Equilibrium Constants: Kc and Kp

For any reaction at equilibrium:

  • 🎯 Kc: Uses molar concentrations (mol/L)
  • 🎯 Kp: Uses partial pressures (for gases)

General reaction:

\[ aA + bB \rightleftharpoons cC + dD \] \[ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}, \quad K_p = \frac{(p_C)^c (p_D)^d}{(p_A)^a (p_B)^b} \]

Relationship between Kp and Kc

\[ K_p = K_c (RT)^{\Delta n} \]

Where:

  • \(\Delta n = (\text{moles of gaseous products}) – (\text{moles of gaseous reactants})\)
  • \(R = 0.0831 \ \text{bar·L·mol}^{-1}\text{·K}^{-1}\)
  • \(T\) = Temperature in Kelvin
  • Pressure must be in bar! ⚠️

Examples:

  1. For \(\ce{H2(g) + I2(g) <=> 2HI(g)}\): \(\Delta n = 0\), so \(K_p = K_c\) 🎈
  2. For \(\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\): \(\Delta n = -2\), so \(K_p = K_c (RT)^{-2}\) 🎈

Homogeneous Equilibria

All reactants/products in the same phase:

  • \(\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\) 🌬️
  • \(\ce{CH3COOC2H5(aq) + H2O(l) <=> CH3COOH(aq) + C2H5OH(aq)}\) 💧

Heterogeneous Equilibria

Involves multiple phases. Pure solids/liquids are excluded from K expressions! 🧊

Examples:

  • \(\ce{H2O(l) <=> H2O(g)}\): \(K_c = [\ce{H2O(g)}]\), \(K_p = p_{\ce{H2O}}\)
  • \(\ce{Ca(OH)2(s) <=> Ca^{2+}(aq) + 2OH^{-}(aq)}\): \(K_c = [\ce{Ca^{2+}}][\ce{OH^{-}}]^2\)

NEET Superstars! 🌟 (High-Yield Concepts)

  1. Kp-Kc relationship: \(K_p = K_c(RT)^{\Delta n}\) and \(\Delta n\) calculation
  2. Heterogeneous equilibria: Omit pure solids/liquids from equilibrium expressions
  3. K calculations: Using equilibrium concentrations (like Problems 6.1-6.3)

Practice Problems

Problem 6.1

Calculate Kc for: \(\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\)
Given at 500K: [N2] = 1.5 × 10-2M, [H2] = 3.0 × 10-2M, [NH3] = 1.2 × 10-2M

\[ K_c = \frac{(1.2 \times 10^{-2})^2}{(1.5 \times 10^{-2})(3.0 \times 10^{-2})^3} = 3.55 \times 10^2 \]

Problem 6.4

For \(\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}\) at 800K, Kc = 4.24
Initial: [CO] = [H2O] = 0.10 M. Find equilibrium concentrations.

\[ \begin{align*} &K_c = \frac{x^2}{(0.10 – x)^2} = 4.24 \\ &x = 0.067 \ M \\ &[\ce{CO2}] = [\ce{H2}] = 0.067 \ M \\ &[\ce{CO}] = [\ce{H2O}] = 0.033 \ M \end{align*} \]

Problem 6.5

For \(\ce{2NOCl(g) <=> 2NO(g) + Cl2(g)}\), Kc = 3.75 × 10-6 at 1069K. Find Kp.

\[ \begin{align*} &\Delta n = 1 \\ &K_p = (3.75 \times 10^{-6}) \times (0.0831 \times 1069) = 0.033 \end{align*} \]

Remember: Always specify the balanced equation when reporting K values! 📝✨