Chemical Equilibrium

1. Dynamic Equilibrium

In a closed system a reversible reaction reaches a state where the forward and backward rates are equal. Concentrations stay constant, but both reactions continue; this is a dynamic equilibrium. It can be approached from either side.

2. Law of Mass Action and $K$

For $aA + bB \rightleftharpoons cC + dD$:

• $\Delta n_g$ = moles of gaseous products − moles of gaseous reactants.
• Pure solids and pure liquids are left out of the expression (their activity is 1).
• Reversing a reaction gives $1/K$; multiplying it by $n$ gives $K^n$; adding reactions multiplies their $K$ values.
• $K$ changes only with temperature.

3. Reaction Quotient $Q$

$Q$ has the same form as $K$ but uses concentrations at any moment.

• $Q < K$: reaction moves forward.
• $Q > K$: reaction moves backward.
• $Q = K$: at equilibrium.

It connects to thermodynamics: $\Delta G = \Delta G^\circ + RT\ln Q$, and at equilibrium $\Delta G^\circ = -RT\ln K$.

4. Le Chatelier's Principle

If a system at equilibrium is disturbed, it shifts so as to partly undo the change.

• Concentration: adding a reactant shifts the equilibrium forward.
• Pressure: raising pressure favours the side with fewer gas moles. No effect if $\Delta n_g = 0$.
• Temperature: heating favours the endothermic direction. It is the only factor that changes $K$.
• Catalyst: speeds up both directions equally; equilibrium is reached sooner but its position is unchanged.
• Inert gas: at constant volume, no effect; at constant pressure, it acts like lowering pressure.

Interactive diagram: open this page in a browser to use it.

5. Degree of Dissociation

For $PCl_5 \rightleftharpoons PCl_3 + Cl_2$ starting with 1 mol and dissociating a fraction $\alpha$, total moles $= 1 + \alpha$, and $K_p = \frac{\alpha^2 P}{1 - \alpha^2}$. Vapour-density data can give $\alpha$: $\alpha = \frac{D - d}{(n-1)d}$, where $D$ is the initial and $d$ the equilibrium vapour density, and $n$ is the number of particles from one molecule.

Practice questions

1. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the relation between Kₚ and K_c is:

   1. Kₚ = K_c
   2. Kₚ = K_c(RT)²
   3. Kₚ = K_c(RT)⁻²
   4. Kₚ = K_c(RT)⁻¹
   Answer

   C. Δ n_g = 2 - 4 = -2, so Kₚ = K_c(RT)⁻².

2. Adding a catalyst to a system at equilibrium:

   1. Increases K
   2. Decreases K
   3. Shifts the equilibrium forward
   4. Does not change the equilibrium position
   Answer

   D. A catalyst lowers the activation energy of both directions equally. Equilibrium is reached faster but neither K nor the composition changes.

3. If K for A ⇌ B is 4, then K for 2B ⇌ 2A is:

   1. 1/16
   2. 1/8
   3. 16
   4. 1/4
   Answer

   A. Reversing gives 1/4; doubling the coefficients squares it: (1/4)² = 1/16.

4. For H₂(g) + I₂(g) ⇌ 2HI(g), increasing the pressure will:

   1. Shift it forward
   2. Shift it backward
   3. Have no effect on the equilibrium
   4. Increase K
   Answer

   C. Δ n_g = 0, so pressure has no effect on the equilibrium position.

5. At a given instant Q_c = 10 while K_c = 100. The reaction will:

   1. Proceed in the forward direction
   2. Proceed in the backward direction
   3. Be at equilibrium
   4. Stop completely
   Answer

   A. Q < K, so more products must form until Q = K.

6. For an exothermic reaction at equilibrium, increasing the temperature:

   1. Increases K
   2. Decreases K
   3. Has no effect on K
   4. Doubles K
   Answer

   B. Heating favours the endothermic (backward) direction, so K decreases.