Question

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.

Answer 1

-3.327 × 10⁶ J/mol

Step-by-step explanation:

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:

`(-1.412 * 10^(4) J  )/(0.7052g) .(166.14g)/(mol) =-3.327 * 10^(6) J/mol`