Question

5.000 g of Compound X with molecular formula C₄H₆ are burned in a constant-pressure calorimeter containing 25.00 kg of water at 25°C. The temperature of the water is observed to rise by 2.161 °C. (You may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) Calculate the standard heat of formation of Compound X at 25°C.round it to 3 significant digits.

Answer 1

Answer:

• ΔHf =  305 kJ/mol

Explanation:

1. Amount of heat absorbed by the water

• Q = m × C × ΔT

Where:

• Q is the heat absorbed: to determine
• m is the mass of water: 25.00 kg = 25,000 g
• Specific heat of water: 4.186 J/g °C
• ΔT is the increase of the temperature: 2.161 ºC

• Q = 25,000 g × 4.186 J/gºC × 2.161 ºC = 226,148.65 J

2. Standard heat of formation of compound X: ΔHf

• ΔH rxn = ∑ΔHf (products) - ∑Δf (reactants)

Reaction:

• C₄H₆ (g) + 11/2 O₂ (g)  →  4CO₂ (g) + 3H₂O (g)

• ΔHrxn = 4ΔHf CO₂ (g) + 3 ΔHf H₂O (g) - ΔHf C₄H₆(g)

• ΔHrxn = Q released in the reaction / number of moles of compound X

• Q released in the reaction = - Q absorbed by water = -226,148.65 J

• number of moles = mass in grams / molar mass

• molar mass of C₄H₆ = 4 × 12.011 g/mol + 6 × 1.008 g/mol = 54.092 g/mol

• number of moles = 5.000 g / 54.092 g/mol = 0.09243511 mol

• ΔHrxn =  - 226,148.65 J / 0.09243511 mol =  -2,446,566.55 J / mol

• ΔHrxn ≈  - 2,446 Kj / mol ≈ - 2,450 kJ / mol

• - 2,450 kJ/mol = 4ΔHf CO₂ (g) + 3 ΔHf H₂O (g) - ΔHf C₄H₆(g)

From tables (at 25ºC):

• ΔHf CO₂ (g) = - 393.5 kJ/mol
• ΔHf H₂O (g) = -241.8 kJ/mol

- 2,450 kJ/mol = 4 (-393.5 kJ/mol) + 3 (-393.5 kJ/mol) - ΔHf C₄H₆

ΔHf C₄H₆ = 2,450 kJ/mol - 2,754.5 = 304.5 kJ/mol ≈ 305 kJ/mol