the concept of dynamic equilibrium for reversible reactions, the equilibrium law expression and equilibrium constant Kc (including the meaning of Q vs Kc and the units of Kc), and the qualitative application of Le Chatelier's principle to predict the effect on equilibrium of changes in concentration, gas pressure (volume), temperature and the addition of a catalyst

What this dot point is asking

VCAA wants the dynamic equilibrium concept (reversible reactions with equal forward and reverse rates), the equilibrium law expression with the equilibrium constant Kc (including units and the meaning of Q vs Kc), and the qualitative application of Le Chatelier's principle to predict shifts when concentration, pressure (volume), or temperature changes, or a catalyst is added.

The answer

The diagram shows the typical concentration-vs-time graph for a reversible reaction reaching equilibrium and Le Chatelier's principle as a shift in response to a stress.

Dynamic equilibrium

A reaction is at dynamic equilibrium when:

1. It is occurring in a closed system (no matter enters or leaves).
2. The forward and reverse reactions are still happening (the equilibrium is "dynamic", not static).
3. The rates of the forward and reverse reactions are equal, so the concentrations of reactants and products stay constant.

A reaction at equilibrium has not stopped; it is occurring in both directions at the same rate. There is no net change in the amount of any species, but molecules are still being interconverted continually.

The equilibrium expression and Kc

For the general equilibrium:

aA + bB ⇌ cC + dD

the equilibrium law expression is:

Kc = [C]^c × [D]^d / ([A]^a × [B]^b)

with square-bracketed concentrations measured at equilibrium in mol L^-1. The product of product concentrations (raised to their coefficients) is on the top, and the product of reactant concentrations (raised to their coefficients) is on the bottom.

Rules:

• Only (g) and (aq) species appear in the expression. Pure (l) and (s) species are omitted because their effective concentrations are constant.
• The value of Kc depends only on temperature for a given reaction.
• Units of Kc depend on the stoichiometry. For aA + bB ⇌ cC + dD, units are (mol L^-1)^((c+d)-(a+b)). Many marking schemes accept "no units" if the convention is applied consistently.

Reading the value of Kc

• Kc >> 1: products dominate at equilibrium (the position lies far to the right).
• Kc << 1: reactants dominate at equilibrium (position lies far to the left).
• Kc ≈ 1: comparable amounts of reactants and products.

Q vs Kc

The reaction quotient Q has the same algebraic form as Kc but uses current concentrations (which may or may not be at equilibrium).

Q = [products] / [reactants] (using current concentrations)

Comparison with Kc tells you which direction the reaction will move:

When Q is computed before equilibrium is reached, the system shifts in whichever direction will move Q towards Kc.

Le Chatelier's principle

Le Chatelier's principle: if a system at equilibrium is disturbed, it shifts to partially counteract the disturbance and restore equilibrium.

Four disturbances are examined in VCE:

1. Change in concentration

• Adding more of a reactant: shift forward (consume the added reactant).
• Removing a product: shift forward (replace the removed product).
• Adding a product: shift reverse.
• Removing a reactant: shift reverse.

Kc is unchanged because temperature is unchanged.

2. Change in pressure (by changing volume)

Only matters for reactions with gases. Compare the moles of gas on each side.

• Increasing pressure (decreasing volume): shift to the side with fewer moles of gas.
• Decreasing pressure (increasing volume): shift to the side with more moles of gas.
• If both sides have the same number of moles of gas, no shift occurs.

Kc is unchanged.

Adding an inert gas at constant volume does not shift the equilibrium because partial pressures (and concentrations) of reactants and products are unchanged.

3. Change in temperature

The only disturbance that changes Kc.

• Increasing temperature: shift in the endothermic direction (the direction that absorbs heat). For an exothermic forward reaction, this means shifting to the left, decreasing Kc.
• Decreasing temperature: shift in the exothermic direction. For an exothermic forward reaction, this means shifting right, increasing Kc.

Treat heat as a "reactant" or "product":

• Exothermic forward: A + B ⇌ C + D + heat. Adding heat shifts left.
• Endothermic forward: A + B + heat ⇌ C + D. Adding heat shifts right.

4. Adding a catalyst

A catalyst lowers the activation energy of both the forward and reverse reactions equally, so equilibrium is reached faster but the position and Kc are unchanged.

