Video Lecture Notes on Chemical Equilibria

Approach to Equilibrium

• Equilibrium constant (K) can be used to understand reaction outcomes.
• K varies significantly across reactions, making it hard to judge a "sensible" value.
• K indicates whether products or reactants are favored:
  • K >> 1: Favors products (reaction goes from left to right).
  • K << 1: Favors reactants (reverse reaction).
  • $K ≈ 1$: Mixture of reactants and products at equilibrium.
• Equilibrium is dynamic with balanced forward and reverse reactions.

Reaction Quotient (Q)

• For a generic reaction: $aA + bB \rightleftharpoons cC + dD$
• Q is the reaction quotient; same expression as K but applies at any time, not just equilibrium.
  • $Q = \frac{[C]^c[D]^d}{[A]^a[B]^b}$
• Q describes the reaction mixture at any given time.
• Q = K only at equilibrium.
• Calculate Q using initial concentrations to see how far from equilibrium.
• If Q < K: Reaction shifts right (towards products).
• If Q > K: Reaction shifts left (towards reactants).
• Q helps predict the direction a reaction will proceed.
• Important for chemical manufacturing to estimate product yield.

Isomerization Reaction Example

• Butane $\rightleftharpoons$ 2-methylpropane (isobutane).
• Equilibrium constant at 298 K: $K = \frac{[Isobutane]}{[Butane]} = 2.5$
• At equilibrium, the system should always be on a line with a slope of 2.5.
• Disturbing the equilibrium (e.g., adding butane) causes the system to adjust back to K.
• The system always drives back to the equilibrium position.
• Disturbances include adding/removing reactants or products.

Gas Phase Reaction Example

• Reaction: $H<em>2 + I</em>2 \rightleftharpoons 2HI$ at fixed temperature (K is fixed).

• Given: $K = 51$

• Initial conditions:

  • $[HI] = 2 \times 10^{-2} \, mol$ in 2L container
  • $[H_2] = 1 \times 10^{-2} \, mol$ in 2L container
  • $[I_2] = 3 \times 10^{-2} \, mol$ in 2L container

• Calculate initial concentrations:

  • $[HI] = \frac{2 \times 10^{-2}}{2} = 1 \times 10^{-2} M$
  • $[H_2] = \frac{1 \times 10^{-2}}{2} = 0.5 \times 10^{-2} M$
  • $[I_2] = \frac{3 \times 10^{-2}}{2} = 1.5 \times 10^{-2} M$

• $Q = \frac{[HI]^2}{[H<em>2][I</em>2]} = \frac{(1 \times 10^{-2})^2}{(0.5 \times 10^{-2})(1.5 \times 10^{-2})} = 1.3$

• Since Q = 1.3 << K = 51, the reaction will drive forward to make more product.

Calculating K

• Two scenarios:
  • Given all equilibrium concentrations: Substitute into the K expression.
  • Given initial concentrations and at least one equilibrium concentration: Use the ICE method.
    • I = Initial
    • C = Change
    • E = Equilibrium

ICE Method Steps

1. Balanced reaction.
2. Tabulate known concentrations (initial and equilibrium).
3. Calculate the change in concentration for a species where initial and final concentrations are known.
4. Use stoichiometry from the balanced reaction to back-calculate the changes for all other species.
5. Calculate equilibrium concentrations for all species.
6. Calculate K.

ICE Method Example

• Reaction: $H<em>2 + I</em>2 \rightleftharpoons 2HI$
• Initial: 1 mole of $H<em>2$ and 1 mole of $I</em>2$, and no $HI$ put into a 5L container.
• At equilibrium: $[HI] = 1.87 \times 10^{-3} M$

• Change in concentration: $[HI]$ increased by $1.87 \times 10^{-3}$. So the change $2x = 1.87 \times 10^{-3}$

• Using Stoichiometry: $H<em>2$ and $I</em>2$ decreased by half the amount of $HI$ formed (x)

  • $x = \frac{1.87 \times 10^{-3}}{2}$. Therefore $[H<em>2] = [I</em>2] = (0.2 - \frac{1.87 \times 10^{-3}}{2}) M$

