Topic 8: Energetics I
Welcome to Energetics I! Whether you love calculations or find them a bit daunting, this guide breaks down every single concept into simple, bite-sized steps. Chemical energetics is all about tracking heat energy during chemical reactions: where energy comes from, where it goes, and how we measure it. Let's dive in!
---1. Fundamentals of Enthalpy
What is Enthalpy Change?
In chemistry, we cannot easily measure the total internal energy of a substance. However, we can measure the energy exchanged with the surroundings during a reaction at constant pressure. This is called the enthalpy change, represented by the symbol \(\Delta H\).
• Enthalpy Change (\(\Delta H\)): The heat energy change measured under conditions of constant pressure.
Standard Conditions (\(\theta\))
To compare reactions fairly across the world, scientists measure enthalpy changes under agreed standard conditions, shown by the standard symbol \(^\theta\):
• Standard Pressure: \(100\text{ kPa}\) (\(1\text{ bar}\))
• Standard Temperature: \(298\text{ K}\) (\(25^\circ\text{C}\))
• Standard Concentration (for solutions): \(1.00\text{ mol dm}^{-3}\)
• Standard State: The physical state (solid, liquid, or gas) that a substance naturally exists in under \(100\text{ kPa}\) and \(298\text{ K}\) (for example, \(\text{H}_2\text{O}\) is a liquid, \(\text{O}_2\) is a gas, and \(\text{C}\) is solid graphite).
Exothermic vs. Endothermic Reactions
Think of chemical bonds as energy stores:
• Exothermic Reactions: Heat energy is released to the surroundings. The temperature of the surroundings increases. Because the chemicals lose energy, the enthalpy change is negative (\(\Delta H < 0\)).
Everyday example: Hand warmers, burning natural gas in a boiler.
• Endothermic Reactions: Heat energy is absorbed from the surroundings. The temperature of the surroundings decreases. Because the chemicals gain energy, the enthalpy change is positive (\(\Delta H > 0\)).
Everyday example: Chemical cold packs used for sports injuries.
Quick Memory Aid: EXothermic = Energy EXits (\(-\)). ENdothermic = Energy enters IN (\(+\)).
Enthalpy Level Diagrams vs. Reaction Profile Diagrams
Examiners frequently test this distinction:
• Enthalpy Level Diagram: Shows only the initial enthalpy of the reactants and the final enthalpy of the products, along with an arrow showing the overall \(\Delta H\) with its sign. Crucial rule: Activation energy (\(E_a\)) is not drawn on an enthalpy level diagram.
• Reaction Profile Diagram: Shows the continuous energy pathway taken by the molecules, including the "energy hump" which represents the activation energy (\(E_a\)).
Key Takeaway: If an exam question asks for an enthalpy level diagram, simply draw horizontal lines for reactants and products with a vertical arrow for \(\Delta H\). Do not draw a curved activation energy hump!
---2. The Four Standard Enthalpies You Must Memorise
In Edexcel chemistry, definitions must be quoted accurately. Make sure to learn these four standard definitions word-for-word:
1. Standard Enthalpy Change of Reaction (\(\Delta_r H^\theta\))
The enthalpy change when a reaction occurs in the molar quantities shown in the chemical equation under standard conditions (\(100\text{ kPa}\), \(298\text{ K}\)), with all reactants and products in their standard states.
2. Standard Enthalpy Change of Formation (\(\Delta_f H^\theta\))
The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions (\(100\text{ kPa}\), \(298\text{ K}\)).
Example: \(2\text{C(s, graphite)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH(l)}\)
Important Fact: By definition, the standard enthalpy change of formation of any pure element in its standard state is zero (\(\Delta_f H^\theta = 0\text{ kJ mol}^{-1}\)), because no chemical change is needed to form an element from itself!
3. Standard Enthalpy Change of Combustion (\(\Delta_c H^\theta\))
The enthalpy change when one mole of a substance is burned completely in oxygen under standard conditions (\(100\text{ kPa}\), \(298\text{ K}\)), with all reactants and products in their standard states.
Example: \(\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}\)
Notice: The substance being burned must always have a stoichiometric balancing coefficient of 1.
4. Standard Enthalpy Change of Neutralisation (\(\Delta_{neut} H^\theta\))
The enthalpy change when one mole of water is formed in the neutralisation reaction between an acid and an alkali under standard conditions (\(100\text{ kPa}\), \(298\text{ K}\)).
