15 - Chemical Equilibrium

Chemical Equilibrium

Assumptions So Far

• Reactions proceed in only one direction.

• Reactions are completed when the limiting reactant is completely consumed.

• However, reactions are actually reversible to some extent.

  • Example: $H<em>2(g) + I</em>2(g) \rightleftharpoons 2HI(g)$

Dynamic Equilibrium

• The condition where the rate of the forward reaction equals the rate of the reverse reaction.

  • Concentrations of products and reactants don't change at the macroscopic level, but the reaction is still occurring on a molecular level.
  • Both products and reactants are present to some extent.

• Equilibrium is NOT:

  • A complete stop of the reaction, as it's still occurring on a molecular level.
  • A state where all concentrations are the same (concentrations are usually different).

Equilibrium Constant (K)

• The ratio at equilibrium of the concentrations of products raised to their stoichiometric coefficients divided by the concentration of reactants raised to their stoichiometric coefficients.

• Gives relative concentrations of products and reactants.

  • General equation: $aA + bB \rightleftharpoons cC + dD$

  • $K<em>c = \frac{[C]^c</em>{eq} [D]^d<em>{eq}}{[A]^a</em>{eq} [B]^b_{eq}}$

    • [A]eq, [B]eq, [C]eq, [D]eq = concentrations of reactants/products at equilibrium

  • The value of K for a reaction reflects the extent to which that reaction goes in the forward direction.

    • Numerically large values of K:

      • The forward reaction is strongly favored.
      • Products are favored over reactants.
      • Relatively more products will be present.

    • Numerically small values of K:

      • The reverse reaction is favored.
      • Reactants will be seen more.

    • Intermediate values of K:

      • Both reactants and products will be present in the equilibrium mixture.
      • No preference in which way is favored.

  • K value depends on stoichiometry and temperature.

  • For a given temperature, the equilibrium constant will always be the same regardless of initial concentrations of reactants and/or products.

Changing Equilibrium Conditions

1. Changing the Direction of the Reaction:

   • Inverts the value of K.

   • Example: $2N<em>2O</em>5(g) \rightleftharpoons 4NO<em>2(g) + O</em>2(g)$

     • $K<em>{forward} = \frac{[NO</em>2]^4 [O<em>2]}{[N</em>2O_5]^2}$
     • $K<em>{reverse} = \frac{[N</em>2O<em>5]^2}{[NO</em>2]^4 [O<em>2]} = \frac{1}{K</em>{forward}}$

2. Multiplying Coefficients:

   • Multiplying the coefficients of a reaction by a certain factor raises K to that factor.

   • Example: $4N<em>2O</em>5(g) \rightleftharpoons 8NO<em>2(g) + 2O</em>2(g)$

     • $K<em>2 = K</em>1^2$

3. Adding Reactions:

   • When adding two or more individual reactions to obtain an overall equation, the corresponding equilibrium constants are multiplied to obtain the equilibrium constant for the overall reaction.

   • Example:

     • Reaction 1: $A \rightleftharpoons 2B$ $K_1$
     • Reaction 2: $2B \rightleftharpoons 3C$ $K_2$
     • Overall: $A \rightleftharpoons 3C$ $K<em>{overall} = K</em>1 \times K_2$

Equilibrium Constant in Terms of Partial Pressures

• For reactions in the gas phase, the equilibrium constant can be expressed in terms of partial pressures of reactants and products at equilibrium.

• Use the ideal gas law: $PV = nRT$

• The concentration of a gas in a mixture is proportional to its partial pressure under constant temperature.

• $K<em>p = \frac{(P</em>C)^c (P<em>D)^d}{(P</em>A)^a (P_B)^b}$

  • Convert partial pressures into atm.

