Question

report your answers to the nearest thousandth digit, in mj / mol (these were originally supposed to be in kj / mol, but all answers are off by a factor of 1000, so to get the correct answer please report in megajoules per mole). determine the molar enthalpy of combustion using the following calorimetric data: mass of calorimeter: 25.5 g mass of calorimeter and water: 325.5 g initial temperature of water: 23.6 ºc final temperature of water: 33.5 ºc mass of fuel burned: 0.284 g type of fuel: benzene c6h6 answer:

Explanation:

Step1: Calculate mass of water

Mass of water = Mass of calorimeter and water - Mass of calorimeter = \( 325.5\ g - 25.5\ g = 300.0\ g \)

Step2: Calculate temperature change

\(\Delta T = \text{Final temperature} - \text{Initial temperature} = 33.5^\circ C - 23.6^\circ C = 9.9^\circ C\)

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

Using \( q = mc\Delta T \), where \( c = 4.184\ J/g^\circ C \) (specific heat of water)
\( q_{water} = 300.0\ g \times 4.184\ J/g^\circ C \times 9.9^\circ C = 300.0 \times 4.184 \times 9.9\ J \)
\( q_{water} = 12458.88\ J = 12.45888\ kJ \) (assuming calorimeter heat capacity is negligible, so total heat absorbed is by water)

Step4: Calculate moles of benzene (fuel)

Molar mass of \( C_6H_6 \): \( 6\times12.01 + 6\times1.008 = 78.11\ g/mol \)
Moles of benzene, \( n = \frac{0.284\ g}{78.11\ g/mol} \approx 0.003636\ mol \)

Step5: Calculate molar enthalpy of combustion (\(\Delta H\))

Heat released by combustion = -Heat absorbed by water (sign indicates exothermic)
\( \Delta H = \frac{-q_{water}}{n} \) (convert \( q_{water} \) to MJ: \( 12.45888\ kJ = 0.01245888\ MJ \))
\( \Delta H = \frac{-0.01245888\ MJ}{0.003636\ mol} \approx -3.426\ MJ/mol \) (wait, but let's check again. Wait, the problem said answers were off by factor of 1000, originally in kJ/mol. Wait, maybe I messed up units. Wait, original calculation: if we calculate in kJ/mol first, then convert to MJ/mol (divide by 1000). Let's recalculate:

\( q_{water} = 300 \times 4.184 \times 9.9 = 300 \times 41.4216 = 12426.48\ J = 12.42648\ kJ \) (maybe my previous calculation had a typo)

Moles: \( 0.284 / 78.11 \approx 0.003636\ mol \)

Enthalpy in kJ/mol: \( -12.42648\ kJ / 0.003636\ mol \approx -3417.6\ kJ/mol \)

Convert to MJ/mol: \( -3417.6\ kJ/mol = -3.4176\ MJ/mol \approx -3.418\ MJ/mol \)? Wait, maybe I made a mistake in mass of water. Wait, mass of calorimeter and water is 325.5 g, calorimeter is 25.5 g, so water is 300 g, correct. Temperature change 9.9, correct. Specific heat 4.184, correct.

Wait, let's recalculate \( q_{water} \):

300 g * 4.184 J/g°C * 9.9°C = 300 * 4.184 * 9.9 = 300 * 41.4216 = 12426.48 J = 12.42648 kJ.

Moles of benzene: 0.284 g / 78.11 g/mol ≈ 0.0036359 mol.

Enthalpy: -12.42648 kJ / 0.0036359 mol ≈ -3417.7 kJ/mol = -3.4177 MJ/mol ≈ -3.418 MJ/mol. But wait, maybe the calorimeter has heat capacity? Wait, the problem doesn't give calorimeter heat capacity, so we assume it's negligible. Alternatively, maybe the original problem's factor of 1000: if the answer was supposed to be in kJ/mol, but is off by 1000, so we need to report in MJ/mol (which is kJ/mol / 1000). Wait, no: the problem says "these were originally supposed to be in kJ / mol, but all answers are off by a factor of 1000, so to get the correct answer please report in megajoules per mole". So if the correct value in kJ/mol is X, then report X / 1000 MJ/mol. Wait, maybe my initial approach was wrong. Let's do it again:

1. Calculate heat absorbed by water: \( q = mc\Delta T = 300g * 4.184 J/g°C * (33.5 - 23.6)°C = 300 * 4.184 * 9.9 = 12426.48 J = 12.42648 kJ \)

2. Moles of benzene: \( n = 0.284g / 78.11g/mol ≈ 0.003636 mol \)

3. Enthalpy of combustion (per mole) in kJ/mol: \( \Delta H = -q / n = -12.42648 kJ / 0.003636 mol ≈ -3417.6 kJ/mol \)

4. Now, the problem says "all answers are off by a factor of 1000, so to get the correct answer please report in megajoules per mole". So divide by 1000: \( -3417.6 kJ/mol = -3.4176 MJ/mol ≈ -3.418 MJ/mol \) (rounded to nearest thousandth)

Wait, but let's check with more precise calculation:

Molar mass of C6H6: 6*12.0096 + 6*1.00784 = 72.0576 + 6.04704 = 78.10464 g/mol

Moles: 0.284 / 78.10464 ≈ 0.003636 mol (same as before)

q_water: 300 * 4.184 * 9.9 = 300 * 41.4216 = 12426.48 J = 12.42648 kJ

ΔH = -12.42648 / 0.003636 ≈ -3417.6 kJ/mol = -3.4176 MJ/mol ≈ -3.418 MJ/mol (to nearest thousandth: 3.418? Wait, 3417.6 / 1000 = 3.4176, so -3.418 when rounded to thousandth? Wait, 3.4176 rounded to thousandth is 3.418 (since the fourth decimal is 6, which rounds up the third decimal 7 to 8? Wait, no: 3.4176, thousandth digit is 7, next digit is 6, so round up: 3.418. So ΔH ≈ -3.418 MJ/mol. But let's check if the calorimeter's heat is considered. Wait, the problem gives mass of calorimeter, but no heat capacity. So we assume calorimeter heat capacity is zero, so all heat goes to water. So the calculation is as above.

Answer:

\boxed{-3.418}