Question

Calculate the enthalpy chamge when 75.0g of ethanol goes from 87.0 C to -15.0 C, given the following information for ethanol: c ethanol = 2.44j/g*C, Delta Hfusion=5.02 kj/mol, delta Hvap=38.56 kj/mol.

Answer 1

There are three steps involved in the process -

cooling of ethanol vapour from 87.0 ^oC to its boiling point (78.5 ^oC) ()+ condensation of vapour ethanol to liquid ethanol ()+ cooling of liquid ethanol from 78.5 ^oC to -15.0 ^oC ()

For cooling of a substance, at a definite phase,

m is mass,, C is heat capacity and si change of temperature.

So,

Frist step, m = 75.0 g , C = 2.44 J / (g . ^oC) and = (87.0 - 78.5) ^oC = 8.5 ^oC

So,

Third step, m = 75.0 g , C = 2.44 J / (g . ^oC) and = { 78.5 - (-15) } ^oC = 93.5 ^oC

So,

For phase change,

n is the number of moles. Molar mass of ethanol = 46 g/mol

So, here, n = (75 / 46) mol = 1.630 mol

So,

So,

Since the overall process is heat releasing or exothermic, so,