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The heat of combustion of ethylene at $18^0 C$ and at constant volume is -335.8Kcal when water is obtained in liquid state. Calculate the heat of combustion at constant pressure and at $18^0 C$.

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: Try to figure out the combustion reaction of ethylene. Combustion is nothing but heating in the presence of air. Air mainly contains oxygen and nitrogen. Think about what is a change in constant volume condition and constant pressure condition.

Complete answer:
We know that combustion reaction is nothing but burning given substance in the presence of air. In air nitrogen and oxygen are present in major quantities. Nitrogen is unreactive due to the high bond dissociation energy of the nitrogen molecule. So oxygen will react with a given substance during the combustion reaction. Combustion of ethylene reaction will be: \[C_{ 2 }H_{ 4_{ \left( g \right) } }\quad +\quad 3O_{ 2_{ \left( g \right) } }\quad →\quad 2CO_{ 2_{ \left( g \right) } }\quad +\quad 2H_{ 2 }O_{ \left( l \right) }\]
Given that reaction is carried out at constant volume heat of combustion is -335.8 Kcal.
We need to calculate the heat of combustion at constant pressure.
Enthalpy change at constant volume $\triangle H\quad =\quad \triangle U$
Enthalpy change at constant pressure $\triangle H\quad =\quad \triangle U\quad -\quad \triangle { n }_{ g }RT$
$\triangle n_{ g }$ for the given reaction will be 2 – 4 = -2
Given the heat of combustion of ethylene at $18^0 C$ and at constant volume is -335.8Kcal.
\[\triangle U\quad =\quad -335.8Kcal\]
R is the universal gas constant, its value is 1.99$calK^{ -1 }\quad mol^{ -1 }$, T is the temperature of the gas in kelvin.
Enthalpy change at constant pressure:
\[\triangle H\quad =\quad \triangle U\quad -\quad \triangle { n }_{ g }RT\quad =\quad -335.8Kcal\quad -\quad (-2)×1.99×291cal\quad =\quad -335.8Kcal\quad +\quad 1158cal\quad =\quad -334.64Kcal\]

Therefore, the heat of combustion at constant pressure and at $18^0 C$ -334.64Kcal.

Note: At constant volume, there will be no expansion of gas and so there will be done by gas. Combustion in the presence of air will be oxidation. In the reaction the number of gaseous moles will decrease so change in gaseous moles is negative.
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