How many kilojoules of heat are required to vaporize 50.0 g of ethanol (MM = 46.07 g). Its molar heat of vaporization is 43.5 kJ/mol.

ΔH = 41.7 kj

Given data:

Mass of ethanol = 50 g

Heat required to vaporize = ?

Boiling point of ethanol = 78.3°C

Molar heat of vaporization = 38.6 kj/mol

Solution:

First of all we will calculate the number of moles.

Number of moles = mass / molar mass

Number of moles = 50 g/ 46.07 g/mol

Number of moles = 1.08 mol

ΔH = 38.6 kj/mol ×1.08 mol

ΔH = 41.7 kj

Explanation:

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.