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.

be sure your answer has a unit symbol, if necessary, and round it to 3 significant digits.

ΔHf =  305 kJ/mol

Explanation:

1. Amount of heat absorbed by the water

Q = m × C × ΔT

Where:

Q is the heat absorbed: to determinem is the mass of water: 25.00 kg = 25,000 gSpecific heat of water: 4.186 J/g °CΔT is the increase of the temperature: 2.161 ºCQ = 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 XQ released in the reaction = - Q absorbed by water = -226,148.65 Jnumber of moles = mass in grams / molar massmolar mass of C₄H₆ = 4 × 12.011 g/mol + 6 × 1.008 g/mol = 54.092 g/molnumber of moles = 5.000 g / 54.092 g/mol = 0.09243511 molΔHrxn =  - 226,148.65 J / 0.09243511 mol =  -2,446,566.55 J / molΔHrxn ≈  - 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/molΔHf 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

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