Chemistry (9701) | Equilibria

AS Level Chemistry (9701) Study Notes: Topic 7 - Equilibria

Welcome to the fascinating world of Chemical Equilibria! Don't worry if this topic seems tricky at first. It’s all about balance—like a chemical seesaw. Understanding equilibrium is crucial because it governs almost all chemical reactions in industry (like making fertilisers) and even in your own body (like regulating blood $\text{pH}$).

In this chapter, we will learn why some reactions stop before completion and how chemists can manipulate conditions to maximise the yield of useful products.

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Section 7.1: Chemical Equilibria - The Dynamic Balance

7.1.1 Reversible Reactions and Dynamic Equilibrium

Most reactions you learned about previously seem to go from reactants to products until one substance runs out. However, many reactions are reversible.

What is a Reversible Reaction?

A reversible reaction is one where the products can react together to reform the original reactants. We show this using a double arrow ($\rightleftharpoons$).

Example: The reaction between sulfur dioxide and oxygen to make sulfur trioxide:

\(2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})\)

Here, the forward reaction makes $\text{SO}_3$, but the reverse reaction breaks down $\text{SO}_3$ into $\text{SO}_2$ and $\text{O}_2$.

What is Dynamic Equilibrium?

When a reversible reaction takes place in a closed system, it eventually reaches a state of dynamic equilibrium.

Think of it like two people on an escalator that goes up (Reactants $\rightarrow$ Products) and another identical one going down (Products $\rightarrow$ Reactants).

• Condition 1: Rate of Forward Reaction = Rate of Reverse Reaction.
  (The rate at which people step onto the up escalator equals the rate at which people step onto the down escalator.)
• Condition 2: Concentration of Reactants and Products Remain Constant.
  (The number of people on the upper floor and the lower floor stays the same, even though they are constantly moving.)

The system is dynamic because the reactions haven't stopped—they are just happening at the same speed in opposite directions.

7.1.2 Le Chatelier's Principle

How do we control where the chemical seesaw tips? We use Le Chatelier's Principle.

Definition of Le Chatelier's Principle:

If a change is made to a system at dynamic equilibrium, the position of equilibrium will shift to minimise (or counteract) this change.

How Changes Affect the Position of Equilibrium:

The position of equilibrium describes whether there are more products or more reactants present when the system reaches its balanced state.

1. Effect of Concentration Change:
   

   If you increase the concentration of a reactant, the system shifts to the right (forward reaction is favoured) to use up the extra reactant.

   If you decrease the concentration of a product (e.g., by removing it as it forms), the system shifts to the right to replace the lost product.

2. Effect of Pressure Change (for Gaseous Systems):
   

   Pressure changes only affect reactions involving gases and where there is a change in the total number of moles of gas.

   If you increase the pressure, the equilibrium shifts to the side with fewer moles of gas.

   If you decrease the pressure, the equilibrium shifts to the side with more moles of gas.

   Trick: Count the gas moles on both sides! For $2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g})$, there are 3 moles on the left and 2 moles on the right. Increasing pressure shifts it right (to 2 moles).

3. Effect of Temperature Change:
   

   Temperature changes are dictated by the reaction's enthalpy change (\(\Delta H\)).

   If the forward reaction is exothermic (\(\Delta H\) is negative, it releases heat):

   • Increase T: Shifts left (endothermic direction) to absorb the excess heat.
   • Decrease T: Shifts right (exothermic direction) to replace the lost heat.

   If the forward reaction is endothermic (\(\Delta H\) is positive, it absorbs heat):

   • Increase T: Shifts right (endothermic direction) to absorb the excess heat.
   • Decrease T: Shifts left (exothermic direction) to release heat.

4. Effect of a Catalyst:
   

   A catalyst has no effect on the position of equilibrium.

   It speeds up the rate of both the forward and reverse reactions equally, meaning the system reaches equilibrium faster, but the final concentrations (the position of equilibrium) remain the same.

7.1.3 Equilibrium Constants: \(K_c\) and \(K_p\)

The equilibrium constant ($K$) is a mathematical representation of the balance at equilibrium. If $K$ is very large ($>10^4$), the equilibrium lies far to the product side. If $K$ is very small ($<10^{-4}$), the equilibrium lies far to the reactant side.

\(K_c\) - The Concentration Constant

For a general reaction:
\(a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}\)

The expression for \(K_c\) is:

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

• The products are always on the top (numerator).
• \([\text{X}]\) represents the concentration of species X in $\text{mol dm}^{-3}$.
• Pure solids and pure liquids are never included in the \(K_c\) expression, as their concentration is considered constant.

\(K_p\) - The Partial Pressure Constant (Gases only!)

When all species are gases, we can use partial pressure instead of concentration.

