Question

determine the molar enthalpy of combustion using the following calorimetric data:
mass of calorimeter: 38.4 g
mass of calorimeter and water: 338.4 g
initial temperature of water: 24.6 ºc
final temperature of water: 32.6 ºc
mass of fuel burned: 0.200 g
type of fuel: ethene c2h4

Explanation:

Step1: Calculate mass of water

Mass of water = Mass of calorimeter and water - Mass of calorimeter = \( 338.4\ g - 38.4\ g = 300\ g \)

Step2: Calculate heat absorbed by water

Using \( Q = mc\Delta T \), where \( c = 4.18\ J/g^\circ C \), \( m = 300\ g \), \( \Delta T = 32.6^\circ C - 24.6^\circ C = 8^\circ C \)
\( Q = 300\ g \times 4.18\ J/g^\circ C \times 8^\circ C = 10032\ J = 10.032\ kJ \)

Step3: Calculate moles of ethene (C₂H₄)

Molar mass of C₂H₄: \( (2\times12.01) + (4\times1.008) = 28.052\ g/mol \)
Moles \( n = \frac{0.200\ g}{28.052\ g/mol} \approx 0.00713\ mol \)

Step4: Calculate molar enthalpy of combustion

Assuming heat absorbed by water is equal to heat released by combustion (with sign for exothermic), \( \Delta H = -\frac{Q}{n} = -\frac{10.032\ kJ}{0.00713\ mol} \approx -1407\ kJ/mol \) (approximate, considering calorimeter heat might be neglected here)

Answer:

Approximately \(-1410\ kJ/mol\) (or more precise \(-1407\ kJ/mol\))