Enthalpy changes; Hess’s Law; energy cycles - Chemistry - HKDSE

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Topic VIII: Chemical Reactions and Energy

Hello! Welcome to one of the most fundamental topics in chemistry: energy changes! Ever wondered why a hand warmer gets hot, or why an instant cold pack feels icy? It's all about chemistry and energy. In this chapter, we'll explore how energy is released or absorbed during chemical reactions. Understanding this helps us design better fuels, create new materials, and even understand biological processes in our own bodies. Don't worry if it seems tricky at first, we'll break it all down together. Let's get started!

1. The Basics: Enthalpy and Energy Changes

Every chemical reaction involves an energy change. The Law of Conservation of Energy states that energy cannot be created or destroyed, only changed from one form to another. In chemistry, we are often interested in the heat energy that is either taken in from the surroundings or released into them.

Meet the Stars: Exothermic vs. Endothermic Reactions

Chemical reactions can be sorted into two main teams based on how they handle heat energy.

Exothermic Reactions: The Heat Givers

These reactions release heat energy into the surroundings.
The surroundings get warmer.
Example: Burning wood in a campfire, the reaction in a hand warmer, or the neutralisation of an acid with an alkali.
Think: EXOthermic = Energy EXITS the system.

Endothermic Reactions: The Heat Takers

These reactions absorb heat energy from the surroundings.
The surroundings get colder.
Example: The reaction in an instant cold pack, photosynthesis, or dissolving some salts like ammonium nitrate in water.
Think: ENdothermic = Energy ENTERS the system.

Introducing Enthalpy Change (ΔH)

Chemists use a special term to describe the heat change of a reaction at constant pressure: enthalpy change, with the symbol ΔH (pronounced 'delta H').

The sign of ΔH is super important!

For an exothermic reaction, heat is released, so the products have less enthalpy than the reactants. The change is negative. ΔH is negative (-).
For an endothermic reaction, heat is absorbed, so the products have more enthalpy than the reactants. The change is positive. ΔH is positive (+).

Visualising Energy: Enthalpy Profile Diagrams

We can draw simple graphs to show these energy changes. These are called enthalpy profile diagrams.

Exothermic Reaction Diagram:

The reactants start with high energy. As the reaction happens, they release energy, and the products end up at a lower energy level. The difference in energy is the negative ΔH.

Reactants → Products + Heat ($$ \Delta H < 0 $$)

Endothermic Reaction Diagram:

The reactants start with low energy. They need to absorb energy from the surroundings to react, so the products end up at a higher energy level. The difference in energy is the positive ΔH.

Reactants + Heat → Products ($$ \Delta H > 0 $$)

Key Takeaway for Section 1

Exothermic: Releases heat, surroundings get hot, ΔH is negative.
Endothermic: Absorbs heat, surroundings get cold, ΔH is positive.
Enthalpy profile diagrams visually show the energy difference between reactants and products.

2. A Deeper Look: Energy and Chemical Bonds

So, where does this energy come from? It's all about breaking and making chemical bonds!

Think of it like building with LEGO bricks:

Bond Breaking: To break a chemical bond, you need to put energy IN. Imagine pulling apart two LEGO bricks that are stuck together – it takes effort! This is an endothermic process.
Bond Forming: When new chemical bonds form, energy is released. Imagine two magnetic LEGO bricks snapping together – they release energy as they become stable. This is an exothermic process.

Every reaction involves both breaking old bonds (in reactants) and forming new bonds (in products). The overall enthalpy change (ΔH) is the net result of these two processes.

If more energy is released forming bonds than is needed to break bonds, the reaction is exothermic (ΔH is -).
If less energy is released forming bonds than is needed to break bonds, the reaction is endothermic (ΔH is +).

Key Takeaway for Section 2

Bond Breaking REQUIRES Energy (Endothermic).
Bond Forming RELEASES Energy (Exothermic).
The overall ΔH depends on the balance between the energy required for bond breaking and the energy released by bond forming.

3. Let's Get Specific: Standard Enthalpy Changes

To compare enthalpy changes fairly, scientists agreed on a set of standard conditions. When a reaction is measured under these conditions, we call it a standard enthalpy change and give it the symbol ΔH^⊖.

