Welcome to Chemical Equilibrium!
Welcome to one of the most exciting and essential topics in AS Level Physical Chemistry: Equilibrium. In industrial chemistry, time is money, and maximizing product yield is everything. Chemical manufacturers don't just want a reaction to happen; they want it to produce as much product as possible, as fast and safely as possible.
Don't worry if this topic feels a bit daunting at first. We will break everything down into bite-sized, logical steps with simple rules, clear examples, and real-world industrial case studies.
1. Reversible Reactions and Dynamic Equilibrium
What is a Reversible Reaction?
In many chemical reactions, reactants turn into products and the reaction stops. These are irreversible reactions. However, some reactions can go in both directions: reactants form products, and products react together to re-form the original reactants.
We represent reversible reactions using the reversible reaction arrow: \( \rightleftharpoons \)
Forward reaction: Reactants \( \rightarrow \) Products
Reverse (backward) reaction: Products \( \rightarrow \) Reactants
What is Dynamic Equilibrium?
Imagine walking up a "down" escalator. If you walk upwards at the exact same speed that the escalator moves downwards, what happens? You stay in the exact same spot! You are still moving, and the escalator is still moving, but your overall position does not change.
This is exactly how a dynamic equilibrium works inside a chemical system. It is established when a reversible reaction takes place in a closed system (where no substances can enter or leave).
The Two Essential Conditions for Dynamic Equilibrium:
1. The rate of the forward reaction is equal to the rate of the reverse reaction.
2. The concentrations of reactants and products remain constant.
Crucial Exam Pitfall: Students often write that the concentrations of reactants and products are equal at equilibrium. This is incorrect! The concentrations are constant (they stay the same over time), but there may be far more product than reactant, or vice versa.
Key Takeaway: Dynamic equilibrium is active (reactions are still happening continuously in both directions), but macroscopic properties (such as color, pressure, and concentration) remain totally unchanged.
2. Le Chatelier's Principle
When a system is at equilibrium, it is balanced. But what happens if we change the conditions? A French chemist named Henri Le Chatelier figured out how systems respond.
Le Chatelier’s Principle Definition:
If a change (stress) is applied to a system at equilibrium, the position of equilibrium will shift to counteract that change.
Think of equilibrium as stubborn: whatever you do to it, it tries to do the exact opposite!
A. Effect of Changing Concentration
• Increase concentration of a reactant: The system opposes this by removing the added reactant. The equilibrium shifts to the right (towards products).
• Increase concentration of a product: The system opposes this by removing the added product. The equilibrium shifts to the left (towards reactants).
• Decrease concentration of a product: The system shifts to the right to replace what was lost.
B. Effect of Changing Pressure
Pressure only affects equilibrium systems that involve gases.
• Increase pressure: The system shifts to the side with fewer moles of gas to reduce the pressure.
• Decrease pressure: The system shifts to the side with more moles of gas to increase the pressure.
• Note: If both sides of the equation have the exact same number of moles of gas, changing the pressure has no effect on the position of equilibrium.
C. Effect of Changing Temperature
To predict the effect of temperature, you must know the enthalpy change (\( \Delta H \)) of the forward reaction.
• Increase temperature: The system attempts to cool down by shifting in the endothermic direction (the direction that absorbs heat).
• Decrease temperature: The system attempts to warm up by shifting in the exothermic direction (the direction that releases heat).
D. Effect of Adding a Catalyst
A catalyst increases the rate of both the forward and reverse reactions equally by providing an alternative pathway with a lower activation energy.
Important Exam Fact: A catalyst does not change the position of equilibrium and does not change the yield of products. It simply allows the system to reach dynamic equilibrium faster.
Key Takeaway: Le Chatelier's Principle tells us that a system shifts to undo any change you impose. Catalysts only speed up the journey to equilibrium; they do not alter the final balance.
3. The Equilibrium Constant (\( K_c \))
While Le Chatelier's Principle tells us the direction in which an equilibrium shifts, the equilibrium constant (\( K_c \)) gives us a mathematical measure of where the equilibrium lies.
Writing the \( K_c \) Expression
For any general reversible reaction:
\( aA + bB \rightleftharpoons cC + dD \)
The equilibrium constant expression is given by:
\( K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \)
Where:
• Square brackets \( [ ] \) denote concentration at equilibrium in \( \text{mol dm}^{-3} \).
• Superscripts (\( a, b, c, d \)) represent the balancing stoichiometric coefficients from the balanced equation.
• Products are always placed in the numerator (top), and reactants in the denominator (bottom).
Note on State Symbols: Only aqueous species and gaseous species are included in the \( K_c \) expression. Pure solids and pure liquids are omitted.
Working Out the Units for \( K_c \)
The units of \( K_c \) are not fixed; they depend entirely on the reaction stoichiometry. You must derive them in each question:
Step-by-step example:
For the reaction: \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \)
1. Write the expression: \( K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} \)
2. Substitute the unit \( \text{mol dm}^{-3} \) for each concentration term:
\( \text{Units} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^3} = \frac{(\text{mol dm}^{-3})^2}{(\text{mol dm}^{-3})^4} \)
3. Cancel out common factors: \( \text{Units} = \frac{1}{(\text{mol dm}^{-3})^2} = (\text{mol dm}^{-3})^{-2} = \text{mol}^{-2}\text{dm}^6 \)
What Affects the Numerical Value of \( K_c \)?