Relation between $K<em>p$ and $K</em>c$

• $K<em>p = K</em>c (RT)^{\Delta n}$

  • R = 0.08206 L atm / (mol K)
  • T = temperature in Kelvin
  • $\Delta n$ = (moles of gas products) - (moles of gas reactants)

• Example:

  • $2NO<em>2(g) + O</em>2(g) \rightleftharpoons 2NO_2(g)$
  • $K_c = 4.67 \times 10^{13}$ at 25°C
  • $\Delta n = 2 - (2+1) = -1$
  • $K_p = 4.67 \times 10^{13} (0.08206 \times 298)^{-1} = 1.91 \times 10^{12}$

Heterogeneous Equilibrium

• Treating pure solids and pure liquids in an equilibrium reaction.

  • Pure solids and pure liquids don't appear in the equilibrium expression.

  • As long as enough pure solid or pure liquid is present, the reaction can take place, but equilibrium is NOT affected.

    • The concentration of pure solid or liquid remains constant.
    • Amount changes, but volume changes proportionally.

  • Example: $CO_2(g) + C(s) \rightleftharpoons 2CO(g)$

    • $K = \frac{[CO]^2}{[CO_2]}$

Calculating Equilibrium Amounts of Products & Reactants

• Must determine the limiting reactant and use it to calculate the amount of products formed and the amount of excess reagent left over.
• For equilibrium reactions, the ICE (Initial, Change, Equilibrium) table must be used.

ICE Table

• The relative changes in concentrations of the reactants and products are determined using stoichiometry.

• Example: $H<em>2(g) + I</em>2(g) \rightleftharpoons 2HI(g)$

  	$H_2(g)$	$I_2(g)$	$2HI(g)$
  Initial (I)	0.952 M	0.952 M	0 M
  Change (C)	-x	-x	+2x
  Equil (E)	0.952-x	0.952-x	2x

• Finding x:

  • Substitute equilibrium concentrations into
    $K<em>c = \frac{[HI]^2}{[H</em>2][I_2]} = \frac{(2x)^2}{(0.952-x)^2} = 54.3$
  • Solve for x:
    $x = 0.749$
  • $[H<em>2] = [I</em>2] = 0.952 - x = 0.952 - 0.749 = 0.203 M$
  • $[HI] = 2x = 2(0.749) = 1.50 M$

• What about the reverse reaction?

  • $2HI(g) \rightleftharpoons H<em>2(g) + I</em>2(g)$

    • The final concentrations and x will be the same, reversal doesn't matter.

The Reaction Quotient (Q)

• How to determine which direction a reaction must proceed to reach equilibrium.

Reaching Equilibrium

• The reaction quotient (Q) is used to determine if a reaction is at equilibrium.

• It is calculated by using the same formula as the equilibrium constant (K) but with initial concentrations instead of equilibrium concentrations.

  • $Q = \frac{[Products]<em>{initial}}{[Reactants]</em>{initial}}$

• Comparing Q to K:

  • If Q = K: The system is already in equilibrium.
    • If Q < K: The system needs more products; the reaction will go to the right (forward direction).
    • If Q > K: The system needs less products; the reaction will go to the left (reverse direction).

• Example:

  • $H<em>2(g) + I</em>2(g) \rightleftharpoons 2HI(g)$ $K = 54.3$ @ 698 K

  • Initial concentrations: [H2] = 0.250 M, [I2] = 0.185 M, [HI] = 2.40 M

    • $Q<em>c = \frac{[HI]^2</em>{initial}}{[H<em>2]</em>{initial} [I<em>2]</em>{initial}} = \frac{(2.40)^2}{(0.250)(0.185)} = 125$
    • Since Q > K, the reaction will go to the left.

Finding Equilibrium Concentrations Using Q

• Use Q to determine the direction of the shift towards equilibrium.

• Use an ICE table to find equilibrium concentrations.

• Example (continued):
  $H<em>2(g) + I</em>2(g) \rightleftharpoons 2HI(g)$

  	H2	I2	2HI
  Initial (I)	0.250	0.185	2.40
  Change (C)	+x	+x	-2x
  Equil (E)	0.250+x	0.185+x	2.40-2x
  *Setting up the equilibrium expression and solving for x:
  $K = \frac{[HI]^2}{[H<em>2][I</em>2]} = \frac{(2.40 - 2x)^2}{(0.250 + x)(0.185 + x)} = 54.3$
  Solving for x, we find x = 0.0865.
  *Substituting x back into the equilibrium expressions:
  $[H_2] = 0.250 + 0.0865 = 0.337 M$
  $[I_2] = 0.185 + 0.0865 = 0.2715 M$
  $[HI] = 2.40 - 2(0.0865) = 2.23 M$

Mathematical Approximations

Small-K Approximation

• When K is very small, the change in concentration (x) is also very small.

