91Ó°ÊÓ

The amount of heat given out when one mole of a substance is burnt in the presence of excess oxygen; the amount of heat generated is called heat of combustion.

The standard conditions are partial pressure of all gases at 1 bar of pressure, 298 K, and concentration of all solutions at 1 M.

The combustion reaction of methanol:

\[{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH(l) + }}\frac{{\bf{3}}}{{\bf{2}}}{{\bf{O}}_{\bf{2}}}{\bf{(g) }} \to {\bf{C}}{{\bf{O}}_{\bf{2}}}{\bf{(g) + 2}}{{\bf{H}}_{\bf{2}}}{\bf{O(l)}}\]

From the above combustion reaction, we can observe that one mole of methanol CH₃OH requires a 3/2 mole of oxygen O₂.

To calculate the amount of heat generated on the combustion of 125 g of methanol, we have to know the enthalpy of combustion of methanol. From Table 5.2, we can get information about the enthalpy of combustion of methanol.

Enthalpy of combustion of methanol, deltaH_c= -726.1 KJ mol^-1.

It means that on combustion of one mole of methanol, 726.1 kJ heat is generated. [negative sign indicates that heat is generated]

Calculate the heat generated when 125 g of methanol is combusted.

First, we will calculate the moles of methanol present, as we already have the information on the amount of heat generated on the combustion of one mole of methanol.

\[\begin{aligned}Formula{\rm{ }}\;use:\\{\rm{No}}\;{\rm{ of}}\;{\rm{ mole = }}\;\frac{{{\rm{weight }}\;{\rm{of}}\;{\rm{ methanol}}}}{{{\rm{Molecular}}\;{\rm{ mass }}\;{\rm{of }}\;{\rm{methanol}}}}\\Calculation:\\{\rm{No}}\;{\rm{ of}}\;{\rm{ mole }}\;{\rm{ = }}\;{\rm{ }}\frac{{125{\rm{ g}}}}{{32{\rm{ g mo}}{{\rm{l}}^{ - 1}}}}{\rm{ }}\;{\rm{ = }}\;{\rm{3}}{\rm{.90625 mol}}\\\left( \begin{aligned}{\rm{Molecular}}\;{\rm{ mass }}\;{\rm{of }}\;{\rm{methanol }}\;{\rm{ = 1 }} \times {\rm{ 12 + 4 }} \times {\rm{ 1 + 16 g}}\\{\rm{ }}\;\;\;\;{\rm{ = 32 g}}\end{aligned} \right){\rm{ }}\end{aligned}\]

As no of the moles of methanol is already calculated and comes out to be\[{\rm{3}}{\rm{.90625 mol}}\], we can easily find out the amount of heat generated.

\(\begin{aligned}{\rm{Enthalpy}}\;{\rm{of}}\;{\rm{combustion}}\;{\rm{per}}\;{\rm{mole}}\;{\rm{ = }}\; - 726.1\;kJ\\{\rm{Hence, }}\\{\rm{Enthalpy}}\;{\rm{ of }}\;{\rm{combustion }}\;{\rm{for}}\;{\rm{ }}3.90625\;mole{\rm{ }}\;{\rm{will}}\;{\rm{ be }}\\{\rm{ = }}3.90625 \times - 726.1kJ\\{\rm{ = }}2836.328{\rm{ k}}J\\\\{\rm{Hence, }}\;{\rm{amount }}\;{\rm{of }}\;{\rm{heat }}\,{\rm{generated }}\;{\rm{will }}\;{\rm{be}}\;{\rm{ }}2836.328{\rm{ k}}J.\end{aligned}\)