Question

Given the standard heat of combustion of methanol CH3OH is 182.6

kcal/mol, Dimethyl ether CH3OCH3 is 347.6 kcal/mol, (methanol/



Dimethyl ether in gas phase, water in liquid phase). Given the heat



of vaporization of water is 10 kcal/mol, methanol is 8.4 kcal/mol,



Dimethyl ether 4.8 kcal/mol. Calculate the reaction of dehydration



of methanol to produce Dimethyl ether. (Indicate the phase of your



components).

Answer 1

Step-by-step explanation:

The given reaction equation showing heat of combustion is as follows.

`2CH_(3)OH(l) + 3O2(g) \rightarrow 2CO_(2)(g) + 4H_(2)O(g)` ....... (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,
`CH_(3)OCH_(3)(g) + 3O_(2)(g) \rightarrow 2CO_(2)(g) + 3H_(2)O(g)` -347.6 kcal/mol

When we reverse the 2nd reaction

`2CO_(2)(g) + 3H_(2)O(g) \rightarrow CH_(3)OCH_(3)(g) + 3O_(2)(g)` heat of reaction +347.6 kcal {sign reversed with reaction} ........ (2)

Now, adding equatio (1) and (2) we get the following.

`2CH_(3)OH(l) \rightarrow CH_(3)OCH_(3)(g) + H_(2)O(g)`

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 =
`100^(o)C`

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 =
`(-17.6 kcal)/(2 mol)`

= - 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,
`(7.6 kcal)/(2 mol)`

= 3.8 Kcal/mol