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).

Explanation:

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

    ....... (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, -347.6 kcal/mol

When we reverse the 2nd reaction

      heat of reaction +347.6 kcal {sign reversed with reaction}  ........ (2)

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

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 =

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 =

                                              = - 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,            

                    = 3.8 Kcal/mol

the answer is true.

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.