What are Standard Conditions?

A pressure of 1 atmosphere (atm).
A temperature of 298 K (which is 25 °C).
For solutions, a concentration of 1.0 mol dm^-3.

Three Key Types You MUST Know

1. Standard Enthalpy Change of Formation (ΔH_f^⊖)

Definition: The enthalpy change when one mole of a compound is formed from its elements in their standard states.
Example: The formation of one mole of liquid water from hydrogen gas and oxygen gas.
$$ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H_f^{\ominus} = -286 \, kJ \, mol^{-1} $$
Important Note: The ΔH_f^⊖ of any element in its standard state (like O₂(g) or C(graphite)) is zero.

2. Standard Enthalpy Change of Combustion (ΔH_c^⊖)

Definition: The enthalpy change when one mole of a substance is completely burned in excess oxygen under standard conditions.
Combustion is always exothermic, so ΔH_c^⊖ is always negative.
Example: The complete combustion of one mole of methane gas.
$$ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \quad \Delta H_c^{\ominus} = -890 \, kJ \, mol^{-1} $$

3. Standard Enthalpy Change of Neutralisation (ΔH_neut^⊖)

Definition: The enthalpy change when one mole of water is formed from the reaction of an acid and an alkali under standard conditions.
Neutralisation is always exothermic, so ΔH_neut^⊖ is always negative.
Example: The reaction between hydrochloric acid and sodium hydroxide.
$$ H^+(aq) + OH^-(aq) \rightarrow H_2O(l) \quad \Delta H_{neut}^{\ominus} \approx -57 \, kJ \, mol^{-1} $$

Key Takeaway for Section 3

Standard conditions allow for fair comparisons (298 K, 1 atm, 1M).
Remember the "one mole" rule in the definitions of formation, combustion, and neutralisation (of water).

4. In the Lab: Measuring Heat - Calorimetry

How do we actually measure these heat changes? We use a technique called calorimetry. In school, this is often done using a simple apparatus like a polystyrene cup (a good insulator) with a lid and a thermometer. This is called a simple calorimeter.

The Magic Formula: q = mcΔT

To calculate the heat change, we use this crucial formula:

$$ q = mc\Delta T $$

Where:

q = the heat energy absorbed or released (in Joules, J).
m = the mass of the substance being heated or cooled (usually water or a solution, in grams, g). For dilute solutions, we often assume the density is 1 g cm^-3, so volume in cm³ = mass in g.
c = the specific heat capacity of the substance (for water, this is 4.2 J g^-1 K^-1). This means it takes 4.2 Joules of energy to raise the temperature of 1 gram of water by 1 Kelvin (or 1 °C).
ΔT = the change in temperature (in K or °C). $$ \Delta T = T_{final} - T_{initial} $$

Step-by-Step Calculation Guide

Here's how you use experimental data to find the enthalpy change (ΔH):

Step 1: Find the heat change (q)

Use the formula $$ q = mc\Delta T $$ to find the heat absorbed by the solution in Joules.

Step 2: Find the number of moles

Calculate the number of moles of the reactant that was the limiting reagent (e.g., the acid in a neutralisation reaction).

Step 3: Calculate the enthalpy change (ΔH)

ΔH is the heat change per mole, usually in kJ mol^-1. So, divide q by the number of moles. Don't forget to convert J to kJ (divide by 1000) and get the sign right!

$$ \Delta H = -\frac{q}{moles} $$

Common Mistake Alert!

The sign of ΔH is opposite to the temperature change of the water.
If the water temperature goes UP (ΔT is +), the reaction was EXOTHERMIC, so ΔH must be NEGATIVE.
If the water temperature goes DOWN (ΔT is -), the reaction was ENDOTHERMIC, so ΔH must be POSITIVE.
This is why we put a negative sign in the final formula: $$ \Delta H = -\frac{q}{moles} $$

Also, remember that simple calorimetry experiments are prone to errors, mainly heat loss to the surroundings, which makes the measured temperature change smaller than the true value.

5. Hess's Law: The Chemist's Clever Shortcut

Some reactions are impossible to measure directly in the lab (e.g., they are too slow, too dangerous, or produce other products). This is where Hess's Law comes to the rescue!