Example: \(\frac{1}{2}\text{H}_2\text{SO}_4\text{(aq)} + \text{NaOH(aq)} \rightarrow \frac{1}{2}\text{Na}_2\text{SO}_4\text{(aq)} + \text{H}_2\text{O(l)}\)
3. Calorimetry: Measuring Heat Changes Experimentally
We measure enthalpy changes experimentally using an instrument called a calorimeter.
The Heat Energy Equation
\(Q = mc\Delta T\)
• \(Q\) = heat energy transferred (in Joules, \(\text{J}\))
• \(m\) = mass of the substance heated or cooled (in grams, \(\text{g}\)). For aqueous solutions, we assume a density of \(1.00\text{ g cm}^{-3}\), so \(1\text{ cm}^3 = 1\text{ g}\).
• \(c\) = specific heat capacity of the liquid (for water or dilute solutions, \(c = 4.18\text{ J g}^{-1}\text{ K}^{-1}\))
• \(\Delta T\) = temperature change (\(T_{\text{final}} - T_{\text{initial}}\) in \(\text{K}\) or \(^\circ\text{C}\))
Converting \(Q\) to Molar Enthalpy Change (\(\Delta H\))
Once \(Q\) is calculated, convert it into \(\text{kJ mol}^{-1}\) using the formula:
\(\Delta H = -\frac{Q}{n \times 1000}\)
• Dividing by \(1000\) converts Joules (\(\text{J}\)) into kilojoules (\(\text{kJ}\)).
• \(n\) = amount in moles of the limiting reactant (or the reactant specified by the definition).
• The Minus Sign: If the temperature increases (\(\Delta T > 0\)), heat was released (exothermic), so \(\Delta H\) must be negative. If temperature drops, \(\Delta H\) is positive.
Two Common Experimental Contexts
Context A: Reactions in Aqueous Solution (e.g., Neutralisation or Displacement)
Performed in an expanded polystyrene cup because polystyrene is a good thermal insulator and has a very low heat capacity.
• Standard Assumptions: Specific heat capacity is \(4.18\text{ J g}^{-1}\text{ K}^{-1}\); solution density is \(1.00\text{ g cm}^{-3}\); heat absorbed by the cup is negligible.
• Cooling Curve Method (Extrapolation): To correct for heat loss to the room, record the temperature of the solution every minute for 3 minutes before adding the second reactant at minute 4. Stir and continue taking readings every minute from minute 5 to 10. Plot temperature against time, draw lines of best fit before and after mixing, and extrapolate back to minute 4 to determine the theoretical maximum temperature change (\(\Delta T\)).
Context B: Enthalpy of Combustion (Spirit Burner and Copper Calorimeter)
A known mass of water in a copper beaker is heated by burning a liquid fuel in a spirit burner.
• Why experimental \(\Delta_c H\) values are less negative than data book values:
1. Heat loss: Significant heat escapes to the surrounding air and draughts rather than entering the water.
2. Incomplete combustion: Soot (carbon) or carbon monoxide forms instead of carbon dioxide, releasing less energy.
3. Fuel evaporation: Fuel evaporates from the wick before and after extinguishing while weighing.
4. Calorimeter heat capacity: Some heat is absorbed by the copper container itself and not transferred to the water.
4. Hess's Law & Enthalpy Cycles
What is Hess's Law?
Hess's Law: The total enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final conditions are the same.
Analogy: Climbing a mountain from base camp to the summit results in the exact same change in elevation, whether you take the direct steep path or a winding path around the side.
Cycle 1: Using Enthalpy of Formation Data (\(\Delta_f H^\theta\))
When given \(\Delta_f H^\theta\) values in an exam question, place the constituent elements in a box at the bottom. The arrows must point UPWARDS from elements to compounds.
\(\Delta_r H^\theta = \sum \Delta_f H^\theta(\text{products}) - \sum \Delta_f H^\theta(\text{reactants})\)
Cycle 2: Using Enthalpy of Combustion Data (\(\Delta_c H^\theta\))
When given \(\Delta_c H^\theta\) values, place the combustion products (\(\text{CO}_2\) and \(\text{H}_2\text{O}\)) at the bottom. The arrows must point DOWNWARDS from reactants and products to combustion products.
