Question

Consider this reaction:

2CH₃OH(l) + 3O₂(g)₂O(l) + 2CO₂(g) H = –1452.8 kJ/mol

a.Is this reaction endothermic or exothermic?

b.H if the equation is multiplied throughout by 2?

c.H if the direction of the reaction is reversed so that the products become the reactants and vice versa?

What is the value of ∆H if water vapor instead of liquid water is formed as the product?

Answer 1

2CH₃OH(l) + 3O₂(g) 4H₂O(l) + 2CO₂(g) ; ∆H = –1452.8 kJ/mol

(a)The value of ∆H is -ve it indicates the heat is evolved in the reaction so it is an exothermic reaction.

(b) If we multiply the equation throughtout by 2 then the value of ∆H also multiplied by 2 then the new

     ∆H is = 2x( –1452.8 kJ/mol)

              = 2905.6 kJ/mol

(c) if the direction of the reaction is reversed so that the products become the reactants then the equation becomes

4H₂O(l) + 2CO₂(g) 2CH₃OH(l) + 3O₂(g) :  

Then the enthalpy change of the reaction is -ve value of the actual value then ∆H' = - ∆H

                                                                                                                                   = - ( –1452.8 kJ/mol)

                                                                                                                                    = +1452.8 kJ/mol

(d) 2CH₃OH(l) + 3O₂(g) 4H₂O(l) + 2CO₂(g)

For the conversion of liquid water to water vapour we have to supply some energy to the liquid.

So an additional amount of energy is added to ∆H

So total enthalpy change is ∆H '' = ∆H + ( +ve value)

                                                    = -ve + (+ve)

                                                   = less -ve value

Therefore the value of ∆H is more