Question

Metaldehyde, ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CHO}}} \right)_4}$ is used as a solid duel for camping stoves. The equation for the complete combustion of metaldehyde is shown.
${\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CHO}}} \right)_{\text{4}}}{\text{(s)}}\,{\text{ + }}\,{\text{10}}{{\text{O}}_{\text{2}}}{\text{(g)}}\, \to {\text{8}}\,{\text{C}}{{\text{O}}_2}{\text{(g)}}\,{\text{ + }}\,{\text{8}}{{\text{H}}_{\text{2}}}{\text{O(l)}}$
${{\Delta H}}_{\text{C}}^{\text{o}}$= standard enthalpy change of combustion.
Which expression will give a correct value for the enthalpy change of formation of metaldehyde?
A. ${{\Delta H}}_{\text{C}}^{\text{o}}{\text{(metaldehyde)}}\, - 8\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(carbon)}} + 8\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(hydrogen)}}$
B. ${{\Delta H}}_{\text{C}}^{\text{o}}{\text{(metaldehyde)}}\, - 8\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(carbon)}} + 16\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(hydrogen)}}$
C. $8\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(carbon)}} + 8\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(hydrogen)}} - {{\Delta H}}_{\text{C}}^{\text{o}}{\text{(metaldehyde)}}\,$
D. $8\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(carbon)}} + 16\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(hydrogen)}} - {{\Delta H}}_{\text{C}}^{\text{o}}{\text{(metaldehyde)}}\,$

Answer 1

Hint: We will write the equation for the formation of formaldehyde from the given equation. Then write the combustion reaction for the formation of each molecule. By adding or subtracting the all equation we will get the equation for the formation of formaldehyde and by adding or subtracting the enthalpy change for each reaction we will get the enthalpy change of formation of metaldehyde.

Complete Step by step answer: When one mole of hydrocarbon is completely burnt in presence of air at standard conditions, the change in enthalpy is known as enthalpy of combustion.
Enthalpy of formation is defined as the change in enthalpy during the formation of one mole of a compound by its elements.
The given reaction for the combustion of metaldehyde and its enthalpy change is as follows:
${\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CHO}}} \right)_{\text{4}}}{\text{(s)}}\,{\text{ + }}\,{\text{10}}{{\text{O}}_{\text{2}}}{\text{(g)}}\, \to {\text{8}}\,{\text{C}}{{\text{O}}_{\text{3}}}{\text{(g)}}\,{\text{ + }}\,{\text{8}}{{\text{H}}_{\text{2}}}{\text{O(l)}}\,...{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(metaldehyde)}}$
So, the reaction for the formation of metaldehyde and its enthalpy change will be the reverse of combustion of metaldehyde so,
\[{\text{8}}\,{\text{C}}{{\text{O}}_2}{\text{(g)}}\,{\text{ + }}\,{\text{8}}{{\text{H}}_{\text{2}}}{\text{O(l)}} \to {\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CHO}}} \right)_{\text{4}}}{\text{(s)}}\,{\text{ + }}\,{\text{10}}{{\text{O}}_{\text{2}}}{\text{(g)}}\,...\, - {{\Delta H}}_{\text{C}}^{\text{o}}{\text{(metaldehyde)}}\]…$(1)$
Carbon dioxide is formed by the combustion of elements of carbon in presence of oxygen. The enthalpy of combustion of carbon will be \[{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(carbon)}}\]. The reaction for the formation of carbon dioxide is as follows:
\[{\text{8}}\,{\text{C(s)}}\,{\text{ + }}\,8{{\text{O}}_{\text{2}}}{\text{(g)}} \to {\text{8}}\,{\text{C}}{{\text{O}}_2}{\text{(g)}}...{\text{8\Delta H}}_{\text{C}}^{\text{o}}{\text{(carbon)}}\]…..$(2)$
Water is formed by the combustion of elements hydrogen in presence of oxygen. The enthalpy of combustion of carbon will be \[{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(hydrogen)}}\]. The reaction for the formation of water is as follows:
\[{\text{8}}\,{{\text{H}}_2}{\text{(g)}}\,{\text{ + }}\,4{{\text{O}}_{\text{2}}}{\text{(g)}} \to {\text{8}}\,{{\text{H}}_2}{\text{O(l)}}..{\text{8\Delta H}}_{\text{C}}^{\text{o}}{\text{(hydrogen)}}\]….$(3)$
On adding equation $(2)$and$(3)$ we will get,
\[{\text{8}}\,{\text{C(s)}}\,{\text{ + 8}}\,{{\text{H}}_2}{\text{(g)}}\,\,{\text{ + }}\,12{{\text{O}}_{\text{2}}}{\text{(g)}} \to {\text{8}}\,{{\text{H}}_2}{\text{O(l)}}\, + \,{\text{8}}\,{\text{C}}{{\text{O}}_2}{\text{(g)}}\]….$(4)$
The enthalpy change is the,
\[{\text{8}}\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(carbon) + }}\,{\text{8\Delta H}}_{\text{C}}^{\text{o}}{\text{(hydrogen)}}\]
On adding the equation$(4)$and $(1)$we get,
\[{\text{8}}\,{\text{C(s)}}\,{\text{ + 8}}\,{{\text{H}}_2}{\text{(g)}}\,\,{\text{ + }}\,2{{\text{O}}_{\text{2}}}{\text{(g)}} \to {\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{CHO}}} \right)_{\text{4}}}{\text{(s)}}\,\]
The enthalpy change is the,
\[{\text{8}}\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(carbon) + }}\,{\text{8\Delta H}}_{\text{C}}^{\text{o}}{\text{(hydrogen)}} - {{\Delta H}}_{\text{C}}^{\text{o}}{\text{(metaldehyde)}}\]
So, the expression for the correct value for the enthalpy change of formation of metaldehyde from elements is\[{\text{8}}\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(carbon) + }}\,{\text{8\Delta H}}_{\text{C}}^{\text{o}}{\text{(hydrogen)}} - {{\Delta H}}_{\text{C}}^{\text{o}}{\text{(metaldehyde)}}\].

Therefore, option (C) $8\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(carbon)}} + 8\,{{\Delta H}}_{\text{C}}^{\text{o}}{\text{(hydrogen)}} - {{\Delta H}}_{\text{C}}^{\text{o}}{\text{(metaldehyde)}}\,$is correct.

Note: On reversing the equation, the sign of enthalpy change also changes. Here, the enthalpy change for the combustion was positive so it became negative on reversing the equation. Oxygen is used for the burning of carbon or hydrogen so, the reaction is known as combustion of carbon or hydrogen, not the combustion of oxygen hence the enthalpy change represents the enthalpy change of the combustion of carbon or hydrogen.