Question

determine the molar enthalpy of combustion using the following calorimetric data:
mass of calorimeter: 22.3 g
mass of calorimeter and water: 222.3 g
initial temperature of water: 21.5 °c
final temperature of water: 36.1 °c
mass of fuel burned: 0.243 g
type of fuel: propane c3h8

Explanation:

Step1: Calculate mass of water

Mass of water = Mass of calorimeter and water - Mass of calorimeter = \(222.3\space g - 22.3\space g = 200\space g\)

Step2: Calculate heat absorbed by water

Using \(Q = mc\Delta T\), where \(m = 200\space g\), \(c = 4.18\space J/g^\circ C\), \(\Delta T = 36.1^\circ C - 21.5^\circ C = 14.6^\circ C\)
\(Q_{water}=200\space g\times4.18\space J/g^\circ C\times14.6^\circ C = 200\times4.18\times14.6\space J = 12133.6\space J\)
Assume calorimeter has same specific heat as water (simplification), heat absorbed by calorimeter \(Q_{calorimeter}=22.3\space g\times4.18\space J/g^\circ C\times14.6^\circ C\approx22.3\times4.18\times14.6\space J\approx1332.5\space J\)
Total heat released \(Q = Q_{water}+Q_{calorimeter}=12133.6 + 1332.5 = 13466.1\space J = 13.4661\space kJ\)

Step3: Calculate moles of propane

Molar mass of \(C_3H_8\): \(3\times12.01 + 8\times1.008 = 36.03 + 8.064 = 44.094\space g/mol\)
Moles \(n=\frac{0.243\space g}{44.094\space g/mol}\approx0.00551\space mol\)

Step4: Calculate molar enthalpy of combustion

\(\Delta H=\frac{-Q}{n}=\frac{-13.4661\space kJ}{0.00551\space mol}\approx - 2444\space kJ/mol\) (negative as combustion is exothermic)

Answer:

Approximately \(-2440\space kJ/mol\) (or more precise \(-2444\space kJ/mol\))