Question

A 0.287-g sample of bianthracene (C28H18) is burned in a bomb calorimeter and the temperature increases from 25.30 °C to 27.50 °C. The calorimeter contains 1.03E3 g of water and the bomb has a heat capacity of 856 J/°C. Based on this experiment, calculate ΔE for the combustion reaction per mole of bianthracene burned (kJ/mol).

Answer #1

heat released by the combusiton of 0.287 g of bianthracene = heat gained by water

= mass of water * T * Cp of water

given mass of water = 1.03 * 10^{3} g = 1030 g

Cp of water = 4 J/g.C

T = 27.5 - 25.3 = 2.2 C

heat gained by water = 1030 g * 4 J/g.C * 2.2 C = 9064 J

heat gained by bomb = 856 J/C * (27.5 - 25.3) C = 1883.2 J

Total heat change = 9064 J + 1883.2 J = 10947.2 J = 10.95 kJ

Mass of bianthracene = 0.287 g

Molar mass of bianthracene = 354 g/mol

No. of moles of bianthracene = 0.287 g/ 354 g/mol = 8.11 *
10^{-4} moles

E = 10.95 kJ /
8.11 * 10^{-4} moles = 13506.3 kJ/mol Answer

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