AP Chem Unit 7 Review | Equilibrium in 10 Minutes!

Main ideas and concepts (AP Chemistry Unit 7: Equilibrium)

• Equilibrium involves reversible processes

  • Examples:
    • Boiling vs. condensing water
    • Precipitation vs. dissolving of precipitates
  • Reversible reactions are shown with a double-headed arrow to indicate both directions occur.

• What “equilibrium” really means

  • The forward reaction rate starts fast, then slows.
  • The reverse reaction rate starts slow, then increases.
  • Equilibrium is reached when:
    • rate(forward) = rate(reverse)
  • The reaction does not stop; concentrations appear constant because the net change is zero.

• How to predict which side equilibrium favors

  • Depends on the relative forward vs. reverse rates:
    • If forward is faster → more products than reactants (equilibrium shifts right)
    • If reverse is faster → more reactants than products (equilibrium shifts left)

• Equilibrium constants: (K_c) and (K_p)

  • Equilibrium constant expressions use the reaction’s stoichiometric coefficients.

  • (K_c) uses concentrations: [ K_c=\frac{[products]^{coeff}}{[reactants]^{coeff}} ]

  • (K_p) uses partial pressures: [ K_p=\frac{(P_{products})^{coeff}}{(P_{reactants})^{coeff}} ]

  • Solids and pure liquids are omitted from (K) calculations.

  • Reaction quotient (Q) is written the same way as (K), but uses current (not necessarily equilibrium) concentrations/pressures.

• Temperature dependence of equilibrium constants

  • All equilibrium constants depend on temperature.
  • Changing temperature is the only way to change the value of (K).

• Interpreting the magnitude of (K)

  • Large (K) → products favored → equilibrium lies to the right
  • Small (K) → reactants favored → equilibrium lies to the left
  • Larger (K) generally means a greater products-to-reactants ratio.

• How (K) changes with manipulating reactions

  • Flip a reaction → (K) becomes the reciprocal
  • Multiply all coefficients by a factor (e.g., double them) → (K) becomes that factor’s power
    • Example: doubled coefficients → (K^2)
  • Add two reactions → the new reaction’s (K) is the product of the individual (K)’s: [ K_{new}=K_1\cdot K_2 ]

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Methodology / step-by-step instructions (ICE box, (Q) vs (K), solubility)

A) Using an ICE box to find equilibrium concentrations

• Use an ICE table (Initial, Change, Equilibrium):
  • I (Initial):
    • Write the starting concentrations of all species.
  • C (Change):
    • Express changes using a variable (commonly (X)).
    • Follow stoichiometry:
      • If products form with reaction extent (X), reactants decrease by amounts matching their coefficients.
    • In the NO example described:
      • [NO] decreases by (2X)
      • [N(_2)] increases by (X) (1-to-2 ratio: N(_2) rises by (X) while NO drops by (2X))
      • [O(_2)] increases by (X)
  • E (Equilibrium):
    • Add initial and change terms to get equilibrium concentrations.
• Plug equilibrium concentrations into the appropriate equilibrium constant expression ((K_c) or (K_p)).
• Solve algebraically for (X).
• Compute equilibrium concentrations for each species using the solved (X).

B) Estimating partial pressure from mole fraction (at equilibrium)

• From a particle/molecule diagram:

  • Determine mole fraction:
    • Example: hydrogen is 6 out of 10 → mole fraction (=6/10=0.6)

  • Use: [ P_i=(\text{mole fraction of } i)\times P_{total} ]

  • Example:

    • Total pressure = 2.0 atm
    • Hydrogen partial pressure = (0.6 \times 2.0 = 1.2) atm

C) Predicting shift direction using (Q) and (K) (when disturbed)

• Compute reaction quotient (Q) from current concentrations/pressures.
• Compare (Q) to (K):
  • If (Q > K):
    • reaction proceeds left (forms more reactants, consumes products)
  • If (Q < K):
    • reaction proceeds right (produces more products)
• A quick visualization mnemonic was mentioned: turn the symbol into a “Pac-man” to visualize direction of progress.

D) Effects of disturbances (Le Châtelier’s principle ideas covered)

• Adding a substance:
  • System reacts to reduce the added substance’s concentration.
  • The opposite side increases.
• Removing a substance:
  • System reacts so the removed side increases.
• Changing volume (gas systems):
  • Decrease volume → shifts toward fewer moles of gas
  • Increase volume → shifts toward more moles of gas
• Temperature changes:
  • Exothermic reaction (heat acts like a product):
    • Adding heat / increasing temperature → shifts toward reactants
    • Lowering temperature → shifts toward products
  • Endothermic reaction:
    • Opposite trend.

E) Solubility product constant (K_{sp}) and common-ion effect (ionic equilibrium)

• General concept

  • For dissolution/dissociation of an ionic compound, the equilibrium constant is (K_{sp}).

• Example: lead(II) bromide

  • Dissociation: [ \text{PbBr}_2(s)\rightleftharpoons \text{Pb}^{2+}+2\text{Br}^- ]

  • Let molar solubility be (X):

    • ([\text{Pb}^{2+}]=X)
    • ([\text{Br}^-]=2X)
  • Substitute into (K_{sp}) expression and solve for (X).
  • Result stated:
    • molar solubility = 0.012 M (convert to g/L if needed).

• Common-ion effect scenario

  • Instead dissolve in 0.10 M NaBr:
    • (\text{Br}^-) is already high.
  • ICE box setup described:
    • Initial ([\text{Pb}^{2+}] = 0)
    • Initial ([\text{Br}^-] = 0.10)
    • Equilibrium:
      • ([\text{Pb}^{2+}] = X)
      • ([\text{Br}^-] = 0.10 + 2X)
  • Because (K_{sp}) is very small, assume (2X) is negligible relative to 0.10:
    • ([\text{Br}^-]\approx 0.10)
  • Solve for (X) → result stated:
    • (X = 6.6 \times 10^{-4}) M
  • Conclusion: adding a common ion reduces solubility.

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Speakers / sources featured

• Jeremy Krug (speaking; creator/instructor of the video and the “10 minute review” series)
• UltimateReviewPacket.com (referenced as a source for practice materials; not a speaking participant)