Chemical Equilibrium

Relationship Between K and Kp [6.3]

• For gas reactions at equilibrium:
  aA{(g)} + bB{(g)} \rightleftharpoons eE{(g)} + dD{(g)}

  • Where A, B, E, and D are gases, and a, b, e, and d are their respective stoichiometric coefficients.

• K_p: Equilibrium constant in terms of pressure.

• Partial pressures (e.g., PA, PB) remain constant at equilibrium.

• Kp = \frac{(PE)^e (PD)^d}{(PA)^a (P_B)^b}

• The relationship between K_p and K (equilibrium constant in terms of concentration) is not always equal and needs conversion.

• Recall:

  • Concentration: [A] = \frac{nA}{V} = CA (where nA is moles of A, V is volume, and CA is concentration).

• Ideal Gas Law: PV = nRT \implies P = \frac{n}{V}RT = CRT

• Therefore, P = C \cdot R \cdot T

• Kp = \frac{(CE \cdot R \cdot T)^e (CD \cdot R \cdot T)^d}{(CA \cdot R \cdot T)^a (C_B \cdot R \cdot T)^b}

• Simplifying the expression:
  K_p = \frac{[E]^e [D]^d}{[A]^a [B]^b} (RT)^{(e+d)-(a+b)}

• K_p = K(RT)^{\Delta n}

  • \Delta n = (e + d) - (a + b) (change in the number of moles of gas).

Why Unitless? Chemical Activity! 6.4

• Both K and K_p appear to have units based on their expressions, but they are actually unitless.

• Equilibrium constants are based on activities, not concentrations or pressures.

• Activity: An effective concentration or pressure.

• a_A = \frac{[A]}{1M}: Activity of A is the concentration of A relative to a reference state of 1 M. Units cancel out.

• aA = \frac{PA}{1 atm}: Activity of A is the pressure of A relative to a reference state of 1 atm. Units cancel out.

• Therefore, K and K_p are unitless because they are based on activities where units cancel out with the reference state.

Example Problem: Calculating Kc and Kp

• Reaction: N2O4(g) \rightleftharpoons 2NO_2(g) at 25^\circ C

• Equilibrium concentrations: [N2O4] = 0.00452 M and [NO_2] = 0.0310 M

• Calculate K_c:

  • Equilibrium expression: Kc = \frac{[NO2]^2}{[N2O4]}

• K_c = \frac{(0.0310)^2}{(0.00452)} = 0.213

• Calculate K_p:

  • Kp = Kc(RT)^{\Delta n}
  • \Delta n = 2 - 1 = 1

• K_p = (0.213) \cdot (0.08206 \frac{L \cdot atm}{mol \cdot K} \cdot 298.15 K) = 5.21

Heterogeneous Chemical Equilibria [6.5]

• Heterogeneous chemical equilibria: Equilibria involving more than one phase (solid, liquid, gas).

• Example: CoCl3(s) + 3H2O(g) \rightleftharpoons CoCl3 \cdot 3H2O(s)

• Equilibrium Constant Expression:
  K = \frac{a{CoCl3 \cdot 3H2O}}{a{CoCl3} \cdot (a{H_2O})^3}

• The activity of a pure solid or pure liquid is equal to 1 (its reference state).

  • a{CoCl3} = \frac{[CoCl3]}{[CoCl3]_{reference}} = 1

• Therefore, the equilibrium expression simplifies to:
  K = \frac{1}{(a{H2O})^3} = \frac{1}{[H_2O]^3}

• Key takeaway: Do not include pure solids and pure liquids in the equilibrium expression.

Example Problem: Calculating K_p for Heterogeneous Equilibrium

• Reaction: CaCO3(s) \rightleftharpoons CaO(s) + CO2(g)

• Kp = P{CO_2}

• Initial conditions: 20.0 g of CaCO_3(s) (MM = 100.09 amu) in a 10.0 L container heated to 800^\circ C.

• At equilibrium: 6.88 g of CaCO_3(s) remains.

• Calculate K_p:

  • Set up an ICE table (Initial, Change, Equilibrium) in terms of moles:
    CaCO3(s) \rightleftharpoons CaO(s) + CO2(g)

    • I: n_i 0 0
    • C: -x +x +x
    • E: n_i - x x x

  • n_i = \frac{20.0 g}{100.09 g/mol} = 0.1998 mol

  • n_e = \frac{6.88 g}{100.09 g/mol} = 0.0687 mol

  • x = ni - ne = 0.1998 mol - 0.0687 mol = 0.131 mol

  • Calculate the pressure of CO2 using the Ideal Gas Law: P{CO2} = \frac{n{CO2} \cdot R \cdot T}{V} P{CO_2} = \frac{(0.131 mol) (0.08206 \frac{L \cdot atm}{mol \cdot K}) (1073.15 K)}{10.0 L} = 1.15 atm

• Therefore, K_p = 1.15

General Approach for Equilibrium Problems Involving Systems Not at Equilibrium

• Consider the reaction: A + B \rightleftharpoons 2C with equilibrium constant K

• To determine the direction a reaction will shift to reach equilibrium, use an ICE table:

  • I: [A]i [B]i [C]_i
  • C: \pm x \pm x \pm 2x
  • E: [A]i \pm x [B]i \pm x [C]_i \pm 2x

• Calculate the reaction quotient, Q:

• Q = \frac{([C]i \pm 2x)^2}{([A]i \pm x)([B]_i \pm x)}

  • Determine the direction of the shift based on comparing Q to K

• After solving for x, plug the value of x back into the 'E' row of the ICE table to find the equilibrium concentrations.

• Solving for x might involve rearrangement or using the quadratic equation.

The Reaction Quotient

• The size of K and the time required to reach equilibrium are not directly related.

• Reaction quotient (Q) is used to determine the direction of the move toward equilibrium.

• Consider the reaction: A \rightleftharpoons 2B with K = 2.0

• Q = \frac{[B]i^2}{[A]i}