Chapter 6 / 6.5 Heterogeneous Equilibrium

Understanding Equilibrium Constants

Let’s explore how to calculate and apply equilibrium constants!

1. Calculating K_c (Concentration Constant)

For the reaction: \[ \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \] K_c is calculated as: \[ K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \] Example: If [PCl₃] = 1.59 M, [Cl₂] = 1.59 M, [PCl₅] = 1.41 M: \[ K_c = \frac{(1.59)^2}{1.41} = 1.79 \]

2. Equilibrium Concentrations from K_c

For: \[ \text{CO(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO}_2\text{(g)} + \text{H}_2\text{(g)} \quad (K_c = 4.24 \text{ at } 800\text{K}) \] Given: Initial [CO] = [H₂O] = 0.10 M
Use an ICE table:

	[CO]	[H₂O]	[CO₂]	[H₂]
Initial (M)	0.10	0.10	0	0
Equilibrium (M)	0.10 – x	0.10 – x	x	x

Equation: \[ K_c = \frac{x^2}{(0.10 – x)^2} = 4.24 \] Solving the quadratic equation gives x = 0.067 M (reject x = 0.194 M as it exceeds initial concentration).
∴ [CO₂] = [H₂] = 0.067 M, [CO] = [H₂O] = 0.033 M.

3. Converting K_c to K_p (Pressure Constant)

For: \[ 2\text{NOCl(g)} \rightleftharpoons 2\text{NO(g)} + \text{Cl}_2\text{(g)} \quad (K_c = 3.75 \times 10^{-6} \text{ at } 1069\text{K}) \] Use: \[ K_p = K_c (RT)^{\Delta n} \] where Δn = (moles of gas products – moles of gas reactants) = (3 – 2) = 1, R = 0.0831 L·bar·mol⁻¹·K⁻¹, T = 1069 K. \[ K_p = (3.75 \times 10^{-6}) \times (0.0831 \times 1069) = 0.033 \]

Heterogeneous Equilibria ⚖️

Definition: Equilibria involving substances in different phases (e.g., solid + gas, solid + liquid).

Key Rules for K Expressions

• Pure solids (s) and liquids (l) are NOT included in K expressions! Their concentrations are constant.
• Only gases (g) and aqueous species (aq) appear in K.

Examples:

1. Decomposition of calcium carbonate: \[ \text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)} \] \[ K_c = [\text{CO}_2] \quad \text{or} \quad K_p = P_{\text{CO}_2} \] At 1100 K: \( K_p = 2.00 \) (when \( P_{\text{CO}_2} = 2.0 \times 10^5 \text{ Pa} \)).
2. Nickel carbonyl formation: \[ \text{Ni(s)} + 4\text{CO(g)} \rightleftharpoons \text{Ni(CO)}_4\text{(g)} \] \[ K_c = \frac{[\text{Ni(CO)}_4]}{[\text{CO}]^4} \]
3. Silver oxide reaction: \[ \text{Ag}_2\text{O(s)} + 2\text{HNO}_3\text{(aq)} \rightleftharpoons 2\text{AgNO}_3\text{(aq)} + \text{H}_2\text{O(l)} \] \[ K_c = \frac{[\text{AgNO}_3]^2}{[\text{HNO}_3]^2} \]

Problem: Finding Partial Pressures

For: \[ \text{CO}_2\text{(g)} + \text{C(s)} \rightleftharpoons 2\text{CO(g)} \quad (K_p = 3.0 \text{ at } 1000\text{K}) \] Given: Initial \( P_{\text{CO}_2} = 0.48 \text{ bar} \), \( P_{\text{CO}} = 0 \text{ bar} \).
ICE table for pressures:

	PCO₂	PCO
Initial (bar)	0.48	0
Equilibrium (bar)	0.48 – x	2x

Equation: \[ K_p = \frac{(2x)^2}{(0.48 – x)} = 3.0 \] Solving: \( 4x^2 + 3x – 1.44 = 0 \) → x = 0.33 bar (reject negative root).
∴ \( P_{\text{CO}} = 2 \times 0.33 = 0.66 \text{ bar} \), \( P_{\text{CO}_2} = 0.48 – 0.33 = 0.15 \text{ bar} \).

Applications of Equilibrium Constants 🧪

Predicting Reaction Extent

• \( K > 10^3 \): Products dominate (reaction nearly complete).
  e.g., \( \text{H}_2 + \text{Cl}_2 \rightleftharpoons 2\text{HCl} \quad (K_c = 4.0 \times 10^{31} \text{ at } 300\text{K}) \)
• \( K < 10^{-3} \): Reactants dominate (reaction barely proceeds).

Key Properties of K

1. K depends only on temperature (not initial concentrations).
2. For reverse reaction: \( K_{\text{reverse}} = \frac{1}{K_{\text{forward}}} \).
3. Scaling the reaction equation scales K:
   e.g., Doubling coefficients → K becomes K².

NEET Quick Tips! 🚀

1. Heterogeneous K expressions: Omit solids/liquids!
2. K_c vs K_p: Use \( K_p = K_c (RT)^{\Delta n} \).
3. Quadratic solutions: Always check validity of roots (e.g., concentrations can’t be negative).