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 +).
Calculating Enthalpy Changes from Bond Enthalpies

The mean bond enthalpy is the average energy required to break one mole of a particular covalent bond in gaseous molecules.

We can estimate the overall enthalpy change of a gaseous reaction using the formula:

\( \Delta H = \sum \text{Bond Enthalpies of bonds broken} - \sum \text{Bond Enthalpies of bonds formed} \)

Note: Calculated values from bond enthalpies are approximate because mean bond enthalpies are averaged over many different compounds rather than measured specifically for one exact molecule.

Key Takeaway for Section 2
  • Bond Breaking REQUIRES Energy (Endothermic).
  • Bond Forming RELEASES Energy (Exothermic).
  • \( \Delta H = \sum \text{BE(broken)} - \sum \text{BE(formed)} \).

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 (1 atm / 101 kPa).
  • 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 (ΔHf⊖)

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

2. Standard Enthalpy Change of Combustion (ΔHc⊖)

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

3. Standard Enthalpy Change of Neutralisation (ΔHneut⊖)

  • Definition: The enthalpy change when one mole of water is formed from the neutralisation reaction between an acid and an alkali under standard conditions.
  • Neutralisation is always exothermic, so ΔHneut⊖ is always negative.
  • Example: The reaction between hydrochloric acid and sodium hydroxide.
  • \( H^+(aq) + OH^-(aq) \rightarrow H_2O(l) \quad \Delta H_{neut}^{\ominus} \approx -57.1 \text{ kJ mol}^{-1} \)
Key Takeaway for Section 3
  • Standard conditions allow for fair comparisons (298 K, 1 atm, 1.0 M).
  • Remember the "one mole" rule in the definitions of formation, combustion, and neutralisation (per mole of water formed).

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 done using a simple apparatus like an expanded polystyrene cup (a good heat insulator) with a lid and a thermometer.

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 solution being heated or cooled (in grams, g). For dilute aqueous solutions, we assume the density is 1.0 g cm-3, so volume in cm3 = mass in g.
  • c = the specific heat capacity of the solution (for water/aqueous solutions, this is 4.2 J g-1 K-1).
  • ΔT = the change in temperature (in K or °C). \( \Delta T = T_{final} - T_{initial} \)
Correcting for Heat Loss: Cooling Curves (Temperature-Time Graphs)

In calorimetry, some heat is inevitably lost to the surroundings as the reaction proceeds, causing the observed maximum temperature to be lower than the true theoretical value. To correct for this:

  • Record the initial temperature of the reactant(s) for a few minutes (e.g. every 30 seconds) to establish a steady baseline.
  • Mix the reactants at a specific time (e.g. at the 3rd minute) without recording temperature at that exact instant.
  • Resume recording the temperature at regular intervals as the mixture heats up and then cools down.
  • Plot a temperature-time graph (cooling curve).
  • Extrapolate the cooling curve back to the time of mixing to determine the theoretical maximum temperature, and calculate the corrected \( \Delta T \).
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 energy in Joules.

Step 2: Find the number of moles

Calculate the number of moles of the limiting reactant.

Step 3: Calculate the enthalpy change (ΔH)

ΔH is the heat change per mole, usually in kJ mol-1. Divide q by the number of moles, convert J to kJ (divide by 1000), and add the correct sign:

\( \Delta H = -\frac{q}{n} \)

Common Mistake Alert!

The sign of ΔH is opposite to the temperature change of the solution.
If the solution temperature goes UP (ΔT is +), the reaction was EXOTHERMIC, so ΔH must be NEGATIVE.
If the solution temperature goes DOWN (ΔT is -), the reaction was ENDOTHERMIC, so ΔH must be POSITIVE.
This is why we apply the sign rule: \( \Delta H = -\frac{q}{n} \)


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 incomplete). 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 chemical reaction is independent of the route taken, provided the initial and final conditions are the same.

Analogy: Hiking a Mountain

Imagine you are climbing a mountain. The total change in your altitude from base camp to the summit is identical whether you take a steep, direct path (Route 1) or a winding, indirect path (Route 2). The initial and final states are all that matter.

Using Hess's Law: The Energy Cycle Method

We use Hess's Law to construct energy cycles (Hess cycles) connecting reactants and products via an alternative route using known standard enthalpy values.

  • Using Standard Enthalpies of Formation: The alternative route goes through the constituent elements in their standard states.
    \( \Delta H^{\ominus}_{reaction} = \sum \Delta H_f^{\ominus}(\text{products}) - \sum \Delta H_f^{\ominus}(\text{reactants}) \)
  • Using Standard Enthalpies of Combustion: The alternative route goes through the common combustion products (e.g. CO2 and H2O).
    \( \Delta H^{\ominus}_{reaction} = \sum \Delta H_c^{\ominus}(\text{reactants}) - \sum \Delta H_c^{\ominus}(\text{products}) \)
Let's Calculate! - A Worked Example

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

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

Step 1: Write the target equation.

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

Step 2: Construct the Hess Cycle using combustion data.

Both sides combust in oxygen to give \( CO_2(g) + 2H_2O(l) \).

Step 3: Calculate enthalpy changes for the indirect route.

  • Combusting reactants: \( \Delta H_{\text{reactants}} = \Delta H_c^{\ominus}[C] + 2 \times \Delta H_c^{\ominus}[H_2] = -394 + 2(-286) = -966 \text{ kJ} \)
  • Combusting products: \( \Delta H_{\text{products}} = \Delta H_c^{\ominus}[CH_4] = -890 \text{ kJ} \)

Step 4: Apply Hess's Law.

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

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

\( \Delta H_f^{\ominus} = -966 - (-890) = -76 \text{ kJ mol}^{-1} \)

Key Takeaway for Section 5
  • Hess's Law: Total enthalpy change is independent of the pathway.
  • Energy Cycles: Ensure arrows and stoichiometric coefficients are carefully accounted for when calculating unknown \( \Delta H \).