Question

A bomb calorimetric experiment was run to determine the enthalpy of combustion of methanol. The reaction is CH3OH(l)+3/2O2(g)→CO2(g)+2H2O(l) The bomb calorimeter has a heat capacity of 250.0 J/K. Burning 0.028 g of methanol resulted in a rise in temperature from 21.50 ∘C to 23.41 ∘C. Calculate the change in internal energy for the combustion of methanol in kJ/mol.

Answer 1

The change in internal energy during the combustion reaction is- 545.71 kJ/mol.

Step-by-step explanation:

First we have to calculate the heat gained by the calorimeter.

`q=c* (T_(final)-T_(initial))`

where,

q = heat gained = ?

c = specific heat =
`250.0 J/^oC`

`T_(i)` = Initial temperature =
`21.50^oC=294.65 K`

`T_(f)` = Final temperature =
`23.41^oC=296.56 K`

Now put all the given values in the above formula, we get:

`q=250.0 J/K* (296.56 -294.65 )K`

`q=477.5 J`

Now we have to calculate the enthalpy change during the reaction.

`\Delta H=-(q)/(n)`

where,

`\Delta H` = enthalpy change = ?

q = heat gained = -477.5 J

n = number of moles methanol =
`\frac{\text{Mass of methanol}}{\text{Molar mass of methanol}}=(0.028 g)/(32 g/mol)=0.000875 mol`

`\Delta H=-(477.5 )/(0.000875 mol)=-545,714.28 J/mol=-545.71 kJ/mol`

Therefore, the change in internal energy during the combustion reaction is- 545.71 kJ/mol.