Answer to Question #139229 in General Chemistry for Flame

Question

1.435 g of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter. The temperature of the water increases from 20.17 oC to 25.84 oC. If the mass of water surrounding the calorimeter is 2000 g and the heat capacity of the bomb calorimeter is 1.80 kJ oC-1, calculate the heat of combustion of naphthalene in kJ mol-1. Consider the heat of reaction is the summation of heat of water and heat of bomb calorimeter.

Answer 1

The temperature of the water increases from 20.17 °C to 25.84 °C

Change of temperature,

∆T = (25.84-20.17) °C

= 5.67 °C

Mass of the water, M = 2000 g

We, know specific heat of the water,

S = 1 Cal/°C

=4.2 J/°C/g

Heat absorbs by the water,

H1 = M×S×∆T

= 2000 g × 4.2 J/°C/g × 5.67 °C

= 47628 J

= 47.628 kJ

Since, the bomb calorimeter is in the tharmal equilibrium with the water. So, change of temperature of the calorimeter,

∆T = (25.84-20.17) °C

= 5.67 °C

Heat capacity of the calorimeter,

S' = 1.80 kJ/°C

Heat absorbs by the calorimeter,

H2 = S' × ∆T

= 1.80 kJ/°C × 5.67 °C

= 10.206 kJ

Total heat produce = H1 + H2

= (47.627 + 10.206) kJ

= 57.833 kJ

Hence, heat of combustion of 1.435 g of naphthalene = 57.833 kJ

Molar mass of napthalene = 128 g/mol

heat of combustion of 1.435 g of naphthalene = 57.833 kJ

heat of combustion of 1 g of napthalene

= (57.833 kJ / 1.435 g)

Heat of combustion of 128 g of napthalene

= (57.833 kJ × 128 g/mol) / 1.435 g

= 5158.623 kJ/mol

Hence, molar heat of combustion of napthalene = 5158.623 kJ mol^-1