Visualising on a concentration vs time graph

A common SAC/exam graph plots concentration on the y-axis and time on the x-axis. Reactant and product concentrations level off when equilibrium is reached. A disturbance shows up as a vertical jump or drop in the concentration of the disturbed species, then a re-equilibration as the system shifts.

For example, adding more H2 to the N2/H2/NH3 equilibrium shows a vertical jump in [H2], then a fall in [H2] and [N2] (consumed) and a rise in [NH3] (produced) until new constant levels are reached.

Examples in context

Example 1. Ammonia synthesis at Incitec Pivot Gibson Island (Brisbane). Incitec Pivot runs the Haber-Bosch process: $\text{N}_2 + 3 \text{H}_2 \rightleftharpoons 2 \text{NH}_3$, $\Delta H = -92 \, \text{kJ/mol}$. At equilibrium at $450^{\circ}\text{C}$ and $200 \, \text{atm}$, partial pressures are typically $\text{N}_2 = 35$, $\text{H}_2 = 105$, $\text{NH}_3 = 60 \, \text{atm}$. $K_p = (60)^2 / (35 \times 105^3) = 3600 / 4.05 \times 10^7 = 8.9 \times 10^{-5}$. Le Chatelier predicts: lower $T$ shifts right (exothermic), but rate too slow; high $P$ shifts right (4 mol $\to$ 2 mol gas); continual removal of $\text{NH}_3$ by liquefaction shifts right. Operating at $200 \, \text{atm}$ instead of $1 \, \text{atm}$ raises yield from $\sim 2\%$ to $\sim 30\%$ per pass.

Example 2. CO$_2$-water equilibrium in Murray-Darling carbonate chemistry. Aquatic chemists at Charles Sturt University, Albury, monitor the system $\text{CO}_2 + \text{H}_2 \text{O} \rightleftharpoons \text{H}_2 \text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$ in the Murray. As river temperature drops from $25^{\circ}\text{C}$ to $10^{\circ}\text{C}$ in winter, $\text{CO}_2$ solubility nearly doubles (exothermic dissolution shifts right). The bicarbonate buffer maintains pH near $8.0$, but algae blooming on summer warmth remove $\text{CO}_2$ by photosynthesis, shifting the equilibrium and raising daytime pH to $9.5$ before recovering at night. $K_{a1}$ of carbonic acid $= 4.3 \times 10^{-7}$ at $25^{\circ}\text{C}$, $2.9 \times 10^{-7}$ at $10^{\circ}\text{C}$, demonstrating temperature dependence of $K$.

Try this

Q1. State Le Chatelier's principle and use it to predict the effect of increasing temperature on the equilibrium $\text{N}_2 \text{O}_4 \rightleftharpoons 2 \text{NO}_2$, $\Delta H = +58 \, \text{kJ/mol}$. [3 marks]

• Cue. A change to equilibrium causes shift partially opposing the change. Endothermic forward, so heat behaves like a reactant; rise in T shifts right; more brown $\text{NO}_2$. $K_c$ increases.

Q2. For $\text{H}_2 + \text{I}_2 \rightleftharpoons 2 \text{HI}$, at equilibrium $[\text{H}_2] = 0.20$, $[\text{I}_2] = 0.10$, $[\text{HI}] = 0.50 \, \text{mol/L}$. (a) Calculate $K_c$. (b) If $0.20 \, \text{mol/L}$ HI is added, predict the direction of shift. (c) Determine $Q$ immediately after the disturbance. [4 marks]

• Cue. (a) $K_c = (0.50)^2 / (0.20 \times 0.10) = 12.5$. (b) Excess HI, shifts left. (c) $Q = (0.70)^2 / (0.20 \times 0.10) = 24.5 > K_c$ confirms left shift.

Q3. For the Haber process $\text{N}_2 + 3 \text{H}_2 \rightleftharpoons 2 \text{NH}_3$, $\Delta H = -92 \, \text{kJ/mol}$. (a) Predict the effect of increased pressure on yield. (b) Predict the effect of higher temperature. (c) Explain why industry compromises at $450^{\circ}\text{C}$. [2+2+2 marks]

• Cue. (a) Shifts right (fewer moles of gas); higher yield. (b) Shifts left (exothermic); lower yield but $K_c$ decreases. (c) Lower T gives higher yield but rate too slow; compromise to balance yield and rate.