In this section, we delve into the practical applications of Hess's Law, focusing on its utility in calculating enthalpy changes of reactions using standard enthalpies of formation (ΔHf⦵) and combustion (ΔHc⦵). Hess's Law is a powerful tool in physical chemistry, providing a pathway to understand and calculate energy changes in chemical reactions.
Understanding Hess's Law
Hess's Law states that the total enthalpy change for a chemical reaction is the same, no matter how many steps the reaction is carried out in. In other words, the enthalpy change is a state function, dependent only on the initial and final states of the system, not on the path taken.
Using ΔHf⦵ to Calculate Enthalpy Changes
The standard enthalpy of formation (ΔHf⦵) is the enthalpy change when one mole of a compound is formed from its elements in their standard states.
Example Calculations
1. Combustion of Methane: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
- Given Data:
- ΔHf⦵[CO2(g)] = -394 kJ/mol
- ΔHf⦵[H2O(l)] = -286 kJ/mol
- ΔHf⦵[CH4(g)] = -75 kJ/mol
- ΔHf⦵[O2(g)] = 0 kJ/mol (as it’s a diatomic element in its standard state)
- Calculation:
- ΔH = [ΔHf⦵ products] - [ΔHf⦵ reactants]
- ΔH = [(-394) + 2(-286)] - [(-75) + 0]
- ΔH = -1301 kJ/mol
2. Formation of Water: H2(g) + 1/2 O2(g) → H2O(l)
- Given Data:
- ΔHf⦵[H2O(l)] = -286 kJ/mol
- ΔHf⦵[H2(g)] = 0 kJ/mol (as it’s a diatomic element in its standard state)
- ΔHf⦵[O2(g)] = 0 kJ/mol (as it’s a diatomic element in its standard state)
- Calculation:
- ΔH = [ΔHf⦵ products] - [ΔHf⦵ reactants]
- ΔH = [-286] - [0 + (1/2)(0)]
- ΔH = -286 kJ/mol
Key Points to Remember
- Ensure all substances are in their standard states.
- Pay careful attention to the stoichiometric coefficients in the balanced equation.
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Using ΔHc⦵ to Calculate Enthalpy Changes
The standard enthalpy of combustion (ΔHc⦵) is the enthalpy change when one mole of a substance is completely burnt in oxygen under standard conditions.
Example Calculations
- Combustion of Ethene: C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l)
- Given Data:
- ΔHc⦵[C2H4(g)] = -1411 kJ/mol
- Calculation:
- Since the combustion of one mole of ethene releases 1411 kJ of energy, the ΔH for the reaction is -1411 kJ/mol.
- Given Data:
Key Points to Remember
- ΔHc⦵ values are always negative, as combustion is an exothermic process.
- Ensure the balanced equation represents the combustion of one mole of the substance.
Applying Equations for Enthalpy Change Calculations
Understanding and applying equations is crucial for calculating enthalpy changes accurately.
General Approach
- Identify the Type of Reaction: Determine whether the data provided is ΔHf⦵ or ΔHc⦵, and whether you are dealing with formation or combustion.
- Write the Balanced Equation: Ensure all reactants and products are in their standard states and the equation is balanced.
- Apply Hess's Law: Use the equation ΔH = [ΔHf⦵ products] - [ΔHf⦵ reactants] or ΔH = ΔHc⦵.
- Perform the Calculation: Be meticulous with units and signs to ensure accuracy.]
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Tips for Success
- Units: Ensure all your enthalpy values are in the same units before performing calculations.
- Signs: Pay careful attention to the signs of your enthalpy values. Remember that exothermic reactions (like combustion) will have negative ΔH values, while endothermic reactions will have positive ΔH values.
- Stoichiometry: Ensure that the stoichiometric coefficients in your balanced equation match those of your calculation. If a coefficient is fractional, remember to divide the enthalpy value by the same number.
By following these steps and tips, and with sufficient practice, you’ll be able to confidently and accurately use Hess's Law to calculate enthalpy changes in a variety of chemical reactions, enhancing your understanding of energy cycles in reactions. Remember, practice is key to mastering these calculations and solidifying your understanding of Hess's Law and its applications.
FAQ
Hess's Law is based on the principle that enthalpy is a state function, meaning it depends only on the initial and final states of a system, not the path taken. While Hess’s Law itself can be applied to reactions under non-standard conditions, the standard enthalpy values (ΔHf⦵ and ΔHc⦵) used in the calculations are specific to standard conditions. To apply Hess’s Law to reactions under non-standard conditions, one would need enthalpy values determined under those specific conditions, which may not always be readily available.
Balancing chemical equations ensures that the law of conservation of mass is upheld, meaning the number of atoms for each element is the same on both sides of the equation. When applying Hess’s Law, it is crucial that the equations for the individual reactions are correctly balanced so that the enthalpy changes associated with each reaction are properly accounted for. If the equations are not balanced, the calculated overall enthalpy change will be incorrect, leading to inaccurate and unreliable results.
The enthalpy change of a reaction provides insight into the energy changes involved, but it does not directly determine spontaneity. Spontaneity is influenced by both enthalpy and entropy changes, as well as temperature, according to Gibbs free energy (ΔG = ΔH - TΔS). A negative ΔG indicates a spontaneous reaction. However, it is possible for a reaction to be endothermic (positive ΔH) and still be spontaneous if it is accompanied by a significant increase in entropy (ΔS).
Standard enthalpy values, such as ΔHf⦵ and ΔHc⦵, are determined under standard conditions (100 kPa and 298 K). However, reactions rarely occur under these exact conditions in real-life scenarios. Temperature, pressure, and concentration can all impact the enthalpy change of a reaction. Thus, while standard enthalpy values provide a useful reference point and allow for the comparison of different reactions, they may not always accurately predict the enthalpy change in practical situations. Care should be taken when applying these values to ensure that the conditions of the reaction are considered.
The sign of the enthalpy change for a reaction indicates whether the reaction is exothermic (releases energy) or endothermic (absorbs energy). If the enthalpy of the products is lower than the enthalpy of the reactants, the reaction releases energy to the surroundings, resulting in a negative ΔH, characteristic of exothermic reactions. Conversely, if the enthalpy of the products is higher than the enthalpy of the reactants, the reaction absorbs energy from the surroundings, resulting in a positive ΔH, indicative of an endothermic reaction. The sign of ΔH is determined by comparing the enthalpy of the reactants and products, often using standard enthalpy values.
Practice Questions
To find the enthalpy change for the reaction, we apply Hess's Law and use the equation ΔH = [ΔHf⦵ products] - [ΔHf⦵ reactants]. The ΔHf⦵ for NH3(g) is -46 kJ/mol, and for N2(g) and H2(g) it is 0 kJ/mol as they are in their standard states. Substituting the values, we get ΔH = [2(-46)] - [0 + 3(0)] = -92 kJ. The enthalpy change for the reaction is -92 kJ, indicating that the synthesis of ammonia is an exothermic process.
To calculate the enthalpy change of the reaction, we use Hess's Law and the equation ΔH = [ΔHf⦵ products] - [ΔHf⦵ reactants]. Substituting the given values, ΔH = [6(-394) + 6(-286)] - [-1273 + 6(0)] = [-2364 - 1716] - [-1273] = -4080 + 1273 = -2807 kJ. Therefore, the enthalpy change for the combustion of glucose is -2807 kJ, indicating an exothermic reaction. This calculation demonstrates the release of energy during the combustion of glucose.
