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.
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Answer:
the final volume of the gas is = 1311.5 mL
Explanation:
Given that:
a sample gas has an initial volume of 61.5 mL
The workdone = 130.1 J
Pressure = 783 torr
The objective is to determine the final volume of the gas.
Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.
Converting the external pressure to atm ; we have
External Pressure :
The workdone W = V
The change in volume ΔV=
ΔV =
ΔV =
ΔV = 1.25 L
ΔV = 1250 mL
Recall that the initial volume = 61.5 mL
The change in volume V is
multiply through by (-), we have:
= 1250 mL + 61.5 mL
= 1311.5 mL
∴ the final volume of the gas is = 1311.5 mL