• We can often neglect x when it is added to or subtracted from a much larger number in the equilibrium expression.

• Example: $2NO<em>2(g) \rightleftharpoons 2NO(g) + O</em>2(g)$ $K = 1.6 \times 10^{-10}$ at 25°C

  • Initial concentration: [NO2] = 0.500 M

  	2NO2	2NO	O2
  Initial (I)	0.500	0	0
  Change (C)	-2x	+2x	+x
  Equil (E)	0.500-2x	2x	x

  • $K = \frac{[NO]^2 [O<em>2]}{[NO</em>2]^2} = \frac{(2x)^2 (x)}{(0.500 - 2x)^2} = 1.6 \times 10^{-10}$

  • Since K is very small, we can assume 0.500 - 2x ≈ 0.500

    • $\frac{4x^3}{(0.500)^2} = 1.6 \times 10^{-10}$
    • Solving for x: x = 2.2 x 10-4 M = [O2]

  • Validating the approximation:

    • 2x should be less than 5% of 0.500
    • $2x = 4.4 \times 10^{-4}$
    • $\frac{4.4 \times 10^{-4}}{0.500} \times 100 = 0.088$

  • If the approximation is not valid, then a 3rd degree polynomial equation must be solved.

Large K Approximation

• When K is very large, we assume the reaction goes to completion.
  Example:
  $N<em>2(g) + 3H</em>2(g) \rightleftharpoons 2NH_3(g)$

  • $K_c = 2.3 \times 10^8$

• Initial amounts: 2.00 mol N2 and 6.00 mol H2

• Since Kc is essentially large we can treat as going to completion, use stoichiometry to calculate products / reactant

• Calculate amount of limiting reactant remaining at equilibrium; solve equilibrium expression

Le Chatelier's Principle

• When a chemical system at equilibrium is subjected to a stress, the system will shift in a direction to relieve the stress.

Types of Stress

• Changing the concentration of a reactant or product.

• Changing the pressure within the reaction container:

  • Adding or removing a gaseous reactant or product.
  • Adding an inert gas.
  • Changing the volume of the reaction container.

• Changing the temperature of the reaction.

• Changing the concentration:

  • When a species is added to a chemical system, a reaction will shift in the direction that consumes the added species.

  • When a species is removed, the reaction will shift in the direction to produce the removed species.

  • Example: $2N<em>2O</em>5(g) \rightleftharpoons 4NO<em>2(g) + O</em>2(g)$

    • Add N2O5 or remove NO2: Reaction shifts to the right.
    • Remove N2O5 or add NO2: Reaction shifts to the left.

Impact of Solids and Liquids on Equilibrium

• Only species that appear in the equilibrium expression impact equilibrium!

Solids

• Adding or removing C(s) from the reaction at equilibrium will not impact equilibrium.

  • Example: $2CO(g) \rightleftharpoons CO_2(g) + C(s)$

Effect of Pressure change on Equilibrium

• Adding or removing a gaseous product or reactant.

  • Changes the partial pressures of the species.
  • Species must appear in the equilibrium expression.

• Adding an inert gas:

  • Adding He(g) to $N<em>2(g) + 3H</em>2(g) \rightleftharpoons 2NH_3(g)$
  • Since an inert gas doesn't appear in the equilibrium expression, adding it won't affect equilibrium.

• Changing the volume of the reaction container:
  $N<em>2(g) + 3H</em>2(g) \rightleftharpoons 2NH_3(g)$

  • The reaction goes in the direction that best utilizes the available space.

  • Increase in volume: reaction toward products

    • Reaction proceeds to generate more gas