Question

A 0.553-g sample of diphenyl phthalate (C₂₀H₁₄O₄) is burned in a bomb calorimeter and the temperature increases from 24.40 °C to 27.57 °C. The calorimeter contains 1.08×10³ g of water and the bomb has a heat capacity of 877 J/°C. The heat capacity of water is 4.184 J g^-1°C^-1. Based on this experiment, calculate ΔE for the combustion reaction per mole of diphenyl phthalate burned.

______ kJ/mol

Answer 1

The heat given by combustion = heat absorbed by bomb calorimeter + heat absorbed by water

Thus

heat absorbed by bomb = heat capacity x differenc ein temperature

= 877J /C x 3.17C

= 2780.09 J

heat absorbed by water = mass x specific heat x difference in temperature

= 1.08x 10³ gx 4.18 J/g.C x3.17C

= 14310.648 J

Total heat given in combustion = 2780.09 J + 14310.648 J

= 17090.738 J

molar mass of diethyl pthalate = 318g/mol

Thus if 0.533 g sample on combustion gives 17090.738 J

1 mol (318g) sample on combustion can give = 318g /molx 17090.738 J/0.533g

=10196725 J

= 10196.73 kJ/mol

Thus the delta E of combustion of mole of diphenylphthalate = 10196.73 kJ/mol