1. Mole Fraction: The ratio of the moles of one gas to the total moles of all gases in the mixture.
\(\text{Mole fraction of gas A} = \frac{\text{moles of A}}{\text{total moles of gas}}\)

2. Partial Pressure: The pressure that a gas would exert if it alone occupied the volume of the whole mixture.
\(\text{Partial pressure of gas A } (P_\text{A}) = \text{Mole fraction of A} \times \text{Total pressure}\)

The expression for \(K_p\) follows the same structure as \(K_c\), using partial pressures ($P$) instead of concentrations:

\(K_p = \frac{(P_\text{C})^c (P_\text{D})^d}{(P_\text{A})^a (P_\text{B})^b}\)

Units for both \(K_c\) and \(K_p\) depend entirely on the reaction stoichiometry (the powers in the expression). You must calculate the units using the definition of the formula.

Calculations involving \(K_c\) and \(K_p\) (7.1.7, 7.1.8)

AS Level calculations typically involve using an ICE table (Initial, Change, Equilibrium) to find the equilibrium moles or concentrations, and then substituting these values into the \(K\) expression.

Remember these steps for \(K_c\) calculations:

1. Initial Moles: Write down the starting moles.
2. Change in Moles: Determine the stoichiometric change (use ratios).
3. Equilibrium Moles: Calculate the final moles (\(I \pm C\)).
4. Equilibrium Concentration: Convert moles to concentration (\(\text{mol}/\text{volume}\)).
5. Substitute concentrations into the \(K_c\) expression.

Note: You are not required to solve quadratic equations for these calculations. The data provided will usually allow for simpler arithmetic solutions.

Section 7.1.10: Industrial Applications of Equilibrium

Industries use Le Chatelier's Principle to maximize product yield while keeping costs (energy) low and the rate of reaction high.

1. The Haber Process (Making Ammonia, \(\text{NH}_3\))

\(\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \quad \Delta H = -92 \text{ kJ mol}^{-1}\) (Exothermic)

Conditions and Why:

• Temperature: Approximately \(400-450^\circ\text{C}\) (moderate).

  Explanation: Since the forward reaction is exothermic, low temperature favours yield (LCP). However, low temperatures result in a very slow rate (Kinetic consideration). A compromise temperature is used to get a reasonable yield at an economically viable rate.

• Pressure: Approximately \(150-350\) atm (high).

  Explanation: There are 4 moles of gas on the left and 2 moles on the right. High pressure shifts the equilibrium to the side with fewer moles (products) (LCP), maximizing yield. High pressure also increases the rate of reaction (Kinetic).

• Catalyst: Iron (Fe)

  Explanation: The catalyst speeds up the reaction rate, allowing the compromise temperature to be used efficiently, but it does not affect the yield.

2. The Contact Process (Making Sulfur Trioxide, \(\text{SO}_3\))

(A crucial intermediate step in making sulfuric acid, \(\text{H}_2\text{SO}_4\))

\(2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) \quad \Delta H = -197 \text{ kJ mol}^{-1}\) (Highly Exothermic)

Conditions and Why:

• Temperature: Approximately \(400-450^\circ\text{C}\) (moderate, similar to Haber).

  Explanation: Low T favours yield (exothermic), but the rate is too slow. A compromise T is needed for an effective rate.

• Pressure: Atmospheric pressure (or slightly higher, typically 1–2 atm).

  Explanation: The forward reaction is favoured by high pressure (3 moles $\rightarrow$ 2 moles). However, because the reaction yield is already very high (>99.5%) even at low pressure, using expensive, high-pressure equipment is not economically justified.

• Catalyst: Vanadium(V) oxide (\(\text{V}_2\text{O}_5\))

  Explanation: Used to ensure a fast reaction rate at the compromise temperature.

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Section 7.2: Brønsted-Lowry Theory of Acids and Bases

Now we switch gears and apply the concept of equilibrium to proton transfer reactions.

7.2.1 & 7.2.2 Common Acids and Alkalis

You must know the names and formulas of these common reagents:

• Common Acids:
  • Hydrochloric acid: \(\text{HCl}\)
  • Sulfuric acid: \(\text{H}_2\text{SO}_4\)
  • Nitric acid: \(\text{HNO}_3\)
  • Ethanoic acid (a weak acid): \(\text{CH}_3\text{COOH}\)
• Common Alkalis (Bases that dissolve in water):
  • Sodium hydroxide: \(\text{NaOH}\)
  • Potassium hydroxide: \(\text{KOH}\)
  • Ammonia (a weak base): \(\text{NH}_3\)

7.2.3 The Brønsted-Lowry Theory

This theory defines acids and bases based on their ability to donate or accept protons (\(\text{H}^+\) ions).

• An Acid is a proton donor.
• A Base is a proton acceptor.

Conjugate Acid-Base Pairs (7.2.1, 7.2.2)

When an acid donates a proton, it forms its conjugate base. When a base accepts a proton, it forms its conjugate acid.

