Question

Calorific value of ethane, in \[KJ/g\] for the reaction \[2{C_2}{H_6} + 7{O_7} \to 4C{O_2} + 6{H_2}O;\Delta {\rm H} = - 745.6Kcal\]
(A) \[ - 12.4\]
(B) $ - 52$
(C) $ - 24.8$
(D) \[ - 104\]

Answer 1

Hint:
Calorific value is the amount of heat energy present in food or full and which is determined by the complete combustion of specified quantity at constant pressure and in normal conditions. It is also called calorific powder. The unit of calorific value is kilojoule per kilogram i.e. $KJ/Kg$.
\[2{C_2}{H_6} + 7{O_2} \to 4C{O_2} + 6{H_2}O;\Delta {\rm H} = - 745.6Kcal\]

Complete step by step answer:
This implies that $2$ moles of ethane reacts with $7$ moles of oxygen gas to produce $4$ moles of carbon dioxide and $6$ moles of water.
We know that heat of combustion of $2$ moles of ethane is $ - 745.6Kcal$
Therefore, $1$ mole of ethane has $\Delta {\rm H} = \dfrac{{ - 745.6}}{2}$
                    $\Delta H = - 372.8Kcal$
We know that, $1Kcal = 4.18KJ$
So, $ - 372.8Kcal = - 372.8 \times 4.18KJ$
$ = - 1558.3KJ$
So, $\Delta H = - 1558.3KJ$
As the formula of calorific value show,
Calorific value = $\dfrac{{Heat{\text{ }}of{\text{ }}combustion}}{{Molecular{\text{ }}weight{\text{ }}of{\text{ }}{C_2}{H_6}}}$
          $ = \dfrac{{ - 1558.3}}{{30}} = - 51.9KJ/g$
So, calorific value = $ - 51.9KJ/g \approx - 52KJ/g$
Which is option (B) $ - 52$.

Additional information:
There are two types of calorific value
1. Higher calorific value (HCV)
2. Lower calorific value (LCV)
Higher calorific value is also known as gross calorific value (GCV)
Lower calorific value is also known as net calorific value (NCV)

Note:
Natural gas has the highest calorific value at $13Kcal/g$ followed by propane and butane at $11.9$ and respectively. And gasoline ranks at fourth at $11.3Kcal/g$.