Answer is: enthalpy of combustion per mole of anthracene is -7073.21 kJ/mol.
Balanced chemical rreaction of anthracene combustion:
C₁₄H₁₀ + 33/2O₂ → 14CO₂ + 5H₂O.
m(C₁₄H₁₀) = 0.455 g; mass of anthracene.
ΔT(H₂O) = 8.63°C.
m(H₂O) = 500.0 g; mass of water.
cp = 4.18 J/g·°C; specific heat of water.
n(C₁₄H₁₀) = m(C₁₄H₁₀) ÷ M(C₁₄H₁₀).
n(C₁₄H₁₀) = 0.455 g ÷ 178.23 g/mol.
n(C₁₄H₁₀) = 0.00255 mol; amount of anthracene.
Q = ΔT(H₂O) · m(H₂O) · cp.
Q = 8.63°C · 500 g · 4.18 J/g·°C.
Q = 18036.7 J ÷ 1000 J/kJ.
Q = 18.037 kJ; heat absorb by water.
ΔH = Q ÷ n(C₁₄H₁₀).
ΔH = 18.037 kJ ÷ 0.00255 mol.
ΔH = 7073.21 kJ/mol.
7073.21 kJ/mole is the enthalpy of combustion per mole of anthracene in a bomb calorimeter.
Enthalpy of combustion for this reaction can be calculated as:
ΔH = Q/n, where
Q = heat absorbed or released
n = no. of moles
First we calculate the no. of moles of anthracene by using the formula:
n = W/M, where
W = given mass = 0.455g
M= molar mass = 178.23 g/mol
n = 0.455g / 178.23 g/mol = 0.00255 mol
Now we calculate the heat absorbed by water because heat absorbed by calorimeter is negligible, as follow:
Q = mcΔT, where
m = mass of water = 500g (given)
c = specific heat of water = 4.18J/g°C (given)
ΔT = change in temperature of water = 8.63°C (given)
Q = 500g × 4.18J/g°C × 8.63°C
Q = 18036.7 J = 18.0367 kJ
Now enthalpy of combustion per mole of anthracene can be calculated, by putting values on the above equation we get:
ΔH = 18.037 kJ / 0.00255 mol.
ΔH = 7073.21 kJ/mol.
Hence, 7073.21 kJ/mol is the enthalpy of combustion.
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