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

Answer:

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

Explanation:

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

$q=c\times (T_{final}-T_{initial})$

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:

$q=250.0 J/K\times (296.56 -294.65 )K$

$q=477.5 J$

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

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = -477.5 J

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

$\Delta H=-\frac{477.5 }{0.000875 mol}=-545,714.28 J/mol=-545.71 kJ/mol$

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