Answer:
The approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
Explanation:
First we have to calculate the heat gained by the calorimeter.
[tex]q=c\times \Delta T[/tex]
where,
q = Heat gained = ?
c = Specific heat = [tex]2.68 kJ/^oC[/tex]
ΔT = The change in temperature = 3.08°C
Now put all the given values in the above formula, we get:
[tex]q=2.68 kJ/^oC\times 3.08^oC[/tex]
[tex]q=8.2544 kJ[/tex]
Now we have to calculate molar enthalpy of combustion of this substance :
[tex]\Delta H_{comb}=-\frac{q}{n}[/tex]
where,
[tex]\Delta H_{comb}[/tex] = enthalpy change = ?
q = heat gained = 8.2544kJ
n = number of moles methane = [tex]\frac{\text{Mass of methane}}{\text{Molar mass of methane }}=\frac{2.00 g}{16.042 g/mol}=0.1247 mole[/tex]
[tex]\Delta H_{comb}=-\frac{8.2544 kJ}{0.1247 mole}=-66.21 kJ/mole\approx -66 kJ/mole[/tex]
Therefore, the approximate molar enthalpy of combustion of this substance is -66 kJ/mole.
