Question

When a certain amount of ethylene was combusted, $6226$ KJ of the heat was evolved. If heat of combustion of ethylene $1411$ KJ, the volume of oxygen (NTP) that entered into the reaction is
(A) $296.5{\text{ }}ml$
(B) $296.5{\text{ }}litres$
(C) $6226{\text{ }}22.4{\text{ }}litres$
(D) $22.4{\text{ }}litres$

Answer 1

Hint : Enthalpy of combustion of chemical reaction is defined as “the change in total enthalpy of reaction when one mole of any substance undergoes complete combustion in the presence of excess of air or oxygen”.

Complete Step By Step Answer:
From the given question we see that $6226$ KJ of energy is released during the complete combustion of ethylene. So, the value of enthalpy of combustion is $6226$ KJ.
Chemical reaction for combustion of ethylene is -
${C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O$
Combustion of one mole of ethylene $\left( {{C_2}{H_4}} \right)$ produces $1411$ KJ of enthalpy then, $6266$ KJ of enthalpy was released by
$1411$ KJ energy is released by $= 1$ mole of $\left( {{C_2}{H_4}} \right)$
$1$ KJ energy is released by $= \dfrac{1}{{1411}}$ mole of $\left( {{C_2}{H_4}} \right)$
so, moles of ethylene produced by $6266$ KJ energy is
$6266$ KJ energy is released by $= \dfrac{1}{{1411}} \times 6266$ mole of $\left( {{C_2}{H_4}} \right)$
$6266$ KJ energy is released by $= 4.412$ moles of $\left( {{C_2}{H_4}} \right)$
It shows that a total $4.412$ moles of ethylene $\left( {{C_2}{H_4}} \right)$ undergoes combustion in $6266$ KJ of enthalpy.
From the reaction we noticed that for every single mole of ethylene three mole of oxygen are required for complete combustion.
Since, volume of oxygen at STP condition is $22.4$ litres the volume of three mole of oxygen is $3 \times 22.4$ .
Volume of three mole of oxygen is $= 62.7$ litres of oxygen
Above calculation shows that one mole of ethylene required $62.7$ litres of oxygen, but we have $4.412$ moles of ethylene in the reaction.
$1$ Mole of $\left( {{C_2}{H_4}} \right)$ need $= 62.7$ litres of ${O_2}$
$4.412$ moles of $\left( {{C_2}{H_4}} \right)$ need $= 62.7$ litres of ${O_2}$
$4.412$ moles of $\left( {{C_2}{H_4}} \right)$ need $= 62.7 \times 4.412$ litres of ${O_2}$
After solving the above equation, we get,
$4.412$ moles of $\left( {{C_2}{H_4}} \right)$ need $= 296.5$ litres of ${O_2}$
So $4.412$ moles of ethylene required $296.5$ litres of oxygen for complete combustion, Hence option (ii) is the correct option.

Note :
The enthalpy of combustion is expressed as $\left( {\Delta {{\rm H}_C}} \right)$ .
This reaction proceeds with release of energy therefore, it is an example of an exothermic reaction. One mole of gas has $22.4$ liter of volume at standard temperature and pressure (STP).