Question

A sample of hexane (C6H14) has a mass of 0.580 g. The sample is burned in a bomb calorimeter that has a mass of 1.900 kg and a specific heat of 3.21 J/giK. What amount of heat is produced during the combustion of hexane if the temperature of the calorimeter increases by 4.542 K? Use .

Answer 1

Answer: The enthalpy of the reaction is -4134.3 kJ/mol

Explanation:

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

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of calorimeter = 1.900 kg = 1900 g  (Conversion factor:  1 kg = 1000 g)

c = heat capacity of calorimeter = 3.21 J/g.K

$\Delta T$ = change in temperature = 4.542 K

Putting values in above equation, we get:

$q=1900g\times 3.21J/g.K\times 4.542K=27701.66J=27.7kJ$

Heat released by the calorimeter will be equal to the heat absorbed 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 hexane = 0.580 g

Molar mass of hexane = 86.18 g/mol

Putting values in above equation, we get:

$\text{Moles of hexane}=\frac{0.580g}{86.18g/mol}=0.0067mol$

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

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

where,

q = amount of heat released = -27.7 kJ

n = number of moles of hexane= 0.0067 moles

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

Putting values in above equation, we get:

$\Delta H_{rxn}=\frac{-27.7kJ}{0.0067mol}=-4134.3kJ/mol$

Hence, the enthalpy of the reaction is -4134.3 kJ/mol