When 10 g of diethyl ether is converted to vapor at its boiling point, about how much heat is absorbed? (c4h10o, δhvap = 15.7 kj/mol, boiling point: 34.6°c) 0.2 kj 3 kj .01 kj 2 j?

ΔH= 3KJ

Explanation:

The total heat absorbed is the total energy in the process, and that is in form of entalpy.

ΔH = q + ΔHvap, where q is the heat necessary for elevate the temperature of dietil ether. Suppose the initial temperature is room temperature (25ºC=298 K), then

q= 10g x2.261 J/gK x(310 K - 298K)= 271.32 J= 0.3 kJ

Then

ΔHvap = 10g C4H10O x (1 mol C4H10O/74.12 g C4H10O) x( 15.7 KJ/ 1 mol C4H10O) = 2.12 KJ

ΔH= 2.5KJ ≈ 3KJ

false

because it does not have the potential to be entirely used up