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

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation:

im pretty sure its b