Understanding Standard Enthalpy Change and Heat Calculations
What is Standard Enthalpy Change ($ΔH^\ominus$)?
Enthalpy change
The enthalpy change (ΔH) of a reaction quantifies the heat exchanged at constant pressure during a chemical or physical change.
- Chemical reactions are accompanied by energy transfers between the system (reactants and products) and the surroundings.
- When this measurement is conducted under standard conditions, defined as 298 K (25°C), 100 kPa, and reactants/products in their standard states, it is referred to as the standard enthalpy change($ΔH^\ominus$).
- Exothermic reactions: Release heat into the surroundings, resulting in a negative $ΔH$ (e.g., combustion of fuels).
- Endothermic reactions: Absorb heat from the surroundings, resulting in a positive $ΔH$ (e.g., melting of ice).
Units of Enthalpy Change
- The standard unit for $\Delta H$ is kilojoules per mole (kJ mol⁻¹).
- This unit reflects the energy change associated with one mole of reactant or product under standard conditions.
- Always ensure your final answer for $\Delta H$ is in kJ mol⁻¹.
- If $Q$ is calculated in joules, convert it to kilojoules by dividing by 1000.
Calculating Heat Transfer $Q$
The heat transferred during a reaction can be calculated using the formula: $$Q = mc\Delta T$$ where:
- $Q$: Heat transferred (in joules, J)
- $m$: Mass of the substance being heated (in grams, g)
- $c$: Specific heat capacity of the substance (in J g⁻¹ K⁻¹)
- $\Delta T$: Temperature change (in kelvin, K, or degrees Celsius, °C)
Breaking Down the Formula
- Mass $m$: This is the mass of the substance (often water in calorimetry experiments) that absorbs or loses heat.
- Specific Heat Capacity $c$: A property of the substance that indicates how much heat is required to raise the temperature of 1 g of the substance by 1 K. For water, $c = 4.18 \, \mathrm{J \ g^{-1} \ K^{-1}}$.
- Temperature Change $\Delta T $: Calculated as $\Delta T = T_{\text{final}} - T_{\text{initial}}$.
Suppose 50 g of water is heated from 20°C to 30°C. How much heat is absorbed by the water?
Solution
$$Q = mc\Delta T$$
$$Q = (50 \, \mathrm{g})(4.18 \, \mathrm{J g^{-1} K^{-1}})(30 - 20 \, \mathrm{K})$$
$$Q = 2090 \, \mathrm{J} \, \text{(or 2.09 kJ)}$$
Connecting Heat ($Q$) to Enthalpy Change ($ΔH$)
The relationship between heat transfer $Q$ and enthalpy change $ \Delta H $ is given by: $$\Delta H = -\frac{Q}{n}$$ where:
- $\Delta H$: Enthalpy change (in kJ mol⁻¹)
- $Q$: Heat released or absorbed (in joules or kilojoules)
- $n$: Number of moles of the limiting reactant (in moles)
Why the Negative Sign?
- The negative sign reflects the system’s perspective.
- If the system releases heat (exothermic reaction), $Q$ is positive, but $\Delta H$ is negative.
- Conversely, if the system absorbs heat (endothermic reaction), $Q$ is negative, but $\Delta H$ is positive.
Students often forget to include the negative sign when calculating $\Delta H$, leading to incorrect conclusions about whether a reaction is exothermic or endothermic.
When 1.15 g of lithium chloride $ \mathrm{LiCl} $ dissolves in 25.0 g of water, the temperature increases by 3.8 K. Calculate the enthalpy change of dissolution for 1 mole of $ \mathrm{LiCl} $.
Solution
Step 1: Calculate $$
$$Q = mc\Delta T$$
$$Q = (25.0 \, \mathrm{g})(4.18 \, \mathrm{J \, g^{-1} \, K^{-1}})(3.8 \, \mathrm{K})$$
$$Q = 397 \, \mathrm{J} \, \text{(or 0.397 kJ)}$$
Step 2: Determine the moles of $\mathrm{LiCl} $
Molar mass of $ \mathrm{LiCl} = 42.39 \, \mathrm{g \, mol^{-1}} $
$$n = \frac{\text{mass}}{\text{molar mass}} = \frac{1.15 \, \mathrm{g}}{42.39 \, \mathrm{g \, mol^{-1}}} = 0.0271 \, \mathrm{mol}$$
Step 3: Calculate $\Delta H $
$$\Delta H = -\frac{Q}{n} = -\frac{0.397 \, \mathrm{kJ}}{0.0271 \, \mathrm{mol}} = -14.6 \, \mathrm{kJ \, mol^{-1}}$$
The enthalpy change of dissolution is $-14.6 \, \mathrm{kJ \, mol^{-1}}$, indicating an exothermic process.
Practical Applications of Enthalpy Calculations
Calorimetry Experiments
- In lab settings, calorimeters (e.g., coffee-cup calorimeters) measure heat changes during reactions.
- These experiments provide valuable insights into reaction energetics, aiding in the design of efficient chemical processes.
- To measure the enthalpy of combustion of a fuel, a spirit burner containing the fuel is placed under a metal calorimeter filled with water.
- The fuel is ignited, and the heat released warms the water.
- By recording the change in temperature of the water and the mass of fuel burned, the heat transferred can be calculated using the formula: $$
Q = mc\Delta T
$$ where:- $ Q $ = heat transferred (in joules or kilojoules)
- $ m $ = mass of water (in kilograms)
- $ c $ = specific heat capacity of water ($ 4.18 \, \text{kJ} \, \text{kg}^{-1} \, \text{K}^{-1} $)
- $ \Delta T $ = temperature change of the water (in kelvin or degrees Celsius)
You burn 0.5 g of ethanol under a calorimeter containing 100 g of water. The temperature of the water rises by 15°C. Calculate the heat transferred to the water.
Given:
- $ m = 0.1 \, \text{kg} $
- $ c = 4.18 \, \text{kJ} \, \text{kg}^{-1} \, \text{K}^{-1} $
- $ \Delta T = 15 \, \text{K} $
Solution
- $$
Q = mc\Delta T = 0.1 \times 4.18 \times 15 = 6.27 \, \text{kJ}
$$ - This means 6.27 kJ of heat was transferred to the water.
Always ensure the thermometer is calibrated and that the reaction is stirred gently to distribute heat evenly.
- What does the sign of $\Delta H$ tell you about the energy transfer in a reaction?
- How could you minimize heat loss in a calorimetry experiment?
- Calculate the enthalpy change for the combustion of 1.00 g of ethanol $ \mathrm{C_2H_5OH} $ if the heat released increases the temperature of 200.0 g of water by 15.0°C. Assume $c_{\text{water}} = 4.18 \, \mathrm{J \, g^{-1} \, K^{-1}} $ and the molar mass of ethanol is $ 46.08 \, \mathrm{g \, mol^{-1}} $.
