including them into the equilibrium constant expression.

• Example:

  • Equilibrium constant is 0.5.
  • Q is calculated as 1.3.
  • Since Q is less than K, the reaction needs to shift towards products.

• Calculating Kc:

  • If the reaction shifts so that three product moles minus two reactive moles are equal to one in the first power.

Chapter 4: The Right Side

• Le Chatelier's Principle:

  • If the equilibrium is stressed, it will shift in the opposite direction to offset the stress and reestablish equilibrium.

• Concentration Changes:

  • Increasing the concentration of a reactant will cause the equilibrium to shift to the right (towards products).
  • Increasing the concentration of a product will cause the equilibrium to shift to the left (towards reactants).
  • Decreasing the concentration of a reactant will cause the equilibrium to shift to the left.
  • Decreasing the concentration of a product will cause the equilibrium to shift to the right.

• Temperature Effects:

  • Endothermic Reaction:
  • Positive delta H (\Delta H > 0) indicates an endothermic reaction (heat is absorbed).
  • Heat can be considered a reactant in this case.
  • Decreasing the temperature will shift the equilibrium to the left.
  • Increasing the temperature will shift the equilibrium to the right.
  • Exothermic Reaction:
  • Negative delta H (\Delta H < 0) indicates an exothermic reaction.

• Pressure Effects:

  • Determine the number of gas moles on each side of the reaction.
  • Increasing the Pressure:
    • Shifts the equilibrium towards the side with fewer gas moles.
  • Decreasing the Pressure:
    • Shifts the equilibrium towards the side with more gas moles.
  • Equal Number of Gas Moles:
    • If both sides have the same number of gas moles, pressure changes have no effect on the equilibrium.

Chapter 5: Increase The Temperature

• Inverse Relationship Between Pressure and Volume:

  • As pressure increases, volume decreases (and vice versa).
  • Volume effects can be evaluated in terms of pressure (they are inverse).

• Decreasing $O_2$ Concentration:

  • If $O_2$ is decreased, the reaction will shift to the right.

• Increasing $SO_2$ Concentration:

  • If $SO_2$ is increased, the reaction will shift to the left.

• Increasing Temperature (Exothermic Reaction):

  • For exothermic reactions (heat is a product, \Delta H < 0), increasing the temperature will shift the equilibrium to the left.

• Pressure Effects (Continued):

  • Increasing pressure shifts the equilibrium to the side with fewer gas moles.

Chapter 6: Conclusion

• Equilibrium Shifts Summary:
  • Removing a reactant: shifts equilibrium towards reactants.
  • Increasing $NH_3$: shifts equilibrium to the left.
  • Increasing $N_2$: shifts equilibrium to the right.