Question

determine the molar enthalpy of combustion using the following calorimetric data:
mass of calorimeter: 25.8 g
mass of calorimeter and water: 125.8 g
initial temperature of water: 20.2 ºc
final temperature of water: 37.8 ºc
mass of fuel burned: 0.147 g
type of fuel: ethene c2h4

Explanation:

Step1: Calculate mass of water

Mass of water = Mass of calorimeter and water - Mass of calorimeter = \(125.8\ \text{g} - 25.8\ \text{g} = 100.0\ \text{g}\)

Step2: Calculate temperature change (\(\Delta T\))

\(\Delta T = \text{Final temperature} - \text{Initial temperature} = 37.8^\circ\text{C} - 20.2^\circ\text{C} = 17.6^\circ\text{C}\)

Step3: Calculate heat absorbed by water (\(q_{\text{water}}\))

Using \(q = mc\Delta T\), where \(c = 4.18\ \text{J/g}^\circ\text{C}\) (specific heat of water), \(m = 100.0\ \text{g}\), \(\Delta T = 17.6^\circ\text{C}\)
\(q_{\text{water}} = 100.0\ \text{g} \times 4.18\ \text{J/g}^\circ\text{C} \times 17.6^\circ\text{C} = 7356.8\ \text{J}\)
Assume heat absorbed by calorimeter is negligible or same as water (simplified), so total heat released by fuel (\(q_{\text{combustion}}\)) is \(-q_{\text{water}} = -7356.8\ \text{J}\) (negative as combustion releases heat)

Step4: Calculate moles of ethene (\(C_2H_4\))

Molar mass of \(C_2H_4\): \(2\times12.01 + 4\times1.008 = 28.052\ \text{g/mol}\)
Moles (\(n\)) = \(\frac{\text{Mass of fuel}}{\text{Molar mass}} = \frac{0.147\ \text{g}}{28.052\ \text{g/mol}} \approx 0.00524\ \text{mol}\)

Step5: Calculate molar enthalpy of combustion (\(\Delta H_{\text{combustion}}\))

\(\Delta H_{\text{combustion}} = \frac{q_{\text{combustion}}}{n} = \frac{-7356.8\ \text{J}}{0.00524\ \text{mol}} \approx -1.404\times10^6\ \text{J/mol} = -1404\ \text{kJ/mol}\) (converted to kJ by dividing by 1000)

Answer:

Approximately \(-1400\ \text{kJ/mol}\) (or more precisely \(-1404\ \text{kJ/mol}\))