• Temperature: Only temperature changes the value of \( K_c \). For an exothermic forward reaction, increasing the temperature decreases \( K_c \). For an endothermic forward reaction, increasing the temperature increases \( K_c \).
• Concentration changes: Do not change \( K_c \).
• Pressure changes: Do not change \( K_c \).
• Catalysts: Do not change \( K_c \).
Key Takeaway: \( K_c \) gives the exact ratio of products to reactants at equilibrium. Memorize this golden rule: Only a change in temperature will alter the numerical value of \( K_c \).
4. Equilibrium in Industrial Processes
In industry, chemical engineers apply equilibrium principles to maximize profit. They need to find the best compromise between reaction rate (how fast ammonia or sulfur trioxide is made) and equilibrium yield (how much is made at equilibrium), while keeping operating costs and safety in mind.
A. The Haber Process (Synthesis of Ammonia)
Ammonia (\( NH_3 \)) is manufactured on a massive scale for fertilizers.
Chemical Equation:
\( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \quad \Delta H = -92\text{ kJ mol}^{-1} \)
Standard Operating Conditions:
• Temperature: \( \approx 450^\circ\text{C} \)
• Pressure: \( \approx 200\text{ atm} \)
• Catalyst: Iron (\( \text{Fe} \))
Industrial Logic & Compromises:
• Temperature Considerations: The forward reaction is exothermic (\( \Delta H = -92\text{ kJ mol}^{-1} \)). According to Le Chatelier's Principle, a low temperature gives the highest equilibrium yield of \( NH_3 \). However, at low temperatures, the reaction rate is far too slow to be commercially viable. Therefore, a compromise temperature of \( 450^\circ\text{C} \) is chosen to achieve a reasonable yield in a short time.
• Pressure Considerations: There are \( 4\text{ moles of gas} \) on the left (\( 1\text{ }N_2 + 3\text{ }H_2 \)) and \( 2\text{ moles of gas} \) on the right (\( 2\text{ }NH_3 \)). Increasing the pressure shifts the equilibrium to the right (fewer moles), giving a higher yield of ammonia at a faster rate. A high pressure of \( 200\text{ atm} \) is used because higher pressures require extremely expensive, thick-walled reaction vessels and high energy costs.
• Catalyst: The iron catalyst speeds up both forward and backward reactions equally, allowing equilibrium to be reached much faster at \( 450^\circ\text{C} \).
B. The Contact Process (Manufacture of Sulfuric Acid)
Sulfuric acid is one of the most widely used industrial chemicals. The key equilibrium step is the oxidation of sulfur dioxide to sulfur trioxide.
Key Equilibrium Step:
\( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -197\text{ kJ mol}^{-1} \)
Standard Operating Conditions:
• Temperature: \( \approx 450^\circ\text{C} \)
• Pressure: \( \approx 1\text{--}2\text{ atm} \) (just above atmospheric pressure)
• Catalyst: Vanadium(V) oxide (\( V_2O_5 \))
Industrial Logic & Compromises:
• Temperature Considerations: The forward reaction is exothermic (\( \Delta H = -197\text{ kJ mol}^{-1} \)). A lower temperature favours a higher yield, but a higher temperature favours a faster rate. A compromise temperature of \( 450^\circ\text{C} \) gives a satisfactory rate without severely reducing the yield.
• Pressure Considerations: There are \( 3\text{ moles of gas} \) on the left (\( 2\text{ }SO_2 + 1\text{ }O_2 \)) and \( 2\text{ moles of gas} \) on the right (\( 2\text{ }SO_3 \)). A high pressure favours the forward reaction. However, even at low pressures of \( 1\text{--}2\text{ atm} \), the yield of \( SO_3 \) is already very high (\( >98\% \)). Building high-pressure equipment is unnecessary and not economically justified.
• Catalyst: The \( V_2O_5 \) catalyst ensures a fast rate of reaction at \( 450^\circ\text{C} \).
Key Takeaway: Industrial conditions balance the thermodynamic yield predicted by Le Chatelier's Principle against the kinetic rate of the reaction and the economic costs of maintaining high temperatures and pressures.
5. Summary & Quick Review of Common Pitfalls
Before your exam, double-check that you have avoided these frequent mistakes:
• Equilibrium vs. Rate: "Position of equilibrium" refers to how much product you get (the yield). "Rate of reaction" refers to how fast you get it. Do not mix them up!
• Catalysts: Catalysts do not increase yield and do not alter the position of dynamic equilibrium. They only reduce the time needed to reach equilibrium.
• Concentration at Equilibrium: Concentrations do not have to be equal to one another; they just remain constant over time.
• Changes in \( K_c \): Only a change in temperature will change the numerical value of \( K_c \). Changing pressure, concentration, or adding a catalyst has zero effect on the value of \( K_c \).