Answer to Question #92423 in Chemistry for Tj

Question

6.40 kJ of energy is needed to raise the temperature of a calorimeter and its contents 1.00 oC. When 2.00 g of propyne (C3H4(g)) is burned in a calorimeter, the temperature of the calorimeter was raised 15.0 oC. What is the molar heat of combustion of propyne?

(Please note that you must include units throughout all of your calculations and provide your answer in scientific notation to the correct number of significant digits.)

Answer 1

According to the combustion equation:

Q₁ = ΔH_comb × N = ΔH_comb × (m_propyne / Mr_propyne),

where Q₁ - heat of propyne combustion, ΔH_comb - molar heat of combustion, N - number of moles, m_propyne - propyne mass, Mr_propyne - propyne molecular weight.

According to the heat equation:

Q₂ = m_cal × c_p × ΔT,

where Q₂ - heat absorbed by a calorimeter and its contents, m_cal - mass of a calorimeter and its contents, c_p - specific heat of calorimeter and its content, ΔT - temperature change.

As Q₁ = Q₂:

ΔH_comb × (m_propyne / Mr_propyne) = m_cal × c_p × ΔT.

From here:

ΔH_comb = (Mr_propyne × m_cal × c_p × ΔT) / m_propyne,

where Mr_propyne = 40 g/mol, ΔT = 15°C, m_propyne = 2 g.

As Q₀ = m_cal × c_p × ΔT_1°C, m_cal × c_p = Q₀ / ΔT_1°C,

where Q₀ - energy needed to raise the temperature of a calorimeter and its contents 1 °C.

As a result:

ΔH_comb = [Mr_propyne × (Q₀ / ΔT_1°C)× ΔT] / m_propyne = [40 g/mol × 6.4 kJ/°C × 15°C] / 2 g = 1920 kJ/mol.

Answer: -1920 kJ/mol.