Question

When 1.0 g of fructose, CH O (s), is burned in a bomb

calorimeter, the temperature of the calorimeter increases by
1.58°C. If the heat capacity of the calorimeter and its
contents is 9.90 kJ/°C, what is the enthalpy of the
combustion of 1 mole of fructose?

Answer 1

The enthalpy of combustion of 1 mole of fructose is 2829 kJ/mol.

Step-by-step explanation:

In bomb calorimetry, the heat produced by the combustion reaction is absorbed by the calorimeter and its contents. The heat released by the combustion of 1.0 g of fructose can be calculated using q = mcΔT, where q is the heat transfer, m is the mass, c is the heat capacity, and ΔT is the change in temperature.

Given that the mass of fructose is 1.0 g, the change in temperature is 1.58 °C, and the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, we can calculate the heat transfer using q = (1.0 g)(9.90 kJ/°C)(1.58 °C) = 15.72 kJ.

To determine the enthalpy of combustion of 1 mole of fructose, we need to convert the mass of fructose to moles. The molar mass of fructose is approximately 180.16 g/mol, so 1.0 g of fructose is equivalent to (1.0 g)/(180.16 g/mol) = 0.00555 mol.

To find the enthalpy of combustion of 1 mole of fructose, we divide the heat transfer by the number of moles: 15.72 kJ / 0.00555 mol = 2829 kJ/mol.