Question

determine the molar enthalpy of combustion using the following calorimetric data:
mass of calorimeter: 42.4 g
mass of calorimeter and water: 542.4 g
initial temperature of water: 20.8 °c
final temperature of water: 34.5 °c
mass of fuel burned: 0.616 g
type of fuel: cyclohexane c6h12

Explanation:

Step1: Calculate mass of water

Mass of water = Mass of calorimeter and water - Mass of calorimeter = \( 542.4\ g - 42.4\ g = 500.0\ g \)

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

\(\Delta T = \text{Final temperature} - \text{Initial temperature} = 34.5\ ^\circ\text{C} - 20.8\ ^\circ\text{C} = 13.7\ ^\circ\text{C}\)

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

Assume specific heat capacity of water (\(c\)) is \(4.18\ J/g^\circ\text{C}\). Using \(q = mc\Delta T\), \(q_{water} = 500.0\ g \times 4.18\ J/g^\circ\text{C} \times 13.7\ ^\circ\text{C}\)
\(q_{water} = 500.0 \times 4.18 \times 13.7 = 28323\ J = 28.323\ kJ\)

Step4: Calculate heat absorbed by calorimeter (assuming calorimeter is made of a material, but if we assume the calorimeter's heat capacity is negligible or if we consider the heat absorbed by calorimeter is included in the system, but usually for simplicity, if not given, we can assume the heat from combustion is absorbed by water and calorimeter. But here, if we assume the calorimeter's specific heat is, say, if we don't have its specific heat, maybe the problem assumes that the heat absorbed by the calorimeter is included or maybe we can proceed with the water's heat. Wait, actually, the molar enthalpy of combustion is the heat released per mole of fuel burned. So the heat released by the fuel (\(q_{combustion}\)) is equal to the heat absorbed by water (and calorimeter, but if calorimeter's mass and specific heat are not given, maybe we assume it's negligible or maybe the problem expects us to use only water. Wait, the mass of calorimeter is given, but its specific heat is not. Maybe the problem assumes that the calorimeter is made of a material with specific heat, but since it's not given, perhaps it's a simplification. Alternatively, maybe the problem expects us to calculate the heat absorbed by water and assume that's equal to the heat released by the fuel (with sign: \(q_{combustion} = -q_{water}\) since combustion releases heat, water absorbs it).

So \(q_{combustion} = -28323\ J\) (negative because fuel releases heat)

Step5: Calculate moles of cyclohexane (\(C_6H_{12}\))

Molar mass of \(C_6H_{12}\): \(6 \times 12.01 + 12 \times 1.008 = 72.06 + 12.096 = 84.156\ g/mol\)
Moles (\(n\)) = Mass / Molar mass = \(0.616\ g / 84.156\ g/mol \approx 0.00732\ mol\)

Step6: Calculate molar enthalpy of combustion (\(\Delta H_{combustion}\))

\(\Delta H_{combustion} = q_{combustion} / n = -28323\ J / 0.00732\ mol \approx -3.87 \times 10^6\ J/mol = -3870\ kJ/mol\) (approximate, considering possible errors in assumptions)

Wait, let's recalculate moles: \(0.616 / 84.156 \approx 0.00732\ mol\)
\(28323\ J / 0.00732\ mol \approx 3,869,262\ J/mol \approx 3870\ kJ/mol\), so \(\Delta H = -3870\ kJ/mol\) (negative because combustion is exothermic)

But let's check the calculations again:

Step3: \(500 \times 4.18 \times 13.7 = 500 \times 4.18 = 2090; 2090 \times 13.7 = 2090 \times 13 + 2090 \times 0.7 = 27170 + 1463 = 28633\ J\) (I think I miscalculated earlier: 13.7*4.18=57.266; 500*57.266=28633 J = 28.633 kJ)

Step5: Molar mass of \(C_6H_{12}\): 6*12.01=72.06, 12*1.008=12.096, total=84.156 g/mol. Moles: 0.616 / 84.156 ≈ 0.00732 mol (correct)

Step6: \(28633\ J / 0.00732\ mol ≈ 3,911,612\ J/mol ≈ 3910\ kJ/mol\), so \(\Delta H = -3910\ kJ/mol\) (approximate)

Wait, maybe the calorimeter's heat is included. Let's assume the calorimeter has a specific heat, but since it's not given, maybe the problem expects us to use the mass of calorimeter and assume its specific heat is, say, if it's made of metal like aluminum (specific heat 0.900 J/g°C). Then heat absorbed by calorimeter: \(q_{calorimeter} = 42.4\ g \times 0.900\ J/g°C \times 13.7\ °C = 42.4 \times 0.9 \times 13.7 ≈ 42.4 \times 12.33 ≈ 523\ J\)
Then total heat absorbed: \(28633\ J + 523\ J = 29156\ J = 29.156\ kJ\)
Then moles: 0.00732 mol
\(\Delta H = -29156\ J / 0.00732\ mol ≈ -3,983,000\ J/mol ≈ -3980\ kJ/mol\)

But since the problem doesn't give the calorimeter's specific heat, maybe it's a simplification and we are to use only water. Let's check standard values: The molar enthalpy of combustion of cyclohexane is about -3919 kJ/mol (varies slightly with conditions), so our calculation is in the ballpark.

So let's proceed with the water's heat (ignoring calorimeter for simplicity, as specific heat of calorimeter is not given):

\(q_{water} = 500\ g \times 4.18\ J/g°C \times 13.7\ °C = 500 \times 4.18 \times 13.7 = 28633\ J = 28.633\ kJ\)

Moles of \(C_6H_{12}\): \(n = \frac{0.616\ g}{84.156\ g/mol} ≈ 0.00732\ mol\)

Molar enthalpy of combustion: \(\Delta H = \frac{q_{combustion}}{n} = \frac{-28633\ J}{0.00732\ mol} ≈ -3910\ kJ/mol\) (rounded to appropriate significant figures: mass of fuel is 0.616 g (3 sig figs), temperature change 13.7 (3 sig figs), mass of water 500.0 (4 sig figs), so maybe 3 sig figs: -3910 kJ/mol or -3.91 × 10³ kJ/mol)

Answer:

The molar enthalpy of combustion of cyclohexane is approximately \(\boxed{-3910\ kJ/mol}\) (or more accurately, around -3920 kJ/mol depending on precise calculations, close to the standard value of ~-3919 kJ/mol)