In an experiment, a 0.7052 g sample of phthalic acid (C8H6O4) is burned completely in a bomb calorimeter. The calorimeter is surrounded by 1.310×103 g of water. During the combustion the temperature increases from 21.77 to 24.01 °C. The heat capacity of water is 4.184 J g-1°C-1. The heat capacity of the calorimeter was determined in a previous experiment to be 823.0 J/°C. Assuming that no energy is lost to the surroundings, calculate the molar heat of combustion of phthalic acid based on these data.

According to the law of conservation of energy, the sum of the energy released by the combustion and the energy absorbed by the calorimeter and the water is zero.

qcomb + qcal + qw = 0

qcomb = - (qcal + qw) [1]

The heat absorbed by the calorimeter (qcal) can be calculated using the following expression:

qcal = Ccal . ΔT = 823.0 J/°C . (24.01°C - 21.77°C) = 1.844 × 10³ J

where,

Ccal is the heat capacity of the calorimeter

ΔT is the difference in temperature

The heat absorbed by the water (qw) can be calculated using the following expression:

qw = cw . mw . ΔT = 4.184 J.g⁻¹.°C⁻¹ . 1.310 × 10³ g . (24.01°C - 21.77°C) = 1.228 × 10⁴ J

where,

cw is the heat capacity of water

mw is the mass of water

ΔT is the difference in temperature

From [1],

qcomb = - (qcal + qw) = -(1.844 × 10³ J + 1.228 × 10⁴ J) = -1.412 × 10⁴ J

The molar heat of combustion of phtalic acid is:

[tex]\frac{-1.412 \times 10^{4} J }{0.7052g} .\frac{166.14g}{mol} =-3.327 \times 10^{6} J/mol[/tex]