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 is -6330 kj/mol
To determine the enthalpy of combustion for biphenyl, we can use the equation:
ΔH = q / n
Where:
ΔH = enthalpy change (in kJ/mol)
q = heat transferred (in kJ)
n = moles of the substance being burned
First, let's calculate the heat transferred, q:
q = mcΔT
Where:
m = mass of biphenyl burned (in grams)
c = heat capacity of the calorimeter (in kJ/°C)
ΔT = change in temperature (final temperature - initial temperature)
Given information:
Mass of biphenyl burned, m = 0.514 g
Heat capacity of the calorimeter, c = 5.86 kJ/°C
Change in temperature, ΔT = (29.4 °C - 25.8 °C) = 3.6 °C
Plugging in the values:
q = (0.514 g) * (5.86 kJ/°C) * (3.6 °C)
q = 10.339 kJ
Next, we need to calculate the moles of biphenyl, n:
Molar mass of biphenyl = (12.01 g/mol * 12) + (1.01 g/mol * 10)
= 154.24 g/mol
moles of biphenyl, n = 0.514 g / 154.24 g/mol
≈ 0.003333 mol
Finally, we can find the enthalpy change, ΔH:
ΔH = 10.339 kJ / 0.003333 mol
ΔH ≈ -3102.031 kJ/mol
Note that the given answer of -6330 kJ/mol is different from the calculated value. Please check your calculations and ensure that the given information is accurate.