Kc and Kp Knowledge-Based Study

Study notes

1. Dynamic Equilibrium and Equilibrium Expressions

   In a reversible chemical reaction at dynamic equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, resulting in constant macroscopic properties such as concentrations and partial pressures.

   For a general reversible reaction taking place in a closed system at constant temperature:

   $$a\text{A}(g) + b\text{B}(g) \rightleftharpoons c\text{C}(g) + d\text{D}(g)$$

   The equilibrium state can be described quantitatively by equilibrium constants:

   • $K_c$ (Concentration Equilibrium Constant): Formulated using the molar concentrations of species at equilibrium:
     $$K_c = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b}$$
   • $K_p$ (Pressure Equilibrium Constant): Formulated using the partial pressures of gaseous species at equilibrium:
     $$K_p = \frac{(P_{\text{C}})^c (P_{\text{D}})^d}{(P_{\text{A}})^a (P_{\text{B}})^b}$$

   Note: Pure solids and pure liquids are omitted from both equilibrium constant expressions because their concentrations and activities remain essentially constant during the reaction.

   Derivation and the Ideal Gas Law

   For gaseous reactants and products, partial pressure can be related to molar concentration using the Ideal Gas Equation, $PV = nRT$, which rearranges to:

   $$P = \left(\frac{n}{V}\right)RT = CRT$$

   where $C$ represents molar concentration in $\text{mol/L}$, $R$ is the ideal gas constant (typically $0.0821\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$), and $T$ is absolute temperature in kelvins ($\text{K}$).

   Substituting this relationship into the $K_p$ expression yields the general relation:

   $$K_p = K_c(RT)^{\Delta n_g}$$

   Here, $\Delta n_g$ is the change in moles of gas between products and reactants:

   $$\Delta n_g = \sum n_{\text{gas, products}} - \sum n_{\text{gas, reactants}} = (c + d) - (a + b)$$

   Key Observations:

   • If $\Delta n_g = 0$, then $(RT)^0 = 1$, which means $K_p = K_c$.
   • If $\Delta n_g > 0$, then $K_p > K_c$ when $RT > 1$.
   • If $\Delta n_g < 0$, then $K_p < K_c$ when $RT > 1$.
   • Equilibrium constants are temperature-dependent; changing temperature changes the numerical values of both $K_c$ and $K_p$.

   Worked Example

   Consider the synthesis of ammonia from its elements:

   $$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$$

   At $500\text{ K}$, the value of $K_c$ is $0.060$. Let us determine the value of $K_p$ at this temperature.

   Step 1: Calculate $\Delta n_g$

   $$\Delta n_g = 2 - (1 + 3) = 2 - 4 = -2$$

   Step 2: Apply the Relationship

   $$K_p = K_c(RT)^{\Delta n_g}$$

   $$K_p = 0.060 \times (0.0821 \times 500)^{-2}$$

   $$K_p = 0.060 \times (41.05)^{-2} = \frac{0.060}{1685.1} \approx 3.56 \times 10^{-5}$$

   This demonstrates how a reaction with fewer gas moles on the product side yields a significantly smaller $K_p$ than $K_c$ at elevated temperatures.