Question

When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from...

When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C.

Find ΔrU and ΔrH for the combustion of biphenyl in kJ mol−1 at 298 K. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.861 kJ °C−1.

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Answer #1

The heat capacity of the bomb calorimeter is(cp) = 5.861 kJ °C−1.

q   = Cp*T

      = 5.861*(29.419-25.823)

       = 21.076156KJ

no of moles of biphenyl   = W/G.M.Wt

                                         = 0.5141/154 = 0.00334moles

q    = 21.076156KJ/0.00334mole = 6310KJ/mole

In bomb calorimeter heat energy is equal to change in internal energy.

ΔrH = 0 KJ/mole

q = - U   = -6310KJ/mole

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