The combustion of 1.760 g of propanol (C3H7OH) increases the temperature of a bomb calorimeter from 298.00K to 302.15K. The heat capacity of the bomb calorimeter is 14.24 kJ/K. Determine ΔH for the combustion of propanol to carbon dioxide gas and liquid water in kJ/mol.

The enthalpy of the reaction is -1709.76 kJ

Explanation:

To calculate the heat absorbed by the calorimeter, we use the equation:

where,

q = heat absorbed

c = heat capacity of calorimeter = 14.24 kJ/K

= change in temperature =

Putting values in above equation, we get:

Heat absorbed by the calorimeter will be equal to the heat released by the reaction.

Sign convention of heat:

When heat is absorbed, the sign of heat is taken to be positive and when heat is released, the sign of heat is taken to be negative.

To calculate the number of moles, we use the equation:

Given mass of propanol = 1.760 g

Molar mass of propanol = 60 g/mol

Putting values in above equation, we get:

To calculate the enthalpy change of the reaction, we use the equation:

where,

q = amount of heat released = -50.096 kJ

n = number of moles = 0.0293 moles

= enthalpy change of the reaction

Putting values in above equation, we get:

Hence, the enthalpy of the reaction is -1709.76 kJ