Table B: Enthalpy of Formation of Reactants and Products
1 answer:
The calculated enthalpy values are as follows:
- Total enthalpy of reactants = -103.85 KJ/mol
- Total enthalpy of products = -2057.68 KJ/mol
- Enthalpy of reaction = -1953.83 kJ/mol
The enthalpy of the reaction is determined as follows:
- Enthalpy of reaction = Total enthalpy of products -Total enthalpy of reactants
- Total enthalpy of reactants = (ΔHf of Reactant 1 x Coefficient) + (ΔHf of Reactant 2 x Coefficient)
- Total enthalpy of products= (ΔHf of Product 1 x Coefficient) + (ΔHf of Product 2 x Coefficient)
Equation of reaction equation: C₃H₈ (g) + 5 O(g) → 4 H₂O(g) + 3CO₂(g)
Total enthalpy of reactants = (-103.85 * 1) + (0 * 5)
Total enthalpy of reactants = -103.85 + 0
Total enthalpy of reactants = -103.85 KJ/mol
Total enthalpy of products = (-393.51 * 4) +(-241.82 * 3)
Total enthalpy of products = (-1574.04) + (-483.64)
Total enthalpy of products = -2057.68 KJ/mol
Enthalpy of reaction = -2057.68 KJ/mol -(-103.85 KJ/mol)
Enthalpy of reaction = -1953.83 kJ/mol
In conclusion, the enthalpy of the reaction is determined from the difference between the total enthalpy of products and reactants.
Learn more about enthalpy of reaction at: brainly.com/question/14047927
#SPJ1
You might be interested in
Answer : The molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.
Solution :
At STP,
22.4 L volume of ethane present in 1 mole of ethane gas
64.28 L volume of ethane present in of ethane gas
And, as we know that
1 mole of ethane molecule contains molecules of ethane
2.869 moles of ethane molecule contains molecules of ethane
Therefore, the molecule of ethane present in 64.28 L of ethane gas at STP is, molecule.