Question

Calculate the enthalpy of formation of acetic acid if the enthalpy of combustion to co2 and h2o is-867kj/mole?

Answer 1

You know that the enthalpy of neutralisation is the enthalpy change per mole of

the substance neutralised. Thus, we have to first calculate the amount (number of

moles) of HCl present in the volume of solution taken for neutralisation in the

experiment. As given in the procedure, when 100 cm3 of 0.5 M HCl in neutralised

using 100 cm3 of 0.5 M NaOH, then, the amount (no. of moles) of HCl present in

the 200 cm3 solution (100 cm3 acid + 100 cm3 base) will be equal to 0.05 mole.

-- - O5 me = 0.5 me To calculate the heat of neutralisation, we have to divide the AH obtained by Eq. 10

Because 1dm = 10cm 7.4 by amount (the number of moles) of hydrochloric acid, i.e.,

and (1 dm13 = (10 cm13 AH(obtained from Eq. 7.4

1

Mneut = amount (No. of moles) of Hkl

1 dm3 = 1000 cm3 or 100 cm3 = - dm3 10 - - AH (obtained from Eq. 7.4)

Thus, to get the amount (number of 0.05

moles) present in 100 cm3 of 0.5 M HCI = ...... J m0l-'

or 200 cm3 of the solution from the Now report your result as given below:

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