Chemical Equilibrium

Relationship Between K and Kp [6.3]

• For gas reactions at equilibrium:
  $aA<em>{(g)} + bB</em>{(g)} \rightleftharpoons eE<em>{(g)} + dD</em>{(g)}$

  • 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., $P<em>A, P</em>B$) remain constant at equilibrium.

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

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

• Recall:

  • Concentration: $[A] = \frac{n<em>A}{V} = C</em>A$ (where $n<em>A$ is moles of A, V is volume, and $C</em>A$ is concentration).

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

• Therefore, $P = C \cdot R \cdot T$

• $K<em>p = \frac{(C</em>E \cdot R \cdot T)^e (C<em>D \cdot R \cdot T)^d}{(C</em>A \cdot R \cdot T)^a (C_B \cdot R \cdot T)^b}$

• 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.

• $a<em>A = \frac{P</em>A}{1 atm}$: 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 $K<em>c$ and $K</em>p$

• Reaction: $N<em>2O</em>4(g) \rightleftharpoons 2NO_2(g)$ at $25^\circ C$

• Equilibrium concentrations: $[N<em>2O</em>4] = 0.00452 M$ and $[NO_2] = 0.0310 M$

• Calculate $K_c$:

  • Equilibrium expression: $K<em>c = \frac{[NO</em>2]^2}{[N<em>2O</em>4]}$

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

• Calculate $K_p$:

  • $K<em>p = K</em>c(RT)^{\Delta n}$
  • $\Delta n = 2 - 1 = 1$

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

Heterogeneous Chemical Equilibria [6.5]

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

• Example: $CoCl<em>3(s) + 3H</em>2O(g) \rightleftharpoons CoCl<em>3 \cdot 3H</em>2O(s)$

• Equilibrium Constant Expression:
  $K = \frac{a<em>{CoCl</em>3 \cdot 3H<em>2O}}{a</em>{CoCl<em>3} \cdot (a</em>{H_2O})^3}$

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

  • $a<em>{CoCl</em>3} = \frac{[CoCl<em>3]}{[CoCl</em>3]_{reference}} = 1$

• Therefore, the equilibrium expression simplifies to:
  $K = \frac{1}{(a<em>{H</em>2O})^3} = \frac{1}{[H_2O]^3}$

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

Example Problem: Calculating $K_p$ for Heterogeneous Equilibrium

• Reaction: $CaCO<em>3(s) \rightleftharpoons CaO(s) + CO</em>2(g)$

• $K<em>p = P</em>{CO_2}$

• 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:
    $CaCO<em>3(s) \rightleftharpoons CaO(s) + CO</em>2(g)$

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

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

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

  • $x = n<em>i - n</em>e = 0.1998 mol - 0.0687 mol = 0.131 mol$

  • Calculate the pressure of $CO<em>2$ using the Ideal Gas Law: $P</em>{CO<em>2} = \frac{n</em>{CO<em>2} \cdot R \cdot T}{V}$$P</em>{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: $[A]<em>i$ $[B]</em>i$ $[C]_i$
  • C: $\pm x$ $\pm x$ $\pm 2x$
  • E: $[A]<em>i \pm x$ $[B]</em>i \pm x$ $[C]_i \pm 2x$

• Calculate the reaction quotient, $Q$:

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

  • 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 $K = 2.0$

• $Q = \frac{[B]<em>i^2}{[A]</em>i}$