Example: Ethanoic acid dissociating in water:

\(\text{CH}_3\text{COOH} \text{ (acid 1)} + \text{H}_2\text{O} \text{ (base 2)} \rightleftharpoons \text{CH}_3\text{COO}^- \text{ (conjugate base 1)} + \text{H}_3\text{O}^+ \text{ (conjugate acid 2)}\)

The pairs are: $(\text{CH}_3\text{COOH}/\text{CH}_3\text{COO}^-)$ and $(\text{H}_3\text{O}^+/\text{H}_2\text{O})$.

7.2.4 Strong vs. Weak Acids and Bases

The terms strong and weak refer to the extent of dissociation (how much the substance breaks up into ions in water).

• Strong Acids/Bases: Fully dissociated in aqueous solution. The reaction goes 100% to completion (forward reaction only).
  Example (Strong Acid): \(\text{HCl}(\text{aq}) \rightarrow \text{H}^+(\text{aq}) + \text{Cl}^-(\text{aq})\)
• Weak Acids/Bases: Partially dissociated in aqueous solution. The reaction establishes an equilibrium where most of the substance remains as the original molecule.
  Example (Weak Acid): \(\text{CH}_3\text{COOH}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{CH}_3\text{COO}^-(\text{aq})\)

7.2.5 & 7.2.6 Qualitative Differences (Strong vs Weak)

Even if a strong acid and a weak acid have the same starting concentration (e.g., $0.10 \text{ mol dm}^{-3}$), they behave very differently because of their degree of dissociation.

Property	Strong Acid (e.g., HCl)	Weak Acid (e.g., \(\text{CH}_3\text{COOH}\))
$\text{pH}$ Value (7.2.5)	Very low $\text{pH}$ (e.g., $\text{pH}=1$ for $0.1 \text{ M}$ solution).	Higher $\text{pH}$ (e.g., $\text{pH}>3$ for $0.1 \text{ M}$ solution).
Electrical Conductivity	High (many ions are present to carry the current).	Low (few ions are present due to partial dissociation).
Reaction with Reactive Metal (e.g., Mg)	Very fast evolution of \(\text{H}_2\) gas.	Slow evolution of \(\text{H}_2\) gas.

Did you know? The differences are all due to the concentration of free \(\text{H}^+\) ions. Strong acids release all their protons, leading to high \([\text{H}^+]\), fast reactions, and low $\text{pH}$.

7.2.7 Neutralization and Salt Formation

Neutralization is the reaction between an acid and a base. The essential reaction that occurs in aqueous solution is the formation of water:

\(\mathbf{\text{H}^+(\text{aq}) + \text{OH}^-(\text{aq}) \rightarrow \text{H}_2\text{O}(\text{l})}\)

When an acid neutralises a base, a salt is always formed (7.2.8). The salt is the ionic compound remaining after the proton (\(\text{H}^+\)) and hydroxide (\(\text{OH}^-\)) ions have reacted.

Example:
\(\text{NaOH} + \text{HNO}_3 \rightarrow \text{NaNO}_3 \text{ (salt)} + \text{H}_2\text{O}\)

7.2.9 & 7.2.10 Titration Curves and Indicators

A $\text{pH}$ titration curve plots the $\text{pH}$ change of a solution as a titrant (acid or base) is added.

Key Features of Titration Curves:

• The equivalence point is when the moles of acid exactly equal the moles of base.
• The steep region (or vertical region) of the curve is where the $\text{pH}$ changes dramatically around the equivalence point.
• The $\text{pH}$ at the equivalence point depends on the strength of the acid and base used (due to salt hydrolysis, although detailed theory is A-Level content, you must know the resultant $\text{pH}$):
  • Strong Acid / Strong Base: Equivalence point at $\mathbf{\text{pH}=7}$.
  • Strong Acid / Weak Base: Equivalence point at $\mathbf{\text{pH}<7}$ (acidic).
  • Weak Acid / Strong Base: Equivalence point at $\mathbf{\text{pH}>7}$ (basic).

Indicator Selection (7.2.10)

An indicator is a weak acid or base whose colour changes over a specific $\text{pH}$ range.

To choose a suitable indicator, its colour change range must fall entirely within the steep region of the titration curve. This ensures that the indicator changes colour immediately when the reaction is complete (at the equivalence point).

Example of sketching a Strong Acid/Strong Base titration (e.g., $\text{HCl}$ titrated by $\text{NaOH}$): Starts very low $\text{pH}$ (1 or 2), steep change around $\text{pH}=7$, ends very high $\text{pH}$ (12 or 13). Methyl orange and phenolphthalein are both suitable because the steep region covers both their ranges.

Example of sketching a Weak Acid/Strong Base titration (e.g., $\text{CH}_3\text{COOH}$ titrated by $\text{NaOH}$): Starts higher $\text{pH}$ (3 or 4), steep change occurs in the basic region (roughly $\text{pH}$ 7-11). Only phenolphthalein is suitable here.