Biphenyl (C1 2H1 0) burned in a bomb calorimeter in
order to determine its heat of combustion. The heat
capacity of the calorimeter was 5.86 kJ/°C When
0.514 g of biphenyl was burned, the temperature of
the calorimeter increased from 25.8 °C to 29.4 °C.
Determine the enthalpy of combustion for biphenyl.
the correct answer provided is -6330kj/mol. I just want help with the solution.
To determine the enthalpy of combustion for biphenyl, we can use the equation:
ΔH = q/n
Where:
ΔH = enthalpy change (in this case, enthalpy of combustion)
q = heat absorbed by the calorimeter
n = number of moles of biphenyl burned
To calculate q, we can use the equation:
q = C * ΔT
Where:
q = heat absorbed (in this case, heat absorbed by the calorimeter)
C = heat capacity of the calorimeter
ΔT = change in temperature
Given:
C = 5.86 kJ/°C
ΔT = 29.4 °C - 25.8 °C = 3.6 °C
Substituting the given values into the equation, we can calculate q:
q = 5.86 kJ/°C * 3.6 °C = 21.096 kJ
Now, let's determine the number of moles of biphenyl used in the combustion:
molar mass of biphenyl (C12H10) = (12.01 g/mol * 12) + (1.008 g/mol * 10) = 154.224 g/mol
Given:
mass of biphenyl = 0.514 g
n = mass/molar mass = 0.514 g / 154.224 g/mol = 0.003334 mol
Now we can substitute the values of q and n into the equation ΔH = q/n:
ΔH = 21.096 kJ / 0.003334 mol ≈ -6330 kJ/mol
Thus, the enthalpy of combustion for biphenyl is approximately -6330 kJ/mol.