\(\Delta_r H^\theta = \sum \Delta_c H^\theta(\text{reactants}) - \sum \Delta_c H^\theta(\text{products})\)
Handy Memory Trick:
• Formation: Products minus Reactants (think "For Public Relations" → \(\text{F} = \text{P} - \text{R}\))
• Combustion: Reactants minus Products (think "Chemical Reaction Progress" → \(\text{C} = \text{R} - \text{P}\))
5. Core Practical 8: Determining an Enthalpy Change Using Hess's Law
Some reactions cannot have their enthalpy changes measured directly. A classic example is the thermal decomposition of solid potassium hydrogencarbonate:
\(2\text{KHCO}_3\text{(s)} \xrightarrow{\Delta H_r} \text{K}_2\text{CO}_3\text{(s)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\)
Why can't we measure this directly? It requires continuous strong heating to decompose, making it impossible to measure the exact temperature change of the solid directly with a thermometer.
The Indirect Method
We react both the reactant and the product separately with excess hydrochloric acid (\(\text{HCl}\)) in an insulated polystyrene cup:
• Reaction 1: \(\text{KHCO}_3\text{(s)} + \text{HCl(aq)} \rightarrow \text{KCl(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\) → gives \(\Delta H_1\)
• Reaction 2: \(\text{K}_2\text{CO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow 2\text{KCl(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}\) → gives \(\Delta H_2\)
Applying the Hess Cycle
By constructing a Hess cycle linking both reactions via the common solution of aqueous potassium chloride (\(\text{KCl(aq)}\)), the overall decomposition enthalpy is calculated as:
\(\Delta H_r = 2\Delta H_1 - \Delta H_2\)
Don't forget: We multiply \(\Delta H_1\) by 2 because the balanced decomposition equation contains \(2\text{ moles}\) of \(\text{KHCO}_3\).
---6. Bond Enthalpies
Definitions
• Bond Enthalpy (Bond Dissociation Enthalpy): The energy required to break one mole of a specific covalent bond in a gaseous molecule to form gaseous atoms under standard conditions.
• Mean Bond Enthalpy: The energy required to break one mole of a specific type of covalent bond, averaged over a wide range of different gaseous molecules.
Calculating Reaction Enthalpy from Bond Enthalpies
Breaking bonds is always endothermic (requires energy, \(+\)). Making bonds is always exothermic (releases energy, \(-\)).
\(\Delta_r H = \sum(\text{Bond enthalpies of bonds broken}) - \sum(\text{Bond enthalpies of bonds made})\)
Memory Aid: Break minus Make (\(\Delta H = \text{Broken} - \text{Made}\)).
Limitations of Mean Bond Enthalpies
Calculations using mean bond enthalpies are only approximations. Discrepancies between calculated values and experimental values arise because:
1. Average environment: Mean bond enthalpies are averaged across many different compounds. In reality, the exact strength of a bond depends slightly on the neighboring atoms in that specific molecule.
2. Physical state assumption: Bond enthalpy definitions strictly apply only to substances in the gaseous state. If any reactant or product is a liquid (like \(\text{H}_2\text{O(l)}\) or \(\text{Br}_2\text{(l)}\)) or a solid, extra energy is involved in state changes (enthalpy of vaporisation or condensation), causing differences.
7. Common Pitfalls to Avoid in the Exam
• Missing the Sign: Always include an explicit \(+\) or \(-\) sign in your final \(\Delta H\) answer (e.g., write \(-57.2\text{ kJ mol}^{-1}\), not just \(57.2\text{ kJ mol}^{-1}\)).
• Wrong Mass in \(Q = mc\Delta T\): When a solid dissolves in water, \(m\) is the mass of the water/solution (e.g. \(50.0\text{ cm}^3 \approx 50.0\text{ g}\)), not the mass of the solid or the mass of fuel in a spirit burner.
• Forgetting to Divide by 1000: \(Q\) is calculated in Joules (\(\text{J}\)), but \(\Delta H\) must be given in kilojoules per mole (\(\text{kJ mol}^{-1}\)). Always divide \(Q\) by \(1000\).
• Limiting Reactant Moles: Always divide \(Q\) by the moles of the reactant that is limiting, not the one in excess.
• State Symbols in Definitions: Never omit the phrases "in their standard states" or "under standard conditions" when writing standard enthalpy definitions!