Question

determine the molar enthalpy of combustion using the following calorimetric data:
mass of calorimeter: 39.0 g
mass of calorimeter and water: 139.0 g
initial temperature of water: 21.8 ºc
final temperature of water: 33.5 ºc
mass of fuel burned: 0.165 g
type of fuel: ethanol c2h5oh

Explanation:

Step 1: Calculate mass of water

Mass of water = Mass of calorimeter and water - Mass of calorimeter = \(139.0\ g - 39.0\ g = 100.0\ g\)

Step 2: Calculate heat absorbed by water

Using \(q = mc\Delta T\), where \(m = 100.0\ g\), \(c = 4.18\ J/g^\circ C\), \(\Delta T = 33.5^\circ C - 21.8^\circ C = 11.7^\circ C\)
\(q_{water} = 100.0\ g \times 4.18\ J/g^\circ C \times 11.7^\circ C = 4890.6\ J\)

Step 3: Assume heat absorbed by calorimeter is negligible (or same as water for simplicity here), so total heat released by fuel \(q_{combustion} = -q_{water}\) (negative as fuel releases heat)

\(q_{combustion} = -4890.6\ J\)

Step 4: Calculate moles of ethanol

Molar mass of \(C_2H_5OH\): \(2\times12.01 + 6\times1.008 + 16.00 = 46.07\ g/mol\)
Moles \(n = \frac{0.165\ g}{46.07\ g/mol} \approx 0.00358\ mol\)

Step 5: Calculate molar enthalpy of combustion (\(\Delta H\))

\(\Delta H = \frac{q_{combustion}}{n} = \frac{-4890.6\ J}{0.00358\ mol} \approx -1.366\times10^6\ J/mol = -1366\ kJ/mol\) (approximate, considering calorimeter heat might be included, but basic calculation)

Answer:

Approximately \(-1370\ kJ/mol\) (or more accurately \(-1.37\times10^3\ kJ/mol\) after precise calculation)