Question

At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ / mol. When 1.697 g of compound A (molar mass = 101.67 g / mol ) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 ∘ C. What is the heat capacity (calorimeter constant) of the calorimeter? C = kJ/°C Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23 ∘ C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?

Answer 1

Answer:

13.85 kJ/°C

-14.89 kJ/g

Explanation:

At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter?

The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:

$1.697g.\frac{1mol}{101.67g} .\frac{(-3039.0kJ)}{mol} =-50.72kJ$

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

Qcomb + Qcal = 0

Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ

The heat capacity (C) of the calorimeter can be calculated using the following expression.

Qcal = C . ΔT

where,

ΔT is the change in the temperature

Qcal = C . ΔT

50.72 kJ = C . 3.661 °C

C = 13.85 kJ/°C

Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?

Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ

The heat of combustion per gram of B is:

$\frac{-56.09 kJ}{3.767g} =-14.89 kJ/g$