• Equilibrium concentrations:

  • $[H_2] = 0.2 - \frac{1.87 \times 10^{-3}}{2} = 0.199 M$
  • $[I_2] = 0.2 - \frac{1.87 \times 10^{-3}}{2} = 0.199 M$
  • $[HI] = 1.87 \times 10^{-3} M$

• Calculate K

  • $K = \frac{[HI]^2}{[H<em>2][I</em>2]} = \frac{(1.87 \times 10^{-3})^2}{(0.199)(0.199)} = 51$

Using ICE Method to Calculate Equilibrium Concentration

• Reaction: $N<em>2(g) + 3H</em>2(g) \rightleftharpoons 2NH_3(g)$
• Given: $K_p$ (pressure equilibrium constant).
• At equilibrium, $P<em>{H</em>2} = 0.928$ bar and $P<em>{N</em>2} = 0.432$ bar.
• Find: $P<em>{NH</em>3}$ at equilibrium.

• $K<em>p = \frac{P</em>{NH<em>3}^2}{P</em>{N<em>2}P</em>{H_2}^3}$

• $x = \sqrt{K<em>p \cdot P</em>{N<em>2} \cdot P</em>{H_2}^3}$

• Solve for x (partial pressure of ammonia).

More ICE Examples

• Reaction: $H<em>2 + I</em>2 \rightleftharpoons 2HI$ in a 2L flask.
• Initial: Two moles of $H<em>2$, four moles of $I</em>2$, equilibrium constant value from previous example is 50.5 at a given temp.
• Find: Equilibrium concentrations of hydrogen, iodine, and hydrogen iodide.

• $K = \frac{[HI]^2}{[H<em>2][I</em>2]}$
• $50.5 = \frac{(2x)^2}{(1-x)(2-x)}$
• Rearrange the equation to get $46.5x^2 -151.5x + 101 = 0$. Now we have a quadratic equation; using the quadratic formula gives:
  • $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Le Chatelier's Principle

• A chemical equilibrium is a dynamic process.
• If a system is disturbed, it readjusts to keep the equilibrium constant constant (unless temperature changes).
• Applies to changes in concentration, pressure, or temperature.
• If a system is disturbed by changing temperature, pressure, or concentration, the system will shift its equilibrium to counteract the effect of the disturbance.
• Nature minimizes the impact of change through this principle.

Effects of Concentration

• Adding a reactant or product:
  *Increases the water water level
  *Equilibrium shifts to the right as it would if we added more reactants to a reversible chemical system.
• Removing a reactant or product:
  *Decreases the water level
  *Equilibrium shifts to the left as it does in a reversible chemical system when we reduce the quantity of the reactants.
• Adding reactant: System consumes it, makes more product, shifts to right, K stays same.
• Removing product: System makes more product, shifts to right, K stays same.
• Adding product: System makes more reactant, shifts to left, K stays same.
• Removing reactant: System makes more reactant, shifts to left, K stays same.

Changing Pressure or Volume

• Consider, $2A(g) \rightleftharpoons B(g)$
• Decrease volume (increase pressure): Favors fewer gas molecules (product B). Reaction adjusts to minimize disturbance.
• Increase volume (decrease pressure): Favors more gas molecules (reactant A). Reaction adjusts to minimize disturbance.
• Decrease Volume/Increase Pressure: Shift towards fewer gas molecules (look at balanced reaction), no change in K.
• Increase Volume/Decrease Pressure: Shift towards more gas molecules, no change in K.
• Adding an inert gas: No change in the partial pressure, so there's no change in the equilibrium position.

Temperature Effects

• K is a function of temperature.
• Reactions are exothermic or endothermic.
• Exothermic: Heat is a product.
• Endothermic: Heat is a reactant.
• Example: $N<em>2O</em>4(g) + Heat \rightleftharpoons 2NO_2(g)$ (endothermic).
• Cooling system: more $N<em>2O</em>4$
• Heating system: more $NO_2$
• Increase temperature: Heat is consumed; shift in the endothermic direction; K changes.
• Decrease temperature: Heat is generated; shift in the exothermic direction; K changes.