Calculate the enthalpy of reaction for the following:
N2 + 3H2 -> 2NH3 they are all gases.
Is this reaction endothermic or exothermic?
Look up delta Hf values in your text or notes, then calculate
deltaHrxn = (n*DHproducts)-(DHreactants).
The sign is - for exothermic; + for endothermic.
Thanks!
I omitted an n in that. It should be
deltaHrxn = (n*DHfproducts) - (n*DHfreactants)
To calculate the enthalpy of reaction (ΔH) for a given reaction, you need to use the bond energy or enthalpy values for each of the compounds involved in the reaction.
In this case, you need to use the bond enthalpy values for N2, H2, and NH3. Bond enthalpy refers to the amount of energy required to break one mole of a specific type of bond in a gaseous molecule.
In general, the enthalpy change of a reaction can be calculated using the following equation:
ΔH = Σ (bond enthalpies of reactants) - Σ (bond enthalpies of products)
For the given reaction: N2 + 3H2 → 2NH3
The reactants are N2 and 3H2, and the product is 2NH3.
Step 1: Find the bond energy values for N2, H2, and NH3. These values can be found in resources like chemical handbooks or online databases. For this example, let's assume the following bond enthalpy values:
N-N (N2) bond energy = 945 kJ/mol
H-H (H2) bond energy = 436 kJ/mol
N-H (NH3) bond energy = 391 kJ/mol
Step 2: Calculate the sum of the bond enthalpies of the reactants and the products:
Reactants:
N2: 1 mol × (N-N bond energy) = 1 mol × 945 kJ/mol
H2: 3 mol × (H-H bond energy) = 3 mol × 436 kJ/mol
Products:
NH3: 2 mol × (N-H bond energy) = 2 mol × 391 kJ/mol
Step 3: Substitute the values into the general enthalpy change equation:
ΔH = Σ (bond enthalpies of reactants) - Σ (bond enthalpies of products)
ΔH = (1 mol × 945 kJ/mol) + (3 mol × 436 kJ/mol) - (2 mol × 391 kJ/mol)
Simplify the equation:
ΔH = 945 kJ + 1308 kJ - 782 kJ
ΔH = 1471 kJ
The enthalpy of reaction (ΔH) for the given reaction is 1471 kJ. The positive sign indicates an endothermic reaction since energy is absorbed or required for the reaction to occur.