What is Hess's Law?

Hess's Law states that the total enthalpy change for a reaction is independent of the route taken.

Analogy: Hiking a Mountain

Imagine you are climbing a mountain. The total change in your altitude from the bottom to the top is the same, whether you take a short, steep path (Route 1) or a long, winding path (Route 2). The start and end points are all that matter.

In chemistry, the "altitude" is enthalpy, and the "paths" are different reaction sequences.

Using Hess's Law: The Energy Cycle Method

We can use Hess's Law to build simple diagrams called energy cycles (or Hess cycles) to find an unknown enthalpy change.

The key is to link the reactants and products of your target reaction through an alternative route where you know all the enthalpy changes (e.g., using known ΔH_f^⊖ or ΔH_c^⊖ values).

Let's say we want to find the ΔH for the reaction A → B.

Route 1 (the direct, unknown route): A → B (ΔH₁)

Route 2 (the indirect, known route): We find a common intermediate, C, that both A and B can be formed from or can form. For example, using enthalpy of formation data, the elements would be the intermediate. Using enthalpy of combustion data, the combustion products (CO₂ and H₂O) would be the intermediate.

A → C (ΔH₂) and B → C (ΔH₃).

According to Hess's Law, the energy change for Route 1 must equal the energy change for Route 2. When constructing the cycle, we follow the arrows:

$$ \Delta H_1 = \Delta H_2 - \Delta H_3 $$ (Note: we subtract ΔH₃ because we are going against its arrow, from C to B).

Let's Calculate! - A Worked Example

Question: Calculate the standard enthalpy change of formation of methane (CH₄), given the following standard enthalpies of combustion:

$$ \Delta H_c^{\ominus}[C(s)] = -394 \, kJ \, mol^{-1} $$
$$ \Delta H_c^{\ominus}[H_2(g)] = -286 \, kJ \, mol^{-1} $$
$$ \Delta H_c^{\ominus}[CH_4(g)] = -890 \, kJ \, mol^{-1} $$

Step 1: Write the target equation.

The formation of methane from its elements:

$$ C(s) + 2H_2(g) \rightarrow CH_4(g) \quad (\Delta H_f^{\ominus} = ?) $$

Step 2: Construct the Hess Cycle using the combustion data.

The reactants (C + 2H₂) and the product (CH₄) all burn in oxygen to form the same combustion products (CO₂ + 2H₂O).

We can draw a cycle where the top arrow is our target reaction, and the downward arrows represent combustion (our indirect route).

Step 3: Calculate the enthalpy changes for the indirect route (Route 2).

Combusting the reactants: This involves burning 1 mole of Carbon and 2 moles of Hydrogen gas.
$$ \Delta H_{reactants} = \Delta H_c^{\ominus}[C] + 2 \times \Delta H_c^{\ominus}[H_2] $$
$$ \Delta H_{reactants} = (-394) + 2 \times (-286) = -394 - 572 = -966 \, kJ $$
Combusting the product: This is just the given ΔH_c^⊖ for methane.
$$ \Delta H_{product} = \Delta H_c^{\ominus}[CH_4] = -890 \, kJ $$

Step 4: Apply Hess's Law.

Route 1 (Target Reaction) + Route 2 (Product Combustion) = Route 2 (Reactant Combustion)

$$ \Delta H_f^{\ominus} + (\Delta H_c^{\ominus}[CH_4]) = (\Delta H_c^{\ominus}[C] + 2 \times \Delta H_c^{\ominus}[H_2]) $$

$$ \Delta H_f^{\ominus} + (-890) = (-966) $$

$$ \Delta H_f^{\ominus} = -966 - (-890) $$

$$ \Delta H_f^{\ominus} = -966 + 890 = -76 \, kJ \, mol^{-1} $$

Key Takeaway for Section 5

Hess's Law: The total enthalpy change is the same no matter which route you take.
Energy Cycles are a powerful visual tool for solving problems.
Be careful with signs and stoichiometry (the numbers in front of the formulas)! If you have to multiply an equation by 2, you must also multiply its ΔH value by 2.