Respuesta :
Explanation:
The given reaction equation showing heat of combustion is as follows.
[tex]2CH_{3}OH(l) + 3O2(g) \rightarrow 2CO_{2}(g) + 4H_{2}O(g)[/tex] ....... (1)
So, for 2 mol methanol the value for heat of combustion is as follows.
2 mol x 182.6 Kcal/mol = -365.2 Kcal
Now, [tex]CH_{3}OCH_{3}(g) + 3O_{2}(g) \rightarrow 2CO_{2}(g) + 3H_{2}O(g)[/tex] -347.6 kcal/mol
When we reverse the 2nd reaction
[tex]2CO_{2}(g) + 3H_{2}O(g) \rightarrow CH_{3}OCH_{3}(g) + 3O_{2}(g)[/tex] heat of reaction +347.6 kcal {sign reversed with reaction} ........ (2)
Now, adding equatio (1) and (2) we get the following.
[tex]2CH_{3}OH(l) \rightarrow CH_{3}OCH_{3}(g) + H_{2}O(g)[/tex]
Now, heat of combustion will also be added.
+347.6 kcal +(-365.2 Kcal)
= -17.6 Kcal
Since, the reaction occurs at STP , water will be in liquid state as B.P = [tex]100^{o}C[/tex]
So, energy needed for water to convert to vapour = 1 mol x 10 Kcal/mol = 10 Kcal, and it will be absorbed by water to form vapor .
Hence, heat of dehydration = [tex]\frac{-17.6 kcal}{2 mol}[/tex]
= - 8.8 Kcal/mol
Therefore, the net calorific value for this after water is evaporated to form gas = 17.6 - 10 = 7.6 Kcal
or, [tex]\frac{7.6 kcal}{2 mol}[/tex]
= 3.8 Kcal/mol
