Measuring enthalpy changes: R1.1.4 Standard enthalpy change IB DP Chemistry Study Notes - New Syllabus 2025
Measuring enthalpy changes – IB DP Chemistry- Study Notes
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Reactivity 1.1.4 – Standard Enthalpy Change of a Reaction (ΔH)
Reactivity 1.1.4 – Standard Enthalpy Change of a Reaction (ΔH)
Definition
The standard enthalpy change of a reaction, denoted as \( \Delta H^\circ \) (read as “delta H naught”), is the amount of heat energy transferred during a chemical reaction that occurs under standard conditions. It measures the energy released or absorbed by the system as bonds are broken and formed during the reaction.
Standard Conditions
A reaction is said to occur under standard conditions if it meets the following criteria:
- Pressure: 100 kPa (1 bar)
- Temperature: 298 K (25°C)
- Concentration (if in solution): 1 mol dm−3
- Physical state: All substances must be in their standard physical states
Under these conditions, the enthalpy change can be meaningfully compared across different reactions.
Standard Enthalpy Changes
| Type | Definition | Symbol | Sign |
|---|---|---|---|
| Standard Enthalpy of Reaction | Energy change when reactants form products as written in the balanced equation under standard conditions | \( \Delta H^\circ_r \) | Exo or Endo |
| Standard Enthalpy of Formation | Energy change when 1 mol of a compound is formed from its elements in their standard states | \( \Delta H^\circ_f \) | Exo or Endo |
| Standard Enthalpy of Combustion | Energy change when 1 mol of a substance burns completely in oxygen under standard conditions | \( \Delta H^\circ_c \) | Always Exothermic |
| Standard Enthalpy of Neutralization | Energy change when 1 mol of water is formed by reacting an acid with a base under standard conditions | \( \Delta H^\circ_{\text{neut}} \) | Always Exothermic |
Units
The enthalpy change is usually expressed in units of kilojoules per mole:
\( \text{kJ mol}^{−1} \)
Why Standard Enthalpy is Useful
Standard enthalpy values provide a consistent reference for comparing the energetics of different chemical reactions. By using \( \Delta H^\circ \), chemists can:
- Predict whether a reaction will release or absorb energy
- Compare the thermodynamic favorability of different reactions
- Use these values in calculations (e.g., Hess’s Law, calorimetry)
System and Surroundings in Enthalpy Change
In thermochemistry, we consider the system (the chemicals undergoing the reaction) and the surroundings (everything else, such as the container or the solution). Energy is transferred between the two:
- Exothermic reaction: System loses energy → surroundings gain heat → temperature increases
- Endothermic reaction: System absorbs energy → surroundings lose heat → temperature decreases
Standard States
Each substance has a standard state, which is the most stable form of the substance at 100 kPa and 298 K.
- For example: O2(g), H2O(l), C(graphite)
- These are used when tabulating standard enthalpy values like enthalpies of formation and combustion
Visualizing Enthalpy Change
The enthalpy change can be represented graphically in an energy profile diagram:
- Y-axis: Potential energy
- X-axis: Reaction progress (reaction coordinate)
- The difference between product and reactant energy levels corresponds to \( \Delta H^\circ \)
Understanding Enthalpy Calculations
The enthalpy change of a chemical process tells us how much energy is transferred as heat during a reaction, usually measured per mole of a specified substance. In laboratory settings, we calculate this by measuring the temperature change of a known mass and applying standard formulas.
1. Heat Energy Equation
\( Q = mc\Delta T \)
- Q: Heat energy (in joules, J)
- m: Mass of substance (in grams, g)
- c: Specific heat capacity (for water, \( 4.18 \, \text{J g}^{-1} \text{K}^{-1} \))
- \( \Delta T \): Temperature change in °C or K
2. Molar Enthalpy Change
\( \Delta H = \frac{-Q}{n} \)
- ΔH: Enthalpy change per mole (in kJ mol−1)
- Q: Heat energy (converted to kJ)
- n: Moles of limiting reactant (mol)
Sign Conventions
- Exothermic reactions: Heat is released, so \( \Delta H < 0 \)
- Endothermic reactions: Heat is absorbed, so \( \Delta H > 0 \)
- Negative sign in the equation accounts for this directionality
Common Assumptions in Lab Calculations
- The solution has the same specific heat capacity as water unless stated otherwise
- The density of aqueous solutions is approximated as \( 1.00 \, \text{g cm}^{−3} \)
- There is no heat loss to the environment (ideal assumption)
Example
Use the following data to calculate the enthalpy change (\( \Delta H \)) of the reaction in kJ mol⁻¹.
