I have compGiven,
The heat capacity of teh calorimeter, c=34.65 kJ/K
The mass of methanol, m=1.761 g=1.761×0.0312=0.055 moles
The rise in the temperature, ΔT=295.55-294.40=1.15 K
The heat lost by the combustion of the methanol is gained by the calorimeter.
Thus the heat gained by the calorimeter is given by,
[tex]\Delta E=c\times\Delta T[/tex]On substituting the known values,
[tex]\begin{gathered} \Delta E=34.56\times10^3\times1.15 \\ =39.74\text{ kJ} \end{gathered}[/tex]Thus the ΔH of the reaction is given by,
[tex]\Delta H=\frac{\Delta E}{m}[/tex]On substituting the known values,
[tex]\begin{gathered} \Delta H=\frac{39.74\times10^3}{0.055} \\ =722.55\frac{kJ}{mol} \end{gathered}[/tex]Thus the ΔH is 722.55 kJ/mol
