The Equilibrium Constant (Kc) (The Law of Mass Action)

1. What is the Equilibrium Constant $(K_c)$?

• The equilibrium constant $(K_c)$ expresses the relationship between the concentrations of reactants and products in a closed system at equilibrium.

• It is derived from the Law of Mass Action, which states that at equilibrium: $K_c = \frac{\text{[Products]}^{\text{coefficients}}}{\text{[Reactants]}^{\text{coefficients}}}$

• The value of $K_c$ is constant for a given reaction at a specific temperature.

2. Writing the $K_c$ Expression

For the general reaction:

$aA + bB \rightleftharpoons cC + dD$

The equilibrium constant expression is:

$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$

• Square brackets [] represent concentration $(mol·dm^{-3}).$
• Coefficients in the balanced equation become exponents in the expression.
• Only gases $(g)$ and aqueous solutions (aqaq) are included. Solids $(s)$ and liquids $(l)$ are excluded.

3. Factors Affecting $K_c$

• Temperature:
  • If temperature increases, $K_c$ changes depending on whether the forward reaction is exothermic or endothermic.
  • Exothermic reaction $(\Delta H < 0): K_c$ decreases when temperature increases.
  • Endothermic reaction $(\Delta H > 0): K_c$ increases when temperature increases.
• What does NOT affect $K_c$?
  • Concentration changes (only shifts equilibrium, does not change $K_c$ ).
  • Pressure changes (only affects gaseous reactions, does not change $K_c$).
  • Adding a catalyst (only speeds up equilibrium, does not change $K_c$ ).

4. The Meaning of $K_c$ Values

• $K_c > 1 →$ More products at equilibrium (reaction favours forward direction).
• $K_c < 1 →$ More reactants at equilibrium (reaction favours reverse direction).
• $K_c \approx 1 →$ Almost equal amounts of reactants and products.

5. Calculating $K_c$

• Use the balanced equation and a table to determine initial, used, and equilibrium concentrations of reactants and products.
• Substitute values into the $K_c$ expression.
• Example: For the reaction:

$2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$

If equilibrium concentrations are:

$[SO_2] = 0.2, \quad [O_2] = 0.1, \quad [SO_3] = 0.3$

Then:

$K_c = \frac{[SO_3]^2}{[SO_2]^2 [O_2]} = \frac{(0.3)^2}{(0.2)^2 (0.1)}$