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 -6330 kj/mol.
I'm looking for help with the solution.
5.86 kJ/C x (29.4-25.8) = q
q/0.514 x (molar mass biphenyl/mol) = ?
To determine the enthalpy of combustion for biphenyl, we need to use the equation:
ΔH = q/m
where ΔH is the enthalpy change, q is the heat transferred, and m is the amount of substance burned.
In this case, we are given the heat capacity of the calorimeter, the mass of biphenyl burned, and the change in temperature.
First, we need to calculate the heat transferred (q) by using the equation:
q = CΔT
where C is the heat capacity of the calorimeter and ΔT is the change in temperature.
Given:
C = 5.86 kJ/°C
ΔT = 29.4 °C - 25.8 °C = 3.6 °C
Plugging in these values, we get:
q = 5.86 kJ/°C * 3.6 °C
q = 21.1 kJ
Now we need to calculate the amount of biphenyl burned (m) in moles. To do this, we need to convert the given mass (0.514 g) to moles. The molar mass of biphenyl is 154.23 g/mol.
m = 0.514 g / 154.23 g/mol
m = 0.00333 mol
Finally, we can calculate the enthalpy of combustion (ΔH) by dividing the heat transferred by the amount of substance burned:
ΔH = q/m
ΔH = 21.1 kJ / 0.00333 mol
ΔH ≈ -6330 kJ/mol
Therefore, the enthalpy of combustion for biphenyl is approximately -6330 kJ/mol.