- Volume of solution: 50.0 g (assume density = 1.00 g/cm³)
- Initial temperature: 25.0°C
- Final temperature: 32.0°C
- Specific heat capacity: \( 4.18 \, \text{J g}^{-1} \, ^\circ \text{C}^{-1} \)
- Moles of limiting reagent: 0.050 mol
▶️Answer/Explanation
Step 1: Use the formula
$ q = mc\Delta T $ where:
- \( m = 50.0 \, \text{g} \)
- \( c = 4.18 \, \text{J g}^{-1} \, ^\circ \text{C}^{-1} \)
- \( \Delta T = 32.0 – 25.0 = 7.0^\circ \text{C} \)
$ q = 50.0 \times 4.18 \times 7.0 = 1463 \, \text{J} $
Step 2: Convert to kJ
$ q = 1.463 \, \text{kJ} $
Step 3: Calculate enthalpy change per mole
$ \Delta H = \frac{-q}{n} = \frac{-1.463}{0.050} = \boxed{-29.3 \, \text{kJ mol}^{-1}} $
Note: The negative sign indicates the reaction is exothermic.
Example
50.0 cm3 of 1.00 mol dm−3 HCl is mixed with 50.0 cm3 of 1.00 mol dm−3 NaOH in a polystyrene cup. The temperature rises from 25.0°C to 31.0°C. Calculate the enthalpy change of neutralization.
▶️Answer/Explanation
- Step 1 – Calculate total volume:
Total = 50.0 + 50.0 = 100.0 cm³ → Assume mass = 100.0 g (same as water) - Step 2 – Calculate heat released (Q):
\( Q = mc\Delta T = 100.0 \times 4.18 \times (31.0 – 25.0) = 100.0 \times 4.18 \times 6.0 = 2508 \, \text{J} = 2.51 \, \text{kJ} \) - Step 3 – Calculate moles of water formed:
Moles HCl = \( \frac{50.0}{1000} \times 1.00 = 0.0500 \, \text{mol} \)
Moles NaOH = 0.0500 mol
So, 0.0500 mol of water formed. - Step 4 – Calculate ΔH:
\( \Delta H = \frac{-Q}{n} = \frac{-2.51}{0.0500} = -50.2 \, \text{kJ mol}^{-1} \) - Final Answer: \( \Delta H = -50.2 \, \text{kJ mol}^{-1} \)
Example
0.500 g of ethanol (C2H5OH) is burned and heats 100.0 g of water from 20.0°C to 30.0°C. Calculate the experimental enthalpy of combustion of ethanol. (Molar mass = 46.0 g mol−1)
▶️Answer/Explanation
- Step 1 – Calculate Q:
\( Q = mc\Delta T = 100.0 \times 4.18 \times (30.0 – 20.0) = 100.0 \times 4.18 \times 10 = 4180 \, \text{J} = 4.18 \, \text{kJ} \) - Step 2 – Calculate moles of ethanol burned:
\( \text{Moles} = \frac{0.500}{46.0} = 0.0109 \, \text{mol} \) - Step 3 – Calculate ΔH:
\( \Delta H = \frac{-4.18}{0.0109} = -383.5 \, \text{kJ mol}^{-1} \) - Final Answer: \( \Delta H = -384 \, \text{kJ mol}^{-1} \) (3 sig. fig.)
Example
Calculate the standard enthalpy change of the reaction:
\( \text{C(s) + O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \)
Given:
- ΔH°f[CO₂(g)] = −393.5 kJ mol⁻¹
- ΔH°f[C(s)] = 0 (element)
- ΔH°f[O₂(g)] = 0 (element)
▶️Answer/Explanation
- Step 1 – Use the standard enthalpy of formation equation:
\( \Delta H^\circ_\text{reaction} = \sum \Delta H^\circ_f(\text{products}) – \sum \Delta H^\circ_f(\text{reactants}) \) - Step 2 – Plug in values:
\( \Delta H^\circ = [ -393.5 ] – [ 0 + 0 ] = -393.5 \, \text{kJ mol}^{-1} \) - Final Answer: \( \Delta H^\circ = -393.5 \, \text{kJ mol}^{-1} \)
This is an exothermic reaction, as expected for combustion.