Question

If 1.785 g of ethanol (CHCHOH) is burned in a constant volume calorimeter causing a temperature increase of 4.32C, then what is the molar heat of combustion in units of kJ/mol of ethanol? (Heat capacity of the calorimeter is 9.49 kJ/C.)

Answer 1

Answer: The enthalpy of the reaction is -1056.44 kJ

Explanation:

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

$q=c\Delta T$

where,

q = heat absorbed

c = heat capacity of calorimeter = 9.49 kJ/°C

$\Delta T$ = change in temperature =  4.32°C

Putting values in above equation, we get:

$q=9.49kJ/^oC\times 4.32^oC=40.99kJ$

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:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of ethanol = 1.785 g

Molar mass of ethanol = 46 g/mol

Putting values in above equation, we get:

$\text{Moles of ethanol}=\frac{1.785g}{46g/mol}=0.0388mol$

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

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = -40.99 kJ

n = number of moles of ethanol = 0.0388 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-40.99kJ}{0.0388mol}=-1056.44kJ/mol$

Hence, the enthalpy of the reaction is -1056.